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Chapter 11: More Operator Overloading

11.1: Overloading `operator[]()'

Here is an example showing the basic use of the class:

    int main()
    {
        IntArray x{ 20 };               // 20 ints

        for (int i = 0; i < 20; i++)
            x[i] = i * 2;               // assign the elements

        for (int i = 0; i <= 20; i++)   // produces boundary overflow
            cout << "At index " << i << ": value is " << x[i] << '\n';
    }

First, the constructor is used to create an object containing 20 ints. The elements stored in the object can be assigned or retrieved. The first for-loop assigns values to the elements using the index operator, the second for-loop retrieves the values but also results in a run-time error once the non-existing value x[20] is addressed. The IntArray class interface is:

    #include <cstddef>

    class IntArray
    {
        int     *d_data;
        size_t d_size;

         public:
            IntArray(size_t size = 1);
            IntArray(IntArray const &other);
            ~IntArray();
            IntArray &operator=(IntArray const &other);

                                                // overloaded index operators:
            int &operator[](size_t index);              // first
            int operator[](size_t index) const;         // second

            void swap(IntArray &other);         // trivial

        private:
            void boundary(size_t index) const;
            int &operatorIndex(size_t index) const;
    };

• One of its constructors has a size_t parameter having a default argument value, specifying the number of int elements in the object.
• The class internally uses a pointer to reach allocated memory. Hence, the necessary tools are provided: a copy constructor, an overloaded assignment operator and a destructor.
• Note that there are two overloaded index operators. Why are there two?

  The first overloaded index operator allows us to reach and modify the elements of non-constant IntArray objects. This overloaded operator's prototype is a function returning a reference to an int. This allows us to use expressions like x[10] as rvalues or lvalues.

  With non-const IntArray objects operator[] can therefore be used to retrieve and to assign values. The return value of the non-const operator[] member is not an int const &, but an int &. In this situation we don't use const, as we must be able to modify the element we want to access when the operator is used as lvalue.

  This whole scheme fails if there's nothing to assign. Consider the situation where we have an IntArray const stable(5). Such an object is an immutable const object. The compiler detects this and refuses to compile this object definition if only the non-const operator[] is available. Hence the second overloaded index operator is added to the class's interface. Here the return value is an int, rather than an int &, and the member function itself is a const member function. This second form of the overloaded index operator is automatically used by the compiler with const objects. It is used for value retrieval instead of value assignment. That, of course, is precisely what we want when using const objects. In this situation members are overloaded only by their const attribute. This form of function overloading was introduced earlier in the C++ Annotations (sections 2.5.4 and 7.7).

  Note the difference between the return types of the two overloaded index operators: int & and int. Here, using a value return type for the const-index operator is OK, because the return type is a simple built-in type: returning a copy of the requested value is likely more efficient than returning a const reference, which must be dereferenced whenever it is used. However, for more complex return types (generally: objects) value return types should be avoided and the const index operator should define a Type const & return type instead of a Type value return type.

  Since IntArray stores values of a primitive type IntArray's operator[] const could also have defined a value return type. However, with objects one usually doesn't want the extra copying that's implied with value return types. In those cases const & return values are preferred for const member functions. So, in the IntArray class an int return value could have been used as well, resulting in the following prototype:

      int IntArray::operator[](size_t index) const;
  

• As there is only one pointer data member, the destruction of the memory allocated by the object is a simple delete[] data.

    #include "intarray.ih"

    IntArray::IntArray(size_t size)
    :
        d_size(size)
    {
        if (d_size < 1)
            throw string("IntArray: size of array must be >= 1");

        d_data = new int[d_size];
    }

    IntArray::IntArray(IntArray const &other)
    :
        d_size(other.d_size),
        d_data(new int[d_size])
    {
        memcpy(d_data, other.d_data, d_size * sizeof(int));
    }

    IntArray::~IntArray()
    {
        delete[] d_data;
    }

    IntArray &IntArray::operator=(IntArray const &other)
    {
        IntArray tmp(other);
        swap(tmp);
        return *this;
    }

    int &IntArray::operatorIndex(size_t index) const
    {
        boundary(index);
        return d_data[index];
    }

    int &IntArray::operator[](size_t index)
    {
        return operatorIndex(index);
    }

    int IntArray::operator[](size_t index) const
    {
        return operatorIndex(index);
    }

    void IntArray::boundary(size_t index) const
    {
        if (index < d_size)
            return;
        ostringstream out;
        out  << "IntArray: boundary overflow, index = " <<
                index << ", should be < " << d_size << '\n';
        throw out.str();
    }

11.2: Overloading the insertion and extraction operators

The class std::ostream defines insertion operators for primitive types, such as int, char *, etc.. In this section we learn how to extend the existing functionality of classes (in particular std::istream and std::ostream) in such a way that they can be used also in combination with classes developed much later in history.

In particular we will show how the insertion operator can be overloaded allowing the insertion of any type of object, say Person (see chapter 9), into an ostream. Having defined such an overloaded operator we're able to use the following code:

    Person kr("Kernighan and Ritchie", "unknown", "unknown");

    cout << "Name, address and phone number of Person kr:\n" << kr << '\n';

                                // declared in `person.h'
    std::ostream &operator<<(std::ostream &out, Person const &person);

                                // defined in some source file
    ostream &operator<<(ostream &out, Person const &person)
    {
        return
            out <<
                "Name:    " << person.name() << ", "
                "Address: " << person.address() << ", "
                "Phone:   " << person.phone();
    }

• The function returns a reference to an ostream object, to enable `chaining' of the insertion operator.
• The two operands of operator<< are passed to the free function as its arguments. In the example, the parameter out was initialized by cout, the parameter person by kr.

In order to overload the extraction operator for, e.g., the Person class, members are needed modifying the class's private data members. Such modifiers are normally offered by the class interface. For the Person class these members could be the following:

    void setName(char const *name);
    void setAddress(char const *address);
    void setPhone(char const *phone);

    void Person::setAddress(char const *address)
    {
        delete[] d_address;
        d_address = strdupnew(address);
    }

    istream &operator>>(istream &in, Person &person)
    {
        string name;
        string address;
        string phone;

        if (in >> name >> address >> phone)    // extract three strings
        {
            person.setName(name.c_str());
            person.setAddress(address.c_str());
            person.setPhone(phone.c_str());
        }
        return in;
    }

11.3: Conversion operators

    class String
    {
        char *d_string;

        public:
            String();
            String(char const *arg);
            ~String();
            String(String const &other);
            String &operator=(String const &rvalue);
            String &operator=(char const *rvalue);
    };

Usually, in classes that are less directly linked to their data than this String class, there will be an accessor member function, like a member char const *String::c_str() const. However, the need to use this latter member doesn't appeal to our intuition when an array of String objects is defined by, e.g., a class StringArray. If this latter class provides the operator[] to access individual String members, it would most likely offer at least the following class interface:

    class StringArray
    {
        String *d_store;
        size_t d_n;

        public:
            StringArray(size_t size);
            StringArray(StringArray const &other);
            StringArray &operator=(StringArray const &rvalue);
            ~StringArray();

            String &operator[](size_t index);
    };

    StringArray sa{ 10 };

    sa[4] = sa[3];  // String to String assignment

    sa[3] = "hello world";

• First, sa[3] is evaluated. This results in a String reference.
• Next, the String class is inspected for an overloaded assignment, expecting a char const * to its right-hand side. This operator is found, and the string object sa[3] receives its new value.

    char const *cp = sa[3];

One possibility is to define an accessor member function c_str():

    char const *cp = sa[3].c_str()

A conversion operator is a kind of overloaded operator, but this time the overloading is used to cast the object to another type. In class interfaces, the general form of a conversion operator is:

    operator <type>() const;

Conversion operators are somewhat dangerous. The conversion is automatically performed by the compiler and unless its use is perfectly transparent it may confuse those who read code in which conversion operators are used. E.g., novice C++ programmers are frequently confused by statements like `if (cin) ...'.

As a rule of thumb: classes should define at most one conversion operator. Multiple conversion operators may be defined but frequently result in ambiguous code. E.g., if a class defines operator bool() const and operator int() const then passing an object of this class to a function expecting a size_t argument results in an ambiguity as an int and a bool may both be used to initialize a size_t.

In the current example, the class String could define the following conversion operator for char const *:

    String::operator char const *() const
    {
        return d_string;
    }

• Conversion operators do not define return types. The conversion operator returns a value of the type specified beyond the operator keyword.
• In certain situations (e.g., when a String argument is passed to a function specifying an ellipsis parameter) the compiler needs a hand to disambiguate our intentions. A static_cast solves the problem.
• With template functions conversion operators may not work immediately as expected. For example, when defining a conversion operator X::operator std::string const() const then cout << X() won't compile. The reason for this is explained in section 21.9, but a shortcut allowing the conversion operator to work is to define the following overloaded operator<< function:

  std::ostream &operator<<(std::ostream &out, std::string const &str)
  {
      return out.write(str.data(), str.length());
  }
  

• If a class X defining a conversion operator also defines an insertion operator accepting an X object the insertion operator is used;
• Otherwise, if the type returned by the conversion operator is insertable then the conversion operator is used;
• Otherwise, a compilation error results. Note that this happens if the type returned by the conversion operator itself defines a conversion operator to a type that may be inserted into a stream.

    #include <iostream>
    #include <string>
    using namespace std;

    struct Insertable
    {
        operator int() const
        {
            cout << "op int()\n";
            return 0;
        }
    };
    ostream &operator<<(ostream &out, Insertable const &ins)
    {
        return out << "insertion operator";
    }
    struct Convertor
    {
        operator Insertable() const
        {
            return Insertable();
        }
    };
    struct Text
    {
        operator int() const
        {
            return 1;
        }
    };
    struct Error
    {
        operator Text() const
        {
            return Text{};
        }
    };

    int main()
    {
        Insertable insertable;
        cout << insertable << '\n';
        Convertor convertor;
        cout << convertor << '\n';
        Error error;
        cout << error << '\n';
    }

Some final remarks regarding conversion operators:

• A conversion operator should be a `natural extension' of the facilities of the object. For example, the stream classes define operator bool(), allowing constructions like if (cin).
• A conversion operator should return an rvalue. It should do so to enforce data-hiding and because it is the intended use of the conversion operator. Defining a conversion operator as an lvalue (e.g., defining an operator int &() conversion operator) opens up a back door, and the operator can only be used as lvalue when explicitly called (as in: x.operator int&() = 5). Don't use it.
• Conversion operators should be defined as const member functions as they don't modify their object's data members.
• Conversion operators returning composed objects should return const references to these objects whenever possible to avoid calling the composed object's copy constructor.

11.4: The keyword `explicit'

Assume a data base class DataBase is defined in which Person objects can be stored. It defines a Person *d_data pointer, and so it offers a copy constructor and an overloaded assignment operator.

In addition to the copy constructor DataBase offers a default constructor and several additional constructors:

• DataBase(Person const &): the DataBase initially contains a single Person object;
• DataBase(istream &in): the data about multiple persons are read from in.
• DataBase(size_t count, istream &in = cin): the data of count persons are read from in, by default the standard input stream.

The above constructors all are perfectly reasonable. But they also allow the compiler to compile the following code without producing any warning at all:

    DataBase db;
    DataBase db2;
    Person person;

    db2 = db;           // 1
    db2 = person;       // 2
    db2 = 10;           // 3
    db2 = cin;          // 4

Implicit promotions are used with statements 2 through 4. Since constructors accepting, respectively a Person, an istream, and a size_t and an istream have been defined for DataBase and since the assignment operator expects a DataBase right-hand side (rhs) argument the compiler first converts the rhs arguments to anonymous DataBase objects which are then assigned to db2.

It is good practice to prevent implicit promotions by using the explicit modifier when declaring a constructor. Constructors using the explicit modifier can only be used to construct objects explicitly. Statements 2-4 would not have compiled if the constructors expecting one argument would have been declared using explicit. E.g.,

    explicit DataBase(Person const &person);
    explicit DataBase(size_t count, std:istream &in);

Having declared all constructors accepting one argument as explicit the above assignments would have required the explicit specification of the appropriate constructors, thus clarifying the programmer's intent:

    DataBase db;
    DataBase db2;
    Person person;

    db2 = db;                   // 1
    db2 = DataBase{ person };   // 2
    db2 = DataBase{ 10 };       // 3
    db2 = DataBase{ cin };      // 4

11.4.1: Explicit conversion operators

For example, a class might define operator bool() const returning true if an object of that class is in a usable state and false if not. Since the type bool is an arithmetic type this could result in unexpected or unintended behavior. Consider:

    void process(bool value);

    class StreamHandler
    {
        public:
            operator bool() const;      // true: object is fit for use
            ...
    };

    int fun(StreamHandler &sh)
    {
        int sx;

        if (sh)                         // intended use of operator bool()
            ... use sh as usual; also use `sx'

        process(sh);                    // typo: `sx' was intended
    }

When defining explicit conversion operators implicit conversions like the one shown in the example are prevented. Such conversion operators can only be used in situations where the converted type is explicitly required (as in the condition clauses of if or while statements), or is explicitly requested using a static_cast. To declare an explicit bool conversion operator in class StreamHandler's interface replace the above declaration by:

        explicit operator bool() const;

Since the C++14 standard istreams define an explicit operator bool() const. As a consequence:

    while (cin.get(ch)) // compiles OK
        ;

    bool fun1()
    {
        return cin;     // 'bool = istream' won't compile as 
    }                   // istream defines 'explicit operator bool'

    bool fun1()
    {
        return static_cast<bool>(cin); // compiles OK
    }

11.5: Overloading the increment and decrement operators

Used as postfix operator, the value's object is returned as an rvalue, temporary const object and the post-incremented variable itself disappears from view. Used as prefix operator, the variable is incremented, and its value is returned as lvalue and it may be altered again by modifying the prefix operator's return value. Whereas these characteristics are not required when the operator is overloaded, it is strongly advised to implement these characteristics in any overloaded increment or decrement operator.

Suppose we define a wrapper class around the size_t value type. Such a class could offer the following (partially shown) interface:

    class Unsigned
    {
        size_t d_value;

        public:
            Unsigned();
            explicit Unsigned(size_t init);

            Unsigned &operator++();
    }

    Unsigned &Unsigned::operator++()
    {
        ++d_value;
        return *this;
    }

To define the postfix operator, an overloaded version of the operator is defined, expecting a (dummy) int argument. This might be considered a kludge, or an acceptable application of function overloading. Whatever your opinion in this matter, the following can be concluded:

• Overloaded increment and decrement operators without parameters are prefix operators, and should return references to the current object.
• Overloaded increment and decrement operators having an int parameter are postfix operators, and should return a value which is a copy of the object at the point where its postfix operator is used.

    Unsigned operator++(int);

    Unsigned Unsigned::operator++(int)
    {
        Unsigned tmp{ *this };
        ++d_value;
        return tmp;
    }

In the above example the statement incrementing the current object offers the nothrow guarantee as it only involves an operation on a primitive type. If the initial copy construction throws then the original object is not modified, if the return statement throws the object has safely been modified. But incrementing an object could itself throw exceptions. How to implement the increment operators in that case? Once again, swap is our friend. Here are the pre- and postfix operators offering the strong guarantee when the member increment performing the increment operation may throw:

    Unsigned &Unsigned::operator++()
    {
        Unsigned tmp{ *this };
        tmp.increment();
        swap(tmp);
        return *this;
    }
    Unsigned Unsigned::operator++(int)
    {
        Unsigned tmp{ *this };
        tmp.increment();
        swap(tmp);
        return tmp;
    }

When calling the increment or decrement operator using its full member function name then any int argument passed to the function results in calling the postfix operator. Omitting the argument results in calling the prefix operator. Example:

    Unsigned uns{ 13 };

    uns.operator++();     // prefix-incrementing uns
    uns.operator++(0);    // postfix-incrementing uns

Both the prefix and postfix increment and decrement operators are deprecated when applied to bool type of variables. In situations where a postfix increment operator could be useful the std::exchange (cf. section 19.1.11) should be used.

11.6: Overloading binary operators

Most binary operators come in two flavors: the plain binary operator (like the + operator) and the compound binary assignment operator (like operator+=). Whereas the plain binary operators return values, the compound binary assignment operators usually return references to the objects for which the operators were called. For example, with std::string objects the following code (annotations below the example) may be used:

    std::string s1;
    std::string s2;
    std::string s3;

    s1 = s2 += s3;                  // 1
    (s2 += s3) + " postfix";        // 2
    s1 = "prefix " + s3;            // 3
    "prefix " + s3 + "postfix";     // 4

• at // 1 the contents of s3 is added to s2. Next, s2 is returned, and its new contents are assigned to s1. Note that += returns s2.
• at // 2 the contents of s3 is also added to s2, but as += returns s2 itself, it's possible to add some more to s2
• at // 3 the + operator returns a std::string containing the concatenation of the text prefix and the contents of s3. This string returned by the + operator is thereupon assigned to s1.
• at // 4 the + operator is applied twice. The effect is:
  1. The first + returns a std::string containing the concatenation of the text prefix and the contents of s3.
  2. The second + operator takes this returned string as its left hand value, and returns a string containing the concatenated text of its left and right hand operands.
  3. The string returned by the second + operator represents the value of the expression.

Now consider the following code, in which a class Binary supports an overloaded operator+:

    class Binary
    {
        public:
            Binary();
            Binary(int value);
            Binary operator+(Binary const &rhs);
    };

    int main()
    {
        Binary b1;
        Binary b2{ 5 };

        b1 = b2 + 3;            // 1
        b1 = 3 + b2;            // 2
    }

    error: no match for 'operator+' in '3 + b2'

In order to understand this remember promotions. As we have seen in section 11.4, constructors expecting single arguments may implicitly be activated when an argument of an appropriate type is provided. We've already encountered this with std::string objects, where NTBSs may be used to initialize std::string objects.

Analogously, in statement // 1, operator+ is called, using b2 as its left-hand side operand. This operator expects another Binary object as its right-hand side operand. However, an int is provided. But as a constructor Binary(int) exists, the int value can be promoted to a Binary object. Next, this Binary object is passed as argument to the operator+ member.

Unfortunately, in statement // 2 promotions are not available: here the + operator is applied to an int-type lvalue. An int is a primitive type and primitive types have no knowledge of `constructors', `member functions' or `promotions'.

How, then, are promotions of left-hand operands implemented in statements like "prefix " + s3? Since promotions can be applied to function arguments, we must make sure that both operands of binary operators are arguments. This implies that plain binary operators supporting promotions for either their left-hand side operand or right-hand side operand should be declared as free operators, also called free functions.

Functions like the plain binary operators conceptually belong to the class for which they implement these operators. Consequently they should be declared in the class's header file. We cover their implementations shortly, but here is our first revision of the declaration of the class Binary, declaring an overloaded + operator as a free function:

    class Binary
    {
        public:
            Binary();
            Binary(int value);
    };

    Binary operator+(Binary const &lhs, Binary const &rhs);

After defining binary operators as free functions, several promotions are available:

• If the left-hand operand is of the intended class type, the right hand argument is promoted whenever possible;
• If the right-hand operand is of the intended class type, the left hand argument is promoted whenever possible;
• No promotions occur when neither operand is of the intended class type;
• An ambiguity occurs when promotions to different classes are possible for the two operands. For example:

      class A;
  
      class B
      {
          public:
              B(A const &a);
      };
  
      class A
      {
          public:
              A();
              A(B const &b);
      };
  
      A operator+(A const &a, B const &b);
      B operator+(B const &b, A const &a);
  
      int main()
      {
          A a;
  
          a + a;
      };
  

  Here, both overloaded + operators are possible candidates when compiling the statement a + a. The ambiguity must be solved by explicitly promoting one of the arguments, e.g., a + B{a}, which enables the compiler to resolve the ambiguity to the first overloaded + operator.

The next step consists of implementing the required overloaded binary compound assignment operators, having the form @=, where @ represents a binary operator. As these operators always have left-hand side operands which are object of their own classes, they are implemented as genuine member functions. Compound assignment operators usually return references to the objects for which the binary compound assignment operators were requested, as these objects might be modified in the same statement. E.g., (s2 += s3) + " postfix".

Here is our second revision of the class Binary, showing the declaration of the plain binary operator as well as the corresponding compound assignment operator:

    class Binary
    {
        public:
            Binary();
            Binary(int value);

            Binary &operator+=(Binary const &rhs);
    };

    Binary operator+(Binary const &lhs, Binary const &rhs);

How should the compound addition assignment operator be implemented? When implementing compound binary assignment operators the strong guarantee should always be kept in mind: if the operation might throw use a temporary object and swap. Here is our implementation of the compound assignment operator:

    Binary &Binary::operator+=(Binary const &rhs)
    {
        Binary tmp{ *this };
        tmp.add(rhs);           // this might throw
        swap(tmp);
        return *this;
    }

It's easy to implement the free binary operator: the lhs argument is copied into a Binary tmp to which the rhs operand is added. Then tmp is returned, using copy elision. The class Binary declares the free binary operator as a friend (cf. chapter 15), so it can call Binary's add member:

    class Binary
    {
        friend Binary operator+(Binary const &lhs, Binary const &rhs);

        public:
            Binary();
            Binary(int value);

            Binary &operator+=(Binary const &other);

        private:
            void add(Binary const &other);
    };

The binary operator's implementation becomes:

    Binary operator+(Binary const &lhs, Binary const &rhs)
    {
        Binary tmp{ lhs };
        tmp.add(rhs);
        return tmp;
    }

If the class Binary is move-aware then it's attractive to add move-aware binary operators. In this case we also need operators whose left-hand side operands are rvalue references. When a class is move aware various interesting implementations are suddenly possible, which we encounter below, and in the next (sub)section. First have a look at the signature of such a binary operator (which should also be declared as a friend in the class interface):

    Binary operator+(Binary &&lhs, Binary const &rhs);

Since the lhs operand is an rvalue reference, we can modify it ad lib. Binary operators are commonly designed as factory functions, returning objects created by those operators. However, the (modified) object referred to by lhs should itself not be returned. As stated in the C++ standard,

A temporary object bound to a reference parameter in a function call persists until the completion of the full-expression containing the call.

The lifetime of a temporary bound to the returned value in a function return statement is not extended; the temporary is destroyed at the end of the full-expression in the return statement.

Alternatively, the binary operator can first create an object by move constructing it from the operator's lhs operand, performing the binary operation on that object and the operator's rhs operand, and then return the modified object (allowing the compiler to apply copy elision). It's a matter of taste which one is preferred.

Here are the two implementations. Because of copy elision the explicitly defined ret object is created in the location of the return value. Both implementations, although they appear to be different, show identical run-time behavior:

                // first implementation: modify lhs
    Binary operator+(Binary &&lhs, Binary const &rhs)   
    {
        lhs.add(rhs);
        return std::move(lhs);
    }
                // second implementation: move construct ret from lhs
    Binary operator+(Binary &&lhs, Binary const &rhs)   
    {
        Binary ret{ std::move(lhs) };
        ret.add(rhs);
        return ret;
    }

    copy operator+          = b1 + b2 
    Copy constructor        = tmp(b1) 
        adding              = tmp.add(b2)
    copy elision            : tmp is returned from b1 + b2
        
    move operator+          = tmp + b3 
    adding                  = tmp.add(b3)
    Move construction       = tmp2(move(tmp)) is returned     

But we're not there yet: in the next section we encounter possibilities for several more interesting implementations, in the context of compound assignment operators.

11.6.1: Member function reference bindings (& and &&)

An expression like

    Binary{} + varB + varC + varD

Now consider the situation where we have a function defining a Binary && parameter, and a second Binary const & parameter. Inside that function these values need to be added, and their sum is then passed as argument to two other functions. We could do this:

    void fun1(Binary &&lhs, Binary const &rhs)
    {
        lhs += rhs;
        fun2(lhs);
        fun3(lhs);
    }

In this example another temporary object is indeed not required: lhs remains in existence until fun1 ends. But different from the binary operators the binary compound assignment operators don't have explicitly defined left-hand side operands. But we still can inform the compiler that a particular member (so, not merely compound assignment operators) should only be used when the objects calling those members is an anonymous temporary object, or a non-anonymous (modifiable or non-modifiable) object. For this we use reference bindings a.k.a. reference qualifiers.

Reference bindings consist of a reference token (&), optionally preceded by const, or an rvalue reference token (&&). Such reference qualifiers are immediately affixed to the function's head (this applies to the declaration and the implementation alike). Functions provided with rvalue reference bindings are selected by the compiler when used by anonymous temporary objects, whereas functions provided with lvalue reference bindings are selected by the compiler when used by other types of objects.

Reference qualifiers allow us to fine-tune our implementations of compund assignment operators like operator+=. If we know that the object calling the compound assignment operator is itself a temporary, then there's no need for a separate temporary object. The operator may directly perform its operation and could then return itself as an rvalue reference. Here is the implementation of operator+= tailored to being used by temporary objects:

    Binary &&Binary::operator+=(Binary const &rhs) &&
    {
        add(rhs);                   // directly add rhs to *this, 
        return std::move(*this);    // return the temporary object itself
    }

    cout << (Binary{} += existingBinary) << '\n';

    Binary &&rref = (Binary{} += existingBinary);
    cout << rref << '\n';

A full-proof alternative implementation of the rvalue-reference bound operator+= returns a move-constructed copy:

    Binary Binary::operator+=(Binary const &rhs) &&
    {
        add(rhs);                   // directly add rhs to *this, 
        return std::move(*this);    // return a move constructed copy
    }

Which implementation to use may be a matter of choice: if users of Binary know what they're doing then the former implementation can be used, since these users will never use the above rref initialization. If you're not so sure about your users, use the latter implementation: formally your users will do something they shouldn't do, but there's no penalty for that.

For the compound assignment operator called by an lvalue reference (i.e., a named object) we use the implementation for operator+= from the previous section (note the reference qualifier):

    Binary &Binary::operator+=(Binary const &other) &
    {
        Binary tmp(*this);
        tmp.add(other);     // this might throw
        swap(tmp);
        return *this;
    }

    operator+=    (&)       = b2 += b3
    Copy constructor        = tmp(b2) 
        adding              = tmp.add(b3)
        swap                = b2 <-> tmp
    return                  = b2

    operator+=    (&)       = b1 += b2
    Copy constructor        = tmp(b1) 
        adding              = tmp.add(b2)
        swap                = b1 <-> tmp
    return                  = b1

When the leftmost object is a temporary then a copy construction and swap call are replaced by the construction of an anonymous object. E.g., with Binary{} += b2 += b3 we observe:

    operator+=    (&)       = b2 += b3
    Copy constructor        = tmp(b2) 
        adding              = tmp.add(b3)
        swap                = b2 <-> tmp
    
    Anonymous object        = Binary{}

    operator+=    (&&)      = Binary{} += b2
        adding              = add(b2)

    return                  = move(Binary{})

For Binary &Binary::operator+=(Binary const &other) & an alternative implementation exists, using one single return statement, but in fact requiring two extra function calls. It's a matter of taste whether you prefer less code to write or executing fewer function calls:

    Binary &Binary::operator+=(Binary const &other) &
    {
        return *this = Binary{*this} += rhs;
    }

11.7: Overloading `operator new(size_t)'

It is possible to define multiple versions of the operator new, as long as each version defines its own unique set of arguments. When overloaded operator new members must dynamically allocate memory they can do so using the global operator new, applying the scope resolution operator ::. In the next example the overloaded operator new of the class String initializes the substrate of dynamically allocated String objects to 0-bytes:

    #include <cstring>
    #include <iosfwd>

    class String
    {
        std::string *d_data;

        public:
            void *operator new(size_t size)
            {
                return memset(::operator new(size), 0, size);
            }
            bool empty() const
            {
                return d_data == 0;
            }
    };

    #include "string.h"
    #include <iostream>
    using namespace std;

    int main()
    {
        String *sp = new String;

        cout << boolalpha << sp->empty() << '\n';   // shows: true
    }

• First, String::operator new was called, allocating and initializing a block of memory, the size of a String object.
• Next, a pointer to this block of memory was passed to the (default) String constructor. Since no constructor was defined, the constructor itself didn't do anything at all.

All member functions (including constructors and destructors) we've encountered so far define a (hidden) pointer to the object on which they should operate. This hidden pointer becomes the function's this pointer.

In the next example of pseudo C++ code, the pointer is explicitly shown to illustrate what's happening when operator new is used. In the first part a String object str is directly defined, in the second part of the example the (overloaded) operator new is used:

    String::String(String *const this);     // real prototype of the default
                                            // constructor

    String *sp = new String;                // This statement is implemented
                                            // as follows:

        String *sp = static_cast<String *>(            // allocation
                        String::operator new(sizeof(String))
                     );
        String::String(sp);                                 // initialization

Following the allocation, the memory is passed (as the this pointer) to the constructor for further processing.

Operator new can have multiple parameters. The first parameter is initialized as an implicit argument and is always a size_t parameter. Additional overloaded operators may define additional parameters. An interesting additional operator new is the placement new operator. With the placement new operator a block of memory has already been set aside and one of the class's constructors is used to initialize that memory. Overloading placement new requires an operator new having two parameters: size_t and char *, pointing to the memory that was already available. The size_t parameter is implicitly initialized, but the remaining parameters must explicitly be initialized using arguments to operator new. Hence we reach the familiar syntactical form of the placement new operator in use:

    char buffer[sizeof(String)];        // predefined memory
    String *sp = new(buffer) String;    // placement new call

    void *operator new(size_t size, char *memory);

    void *String::operator new(size_t size, char *memory)
    {
        return memset(memory, 0, size);
    }

        // use:
    String *next(String **pointers, size_t *idx)
    {
        return new(pointers, (*idx)++) String;
    }

        // implementation:
    void *String::operator new(size_t size, String **pointers, size_t idx)
    {
        return pointers[idx] = ::operator new(size);
    }

11.8: Overloading `operator delete(void *)'

Operator delete must define a void * parameter. A second overloaded version defining a second parameter of type size_t is related to overloading operator new[] and is discussed in section 11.9.

Overloaded operator delete members return void.

The `home-made' operator delete is called when deleting a dynamically allocated object after executing the destructor of the associated class. So, the statement

    delete ptr;

    ptr->~String(); // call the class's destructor

                    // and do things with the memory pointed to by ptr
    String::operator delete(ptr);

    void String::operator delete(void *ptr)
    {
        // any operation considered necessary, then, maybe:
        ::delete ptr;
    }

    void operator delete(void *ptr);

11.9: Operators `new[]' and `delete[]'

As it is possible to overload new[] and delete[] as well as operator new and operator delete, one should be careful in selecting the appropriate set of operators. The following rule of thumb should always be applied:

If new is used to allocate memory, delete should be used to deallocate memory. If new[] is used to allocate memory, delete[] should be used to deallocate memory.

By default these operators act as follows:

• operator new is used to allocate a single object or primitive value. With an object, the object's constructor is called.
• operator delete is used to return the memory allocated by operator new. Again, with class-type objects, the class's destructor is called.
• operator new[] is used to allocate a series of primitive values or objects. If a series of objects is allocated, the class's default constructor is called to initialize each object individually.
• operator delete[] is used to delete the memory previously allocated by new[]. If objects were previously allocated, then the destructor is called for each individual object. Be careful, though, when pointers to objects were allocated. If pointers to objects were allocated the destructors of the objects to which the allocated pointers point won't automatically be called. A pointer is a primitive type and so no further action is taken when it is returned to the common pool.

11.9.1: Overloading `new[]'

    void *operator new[](size_t size);

    void *operator new[](size_t size)
    {
        return ::operator new[](size);
        // alternatively:
        // return ::operator new(size);
    }

Once the overloaded operator new[] has been defined, it is automatically used in statements like:

    String *op = new String[12];

    void *String::operator new[](size_t size, char *memory)
    {
        return memset(memory, 0, size);
    }

    char buffer[12 * sizeof(String)];
    String *sp = new(buffer) String[12];

11.9.2: Overloading `delete[]'

    void operator delete[](void *memory);

There are some subtleties to be aware of when implementing operator delete[]. Although the addresses returned by new and new[] point to the allocated object(s), there is an additional size_t value available immediately before the address returned by new and new[]. This size_t value is part of the allocated block and contains the actual size of the block. This of course does not hold true for the placement new operator.

When a class defines a destructor the size_t value preceding the address returned by new[] does not contain the size of the allocated block, but the number of objects specified when calling new[]. Normally that is of no interest, but when overloading operator delete[] it might become a useful piece of information. In those cases operator delete[] does not receive the address returned by new[] but rather the address of the initial size_t value. Whether this is at all useful is not clear. By the time delete[]'s code is executed all objects have already been destroyed, so operator delete[] is only to determine how many objects were destroyed but the objects themselves cannot be used anymore.

Here is an example showing this behavior of operator delete[] for a minimal Demo class:

    struct Demo
    {
        size_t idx;
        Demo()
        {
            cout << "default cons\n";
        }
        ~Demo()
        {
            cout << "destructor\n";
        }
        void *operator new[](size_t size)
        {
            return ::operator new(size);
        }
        void operator delete[](void *vp)
        {
            cout << "delete[] for: " << vp << '\n';
            ::operator delete[](vp);
        }
    };

    int main()
    {
        Demo *xp;
        cout << ((int *)(xp = new Demo[3]))[-1] << '\n';
        cout << xp << '\n';
        cout << "==================\n";
        delete[] xp;
    }
    // This program displays (your 0x?????? addresses might differ, but
    // the difference between the two should be sizeof(size_t)):
    //  default cons
    //  default cons
    //  default cons
    //  3
    //  0x8bdd00c
    //  ==================
    //  destructor
    //  destructor
    //  destructor
    //  delete[] for: 0x8bdd008

        delete[] new String[5];

Operator delete[] may also be overloaded using an additional size_t parameter:

    void operator delete[](void *p, size_t size);

    void String::operator delete[](void *p, size_t size)
    {
        cout << "deleting " << size << " bytes\n";
        ::operator delete[](ptr);
    }

Additional overloads of operator delete[] may be defined, but to use them they must explicitly be called as static member functions (cf. chapter 8). Example:

        // declaration:
    void String::operator delete[](void *p, ostream &out);
        // usage:
    String *xp = new String[3];
    String::operator delete[](xp, cout);

11.9.3: The `operator delete(void *, size_t)' family

Since the C++14 standard the global void operator delete(void *, size_t size) and void operator delete[](void *, size_t size) functions can also be overloaded.

When a global sized deallocation function is defined, it is automatically used instead of the default, non-sized deallocation function. The performance of programs may improve if a sized deallocation function is available (cf. http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3663.html).

11.9.4: `new[]', `delete[]' and exceptions

To begin, new[] might throw while trying to allocate the required memory. In this case a bad_alloc is thrown and we don't leak as nothing was allocated.

Having allocated the required memory the class's default constructor is going to be used for each of the objects in turn. At some point a constructor might throw. What happens next is defined by the C++ standard: the destructors of the already constructed objects are called and the memory allocated for the objects themselves is returned to the common pool. Assuming that the failing constructor offers the basic guarantee new[] is therefore exception safe even if a constructor may throw.

The following example illustrates this behavior. A request to allocate and initialize five objects is made, but after constructing two objects construction fails by throwing an exception. The output shows that the destructors of properly constructed objects are called and that the allocated substrate memory is properly returned:

    #include <iostream>
    using namespace std;

    static size_t count = 0;

    class X
    {
        int x;

        public:
            X()
            {
                if (count == 2)
                    throw 1;
                cout << "Object " << ++count << '\n';
            }
            ~X()
            {
                cout << "Destroyed " << this << "\n";
            }
            void *operator new[](size_t size)
            {
                cout << "Allocating objects: " << size << " bytes\n";
                return ::operator new(size);
            }
            void operator delete[](void *mem)
            {
                cout << "Deleting memory at " << mem << ", containing: " <<
                    *static_cast<int *>(mem) << "\n";
                ::operator delete(mem);
            }
    };

    int main()
    try
    {
        X *xp = new X[5];
        cout << "Memory at " << xp << '\n';
        delete[] xp;
    }
    catch (...)
    {
        cout << "Caught exception.\n";
    }
    // Output from this program (your 0x??? addresses might differ)
    //  Allocating objects: 24 bytes
    //  Object 1
    //  Object 2
    //  Destroyed 0x8428010
    //  Destroyed 0x842800c
    //  Deleting memory at 0x8428008, containing: 5
    //  Caught exception.

11.10: Function Objects

Function objects are important when using generic algorithms. The use of function objects is preferred over alternatives like pointers to functions. The fact that they are important in the context of generic algorithms leaves us with a didactic dilemma. At this point in the C++ Annotations it would have been nice if generic algorithms would already have been covered, but for the discussion of the generic algorithms knowledge of function objects is required. This bootstrapping problem is solved in a well known way: by ignoring the dependency for the time being, for now concentrating on the function object concept.

Function objects are objects for which operator() has been defined. Function objects are not just used in combination with generic algorithms, but also as a (preferred) alternative to pointers to functions.

Function objects are frequently used to implement predicate functions. Predicate functions return boolean values. Predicate functions and predicate function objects are commonly referred to as `predicates'. Predicates are frequently used by generic algorithms such as the count_if generic algorithm, covered in chapter 19, returning the number of times its function object has returned true. In the standard template library two kinds of predicates are used: unary predicates receive one argument, binary predicates receive two arguments.

Assume we have a class Person and an array of Person objects. Further assume that the array is not sorted. A well known procedure for finding a particular Person object in the array is to use the function lsearch, which performs a lineair search in an array. Example:

    Person &target = targetPerson();    // determine the person to find
    Person *pArray;
    size_t n = fillPerson(&pArray);

    cout << "The target person is";

    if (!lsearch(&target, pArray, &n, sizeof(Person), compareFunction))
        cout << " not";
    cout << "found\n";

The comparison function must be available, as its address is one of the arguments of lsearch. It must be a real function having an address. If it is defined inline then the compiler has no choice but to ignore that request as inline functions don't have addresses. CompareFunction could be implemented like this:

    int compareFunction(void const *p1, void const *p2)
    {
        return *static_cast<Person const *>(p1)     // lsearch wants 0
                !=                                  // for equal objects
                *static_cast<Person const *>(p2);
    }

On average n / 2 times at least the following actions take place:

1. The two arguments of the compare function are pushed on the stack;
2. The value of the final parameter of lsearch is determined, producing compareFunction's address;
3. The compare function is called;
4. Then, inside the compare function the address of the right-hand argument of the Person::operator!= argument is pushed on the stack;
5. Person::operator!= is evaluated;
6. The argument of the Person::operator!= function is popped off the stack;
7. The two arguments of the compare function are popped off the stack.

    Person const *PersonSearch(Person *base, size_t nmemb,
                               Person const &target);

    Person &target = targetPerson();
    Person *pArray;
    size_t n = fillPerson(&pArray);

    cout << "The target person is";

    if (!PersonSearch(pArray, n, target))
        cout << " not";

    cout << "found\n";

    Person const *PersonSearch(Person *base, size_t nmemb,
                                Person const &target)
    {
        for (int idx = 0; idx < nmemb; ++idx)
            if (target(base[idx]))
                return base + idx;
        return 0;
    }

    bool Person::operator()(Person const &other) const
    {
        return *this == other;
    }

    bool operator()(Person const &other) const;

1. The address of the right-hand argument of the Person::operator== argument is pushed on the stack;
2. The operator== function is evaluated (which probably also is a semantic improvement over calling operator!= when looking for an object equal to a specified target object);
3. The argument of Person::operator== argument is popped off the stack.

Function objects may truly be defined inline. Functions that are called indirectly (i.e., using pointers to functions) can never be defined inline as their addresses must be known. Therefore, even if the function object needs to do very little work it is defined as an ordinary function if it is going to be called through pointers. The overhead of performing the indirect call may annihilate the advantage of the flexibility of calling functions indirectly. In these cases using inline function objects can result in an increase of a program's efficiency.

An added benefit of function objects is that they may access the private data of their objects. In a search algorithm where a compare function is used (as with lsearch) the target and array elements are passed to the compare function using pointers, involving extra stack handling. Using function objects, the target person doesn't vary within a single search task. Therefore, the target person could be passed to the function object's class constructor. This is in fact what happens in the expression target(base[idx]) receiving as its only argument the subsequent elements of the array to search.

11.10.1: Constructing manipulators

Actually the construction of a manipulator is rather simple. To start, a definition of the manipulator is needed. Let's assume we want to create a manipulator w10 which sets the field width of the next field to be written by the ostream object to 10. This manipulator is constructed as a function. The w10 function needs to know about the ostream object in which the width must be set. By providing the function with an ostream & parameter, it obtains this knowledge. Now that the function knows about the ostream object we're referring to, it can set the width in that object.

Next, it must be possible to use the manipulator in an insertion sequence. This implies that the return value of the manipulator must be a reference to an ostream object also.

From the above considerations we're now able to construct our w10 function:

    #include <ostream>
    #include <iomanip>

    std::ostream &w10(std::ostream &str)
    {
        return str << std::setw(10);
    }

        #include <iostream>
        #include <iomanip>

        using namespace std;

        extern ostream &w10(ostream &str);

        int main()
        {
            w10(cout) << 3 << " ships sailed to America\n";
            cout << "And " << w10 << 3 << " more ships sailed too.\n";
        }

    ostream& operator<<(ostream &(*func)(ostream &str))
    {
        return (*func)(*this);
    }

    ios_base &operator<<(ios_base &(*func)(ios_base &base))
    {
        (*func)(*this);
        return *this;
    }

The above procedure does not work for manipulators requiring arguments. It is of course possible to overload operator<< to accept an ostream reference and the address of a function expecting an ostream & and, e.g., an int, but while the address of such a function may be specified with the <<-operator, the arguments itself cannot be specified. So, one wonders how the following construction has been implemented:

    cout << setprecision(3)

11.10.1.1: Manipulators requiring arguments

//The problem appears to be that you can't call a function in an //insertion sequence: when using multiple operator<< operators in one statement //the compiler calls the functions, saves their return values, and then //uses their return values in the insertion sequence. That invalidates the //ordering of the arguments passed to your <<-operators. // //So, one might consider constructing another overloaded operator<< accepting //the address of a function receiving not just the ostream reference, but a //series of other arguments as well. But this creates the problem that it isn't //clear how the function should receive its arguments: you can't just call it //since that takes us back to the above-mentioned problem. Merely passing its //address is fine, but then no arguments can be passed to the function.

Manipulators, maybe requiring arguments, can also be defined without using macros. One solution, suitable for modifying globally available objects (like cin, or cout) is based on using anonymous objects:

• First, a class is defined, e.g. Align, whose constructor expects the arguments configuring the required manipulation. In our example representing, respectively, a field width and an alignment type.
• The class also supports an overloaded insertion (or extraction) operator. E.g.,

      ostream &operator<<(ostream &ostr, Align const &align)
  

• Next, the (anonymous) object is inserted into a stream. The insertion operator passes the stream to Align::align, allowing that member to configure (and return) the provided stream.

    #include <iostream>
    #include <iomanip>

    class Align
    {
        unsigned d_width;
        std::ios::fmtflags d_alignment;

        public:
            Align(unsigned width, std::ios::fmtflags alignment);
            std::ostream &operator()(std::ostream &ostr) const;
    };

    Align::Align(unsigned width, std::ios::fmtflags alignment)
    :
        d_width(width),
        d_alignment(alignment)
    {}

    std::ostream &Align::operator()(std::ostream &ostr) const
    {
        ostr.setf(d_alignment, std::ios::adjustfield);
        return ostr << std::setw(d_width);
    }

    std::ostream &operator<<(std::ostream &ostr, Align &&align)
    {
        return align(ostr);
    }

    using namespace std;

    int main()
    {
        cout
            << "`" << Align{ 5, ios::left } << "hi" << "'"
            << "`" << Align{ 10, ios::right } << "there" << "'\n";
    }

    /*
        Generated output:

        `hi   '`     there'
    */

When (local) objects must be manipulated, then the class that must provide manipulators may define function call operators receiving the required arguments. E.g., consider a class Matrix that should allow its users to specify the value and line separators when inserting the matrix into an ostream.

Two data members (e.g., char const *d_valueSep and char const *d_lineSep) are defined (and initialized to acceptable values). The insertion function inserts d_valueSep between values, and d_lineSep at the end of inserted rows. The member operator()(char const *valueSep, char const *lineSep) simply assigns values to the corresponding data members.

Given an object Matrix matrix, then at this point matrix(" ", "\n") can be called. The function call operator should probably not insert the matrix, as the responsibility of manipulators is to manipulate, not to insert. So, to insert a matrix a statement like

        cout << matrix(" ", "\n") << matrix << '\n';

The return value of the function call operator remains to be specified. The return value should be insertable, but in fact should not insert anything at all. An empty NTBS could be returned, but that's a bit kludge-like. Instead the address of a manipulator function, not performing any action, can be returned. Here's the implementation of such an empty manipulator:

        // static       (alternatively a free function could be used)
        std::ostream &Matrix::nop(std::ostream &out)
        {
            return out;
        }

        std::ostream &( 
            *Matrix::operator()(char const *valueSep, char const *lineSep) ) 
                                                            (std::ostream &)
        {
            d_valueSep = valueSep;
            d_lineSep = lineSep;
            return nop;
        }

11.11: Lambda expressions

Frequently the function or function object is not readily available, and it must be defined in or near the location where it is used. This is commonly realized by defining a class or function in the anonymous namespace (say: class or function A), passing an A to the code needing A. If that code is itself a member function of the class B, then A's implementation might benefit from having access to the members of class B.

This scheme usually results in a significant amount of code (defining the class), or it results in complex code (to make available software elements that aren't automatically accessible to A's code). It may also result in code that is irrelevant at the current level of specification. Nested classes don't solve these problems either. Moreover, nested classes can't be used in templates.

lambda expressions solve these problems. A lambda expression defines an anonymous function object which may immediately be passed to functions expecting function object arguments, as explained in the next few sections.

According to the C++ standard, lambda expressions provide a concise way to create simple function objects. The emphasis here is on simple: a lambda expression's size should be comparable to the size of inline-functions: just one or maybe two statements. If you need more code, then encapsulate that code in a separate function which is then called from inside the lambda expression's compound statement, or consider designing a separate function object.

11.11.1: Lambda expressions: syntax

Lambda expressions are used inside blocks, classes or namespaces (i.e., pretty much anywhere you like). Their implied closure type is defined in the smallest block, class or namespace scope containing the lambda expression. The closure object's visibility starts at its point of definition and ends where its closure type ends.

The closure type defines a (const) public inline function call operator. Here is an example of a lambda expression:

    []                      // the `lambda-introducer'
    (int x, int y)          // the `lambda-declarator'
    {                       // a normal compound-statement
        return x * y;
    }

    [](int x, int y) mutable
    ...

Declarator specifiers can be mutable, constexpr, or both. A constexpr lambda-expression is itself a constexpr, which may be compile-time evaluated if its arguments qualify as const-expressions. Moreover, if a lambda-expression is defined inside a constexpr function then the lambda-expression itself must qualify as a constexpr, and explicitly specifying the constexpr declarator specifier is not required. The following function definitions, therefore, are identical:

    int constexpr change10(int n)
    {
        return [n] 
               { 
                   return n > 10 ? n - 10 : n + 10; 
               }();
    }
    
    int constexpr change10(int n)
    {
        return [n] () constexpr 
               { 
                   return n > 10 ? n - 10 : n + 10; 
               }();
    }

A closure object as defined by the previous lambda expression could be used e.g., in combination with the accumulate (cf. section 19.1.1) generic algorithm to compute the product of a series of int values stored in a vector:

    cout << accumulate(vi.begin(), vi.end(), 1,
                [](int x, int y) { return x * y; });

• the lambda expression does not contain a return statement (i.e., a void lambda expression);
• the lambda expression contains a single return statement; or
• the lambda expression contains multiple return statements returning values of identical types (e.g., all int values).

If there are multiple return statements returning values of different types then the lambda expression's return type must specified be explicitly using a late-specified return type, (cf. section 3.3.6):

    [](int x, int y) -> int
    {
        return y < 0 ?
                    x / static_cast<double>(y)
                :
                    z + x;
    }

Variables that are visible at the location of a lambda expression can be accessed by the lambda expression. How these variables are accessed depends on the contents of the lambda-introducer (the area between the square brackets, called the lambda-capture). The lambda-capture allows passing a local context to lambda expressions.

Visible global and static variables as well as local variables defined in the lambda expression's compound statement itself can directly be accessed and, when applicable, be modified. Example:

    int global;
    
    void fun()
    {
        []()  // [] may contain any specification
        { 
            int localVariable = 0;
            localVariable = ++global; 
        };
    }

If a lambda expression is defined inside a function then the lambda expression may access all the function's local variables which are visible at the lambda expression's point of definition.

An initial & character in the lambda-capture accesses these local variables by reference. These variables can then be modified from within the lambda expression.

An initial = character in the lambda-capture creates a local copy of the referred-to local variables. Note that in this case the values of these local copies can only be changed by the lambda expression if the lambda expression is defined using the mutable keyword. E.g.,

    struct Class
    {
        void fun()
        {
            int var = 0;
            [=]() mutable
            {
                ++var;  // modifies the local
            }           // copy, not fun's var
        }
    }

When defining lambda-expressions in (non-static) member functions, the lambda-capture may also contain *this. Even when not specified, lambda expressions implicitly capture their this pointers, and class members are always accessed relative to this. But when members are called asynchronously a problem may arise, because the asynchronously called lambda function may refer to members of an object whose lifetime ended shortly after asynchronously calling the lambda function. This potentially arising problem is solved by using `, *this' in the lambda-capture if it starts with =, e.g., [=, *this] (in addition, variables may still also be captured, as usual). When specifying `, *this' the object to which this refers is explicitly captured: if the object's scope ends it is not immediately destroyed, but it is captured by the lambda-expression for the duration of that expression. In order to use the `, *this' specification, the object must directly be available. Consider the following example:

    struct s2 
    {
        double ohseven = .007;
        auto f() 
        {
            return [this] 
                    {
                        return [*this] 
                                {
                                    return ohseven; // OK
                                }
                    }();
        }
        auto g() 
        {
            return [] 
                    {
                      return [*this] 
                        { 
                            // error: *this not captured by outer
                            // lambda-expression 
                        }; 
                    }();
        }
    };

Fine-tuning lambda-captures is also possible. With an initial =, comma-separated &var specifications indicate that the mentioned local variables should be processed by reference, rather than as copies; with an initial &, comma separated var specifications indicate that local copies should be used of the mentioned local variables. Again, these copies have immutable values unless the lambda expression is provided with the mutable keyword.

Another fine-tuning consists of using this in the lambda-capture: it also allows the lambda-expression to access the surrounding class members. Example:

    class Data
    {
        std::vector<std::string> d_names;
        public:
            void show() const
            {
                int count = 0;
                std::for_each(d_names.begin(), d_names.end(),
                    [this, &count](std::string const &name)
                    {
                        std::cout << ++count << ' ' <<
                            capitalized(name) << '\n';
                    }
                );
            }
        private:
            std::string capitalized(std::string name);
    };

Although lambda expressions are anonymous function objects, they can be assigned to variables. Often, the variable is defined using the keyword auto. E.g.,

    auto sqr = [](int x)
               {
                   return x * x;
               };

11.11.2: Using lambda expressions

First we consider named lambda expressions. Named lambda expressions nicely fit in the niche of local functions: when a function needs to perform computations which are at a conceptually lower level than the function's task itself, then it's attractive to encapsulate these computations in a separate support function and call the support function where needed. Although support functions can be defined in anonymous namespaces, that quickly becomes awkward when the requiring function is a class member and the support function also must access the class's members.

In that case a named lambda expression can be used: it can be defined inside a requiring function, and it may be given full access to the surrounding class. The name to which the lambda expression is assigned becomes the name of a function which can be called from the surrounding function. Here is an example, converting a numeric IP address to a dotted decimal string, which can also be accessed directly from an Dotted object (all implementations in-class to conserve space):

    class Dotted
    {
        std::string d_dotted;
        
        public:
            std::string const &dotted() const
            {
                return d_dotted;
            }
            std::string const &dotted(size_t ip)
            {
                auto octet = 
                    [](size_t idx, size_t numeric)
                    {
                        return to_string(numeric >> idx * 8 & 0xff);
                    };

                d_dotted = 
                        octet(3, ip) + '.' + octet(2, ip) + '.' +
                        octet(1, ip) + '.' + octet(0, ip);

                return d_dotted;
            }
    };

Next we consider the use of generic algorithms, like the for_each (cf. section 19.1.18):

    void showSum(vector<int> const &vi)
    {
        int total = 0;
        for_each(
            vi.begin(), vi.end(),
            [&](int x)
            {
                total += x;
            }
        );
        std::cout << total << '\n';
    }

But although generic algorithms are extremely useful, there may not always be one that fits the task at hand. Furthermore, an algorithm like for_each looks a bit unwieldy, now that the language offers range-based for-loops. So let's try this, instead of the above implementation:

    void showSum(vector<int> const &vi)
    {
        int total = 0;
        for (auto el: vi)
            [&](int x)
            {
                total += x;
            };

        std::cout << total << '\n';
    }

When a generic algorithm is given a lambda function, its implementation instantiates a reference to a function. The referenced function is thereupon called from within the generic algorithm. But, in the above example the range-based for-loop's nested statement merely represents the definition of a lambda function. Nothing is actually called, and hence total remains equal to 0.

Thus, to make the above example work we not only must define the lambda expression, but we must also call the lambda function. We can do this by giving the lambda function a name, and then call the lambda function by its given name:

    void showSum(vector<int> const &vi)
    {
        int total = 0;
        for (auto el: vi)
        {
            auto lambda = [&](int x)
                            {
                                total += x;
                            };

            lambda(el);
        }
        std::cout << total << '\n';
    }

In fact, there is no need to give the lambda function a name: the auto lambda definition represents the lambda function, which could also directly be called. The syntax for doing this may look a bit weird, but there's nothing wrong with it, and it allows us to drop the compound statement, required in the last example, completely. Here goes:

    void showSum(vector<int> const &vi)
    {
        int total = 0;
        for (auto el: vi)
            [&](int x)
            {
                total += x;
            }(el);          // immediately append the 
                            // argument list to the lambda
                            // function's definition
        std::cout << total << '\n';
    }

Lambda expressions can also be used to prevent spurious returns from condition_variable's wait calls (cf. section 20.5.3).

The class condition_variable allows us to do so by offering wait members expecting a lock and a predicate. The predicate checks the data's state, and returns true if the data's state allows the data's processing. Here is an alternative implementation of the down member shown in section 20.5.3, checking for the data's actual availability:

    void down()
    {
        unique_lock<mutex> lock(sem_mutex);
        condition.wait(lock, 
            [&]()
            {
                return semaphore != 0
            }
        );
        --semaphore;
    }

Lambda expression are primarily used to obtain functors that are used in a very localized section of a program. Since they are used inside an existing function we should realize that once we use lambda functions multiple aggregation levels are mixed. Normally a function implements a task which can be described at its own aggregation level using just a few sentences. E.g., ``the function std::sort sorts a data structure by comparing its elements in a way that is appropriate to the context where sort is called''. By using an existing comparison method the aggregation level is kept, and the statement is clear by itself. E.g.,

    sort(data.begin(), data.end(), greater<DataType>());

    sort(data.begin(), data.end(), 
        [&](DataType const &lhs, DataType const &rhs)
        {
            return lhs.greater(rhs);
        }
    );

On the other hand: lambda expressions also simplify code because the overhead of defining tailor-made functors is avoided. The advice, therefore, is to use lambda expressions sparingly. When they are used make sure that their sizes remain small. As a rule of thumb: lambda expressions should be treated like in-line functions, and should merely consist of one, or maybe occasionally two expressions.

A special group of lambda expressions is known as generic lambda expressions. As generic lambda expressions are in fact class templates, their coverage is postponed until chapter 22.

11.12: The case of [io]fstream::open()

    fstream out;
    out.open("/tmp/out", ios::out);

    fstream out;
    out.open("/tmp/out", ios::in | ios::out);

When trying to combine enum values using a `home made' enum we may run into problems. Consider the following:

    enum Permission
    {
        READ =      1 << 0,
        WRITE =     1 << 1,
        EXECUTE =   1 << 2
    };

    void setPermission(Permission permission);

    int main()
    {
        setPermission(READ | WRITE);
    }

The question is of course: why is it OK to combine ios::openmode values passing these combined values to the stream's open member, but not OK to combine Permission values.

Combining enum values using arithmetic operators results in int-typed values. Conceptually this never was our intention. Conceptually it can be considered correct to combine enum values if the resulting value conceptually makes sense as a value that is still within the original enumeration domain. Note that after adding a value READWRITE = READ | WRITE to the above enum we're still not allowed to specify READ | WRITE as an argument to setPermission.

To answer the question about combining enumeration values and yet stay within the enumeration's domain we turn to operator overloading. Up to this point operator overloading has been applied to class types. Free functions like operator<< have been overloaded, and those overloads are conceptually within the domain of their class.

As C++ is a strongly typed language realize that defining an enum is really something beyond the mere association of int-values with symbolic names. An enumeration type is really a type of its own, and as with any type its operators can be overloaded. When writing READ | WRITE the compiler performs the default conversion from enum values to int values and applies the operator to ints. It does so when it has no alternative.

But it is also possible to overload the enum type's operators. Thus we may ensure that we'll remain within the enum's domain even though the resulting value wasn't defined by the enum. The advantage of type-safety and conceptual clarity is considered to outweigh the somewhat peculiar introduction of values hitherto not defined by the enum.

Here is an example of such an overloaded operator:

    Permission operator|(Permission left, Permission right)
    {
        return static_cast<Permission>(static_cast<int>(left) | right);
    }

Operators like the above were defined for the ios::openmode enumeration type, allowing us to specify ios::in | ios::out as argument to open while specifying the corresponding parameter as ios::openmode as well. Clearly, operator overloading can be used in many situations, not necessarily only involving class-types.

11.13: User-defined literals

A user-defined literal is defined by a function (see also section 23.3) that must be defined at namespace scope. Such a function is called a literal operator. A literal operator cannot be a class member function. The names of a literal operator must start with an underscore, and a literal operator is used (called) by suffixing its name (including the underscore) to the argument that must be passed to it . Assuming _NM2km (nautical mile to km) is the name of a literal operator, then it could be called as 100_NM2km, producing, e.g., the value 185.2.

Using Type to represent the return type of the literal operator its generic declaration looks like this:

    Type operator "" _identifier(parameter-list);

• unsigned long long int. It is used as, e.g., 123_identifier. The argument to this literal operator can be decimal constants, binary constants (initial 0b), octal constants (initial 0) and hexadecimal constants (initial 0x);
• long double. It is used as, e.g., 12.25_NM2km;
• char const *text. The text argument is an NTBS. It is used as, e.g., 1234_pental. The argument must not be given double quotes, and must represent a numeric constant, as also expected by literal operators defining unsigned long long int parameters.
• char const *text, size_t len. Here, the compiler determines len as if it had called strlen(text). It is used as, e.g., "hello"_nVowels;
• wchar_t const *text, size_t len, same as the previous one, but accepting a string of wchar_t characters. It is used as, e.g., L"1234"_charSum;
• char16_t const *text, size_t len, same as the previous one, but accepting a string of char16_t characters. It is used as, e.g., u"utf 16"_uc;
• char32_t const *text, size_t len, same as the previous one, but accepting a string of char32_t characters. It is used as, e.g., U"UTF 32"_lc;

A literator operator can define any return type. Here is an example of a definition of the _NM2km literal operator:

    double operator "" _NM2km(char const *nm)
    {
        return std::stod(nm) * 1.852;
    }

    double value = 120_NM2km;   // example of use

    double constexpr operator "" _NM2km(long double nm)
    {
        return nm * 1.852;
    }

    double value = 450.5_NM2km;   // example of use

A numeric constant can also be processed completely at compile-time. Section 23.3 provides the details of this type of literal operator.

Arguments to literal operators are themselves always constants. A literal operator like _NM2km cannot be used to convert, e.g., the value of a variable. A literal operator, although it is defined as a function, cannot be called like a function. The following examples therefore result in compilation errors:

    double speed;

    speed_NM2km;        // no identifier 'speed_NM2km'
    _NM2km(speed);      // no function _NM2km
    _NM2km(120.3);      // no function _NM2km

11.14: Overloadable operators

    +       -       *       /       %       ^       &       |
    ~       !       ,       =       <       >       <=      >=
    ++      --      <<      >>      ==      !=      &&      ||
    +=      -=      *=      /=      %=      ^=      &=      |=
    <<=     >>=     []      ()      ->      ->*     new     new[]
    delete  delete[]

`Textual' alternatives of operators are also overloadable (e.g., operator and). However, note that textual alternatives are not additional operators. So, within the same context operator&& and operator and can not both be overloaded.

Several of these operators may only be overloaded as member functions within a class. This holds true for the '=', the '[]', the '()' and the '->' operators. Consequently, it isn't possible to redefine, e.g., the assignment operator globally in such a way that it accepts a char const * as an lvalue and a String & as an rvalue. Fortunately, that isn't necessary either, as we have seen in section 11.3.

Finally, the following operators cannot be overloaded:

    .       .*      ::      ?:      sizeof  typeid

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