C++ named requirements: FunctionObject

A FunctionObject type is the type of an object that can be used on the left of the function call operator.

Requirements

The type T satisfies FunctionObject if

• The type T satisfies std::is_object, and

Given

• f, a value of type T or const T
• args, suitable argument list, which may be empty

The following expressions must be valid:

pub	Expression	Requirements
pub	f(args)	performs a function call

Notes

Functions and references to functions are not function object types, but can be used where function object types are expected due to function-to-pointer implicit conversion.

Standard library

• All pointers to functions satisfy this requirement.
• All function objects defined in <functional&gt;
• Some return types of functions of <functional&gt;

Example

demonstrates different types of function objects

#include <iostream>
 
void foo(int x) { std::cout << "foo(" << x << ")\n"; }
 
int main()
{
    void(*fp)(int) = foo;
    fp(1); // calls foo using the pointer to function
 
    struct S {
        void operator()(int x) const { std::cout << "S::operator(" << x << ")\n"; }
    } s;
    s(2); // calls s.operator()
    s.operator()(3); // the same
 
    auto lam = [](int x) { std::cout << "lambda(" << x << ")\n"; };
    lam(4); // calls the lambda
    lam.operator()(5); // the same
 
    struct T {
        static void bar(int x) { std::cout << "T::bar(" << x << ")\n"; }
        using FP = void (*)(int);
        operator FP() const { return bar; }
    } t;
    t(6); // t is converted to a function pointer
    static_cast<void (*)(int)>(t)(7); // the same
}

Result

foo(1)
S::operator(2)
S::operator(3)
lambda(4)
lambda(5)
T::bar(6)
T::bar(7)