C++ named requirements: FunctionObject - cppreference.com (original) (raw)

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

[edit] 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:

[edit] 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.

[edit] Standard library

• All pointers to functions satisfy this requirement.
• All function objects defined in .
• Some return types of functions of .

[edit] Example

Demonstrates different types of function objects.

#include #include   void foo(int x) { std::cout << "foo(" << x << ")\n"; } void bar(int x) { std::cout << "bar(" << x << ")\n"; }   int main() { void(*fp)(int) = foo; fp(1); // calls foo using the pointer to function   std::invoke(fp, 2); // all FunctionObject types are Callable   auto fn = std::function(foo); // see also the rest of fn(3); fn.operator()(3); // the same effect as fn(3)   struct S { void operator()(int x) const { std::cout << "S::operator(" << x << ")\n"; } } s; s(4); // calls s.operator() s.operator()(4); // the same as s(4)   auto lam = [](int x) { std::cout << "lambda(" << x << ")\n"; }; lam(5); // calls the lambda lam.operator()(5); // the same as lam(5)   struct T { using FP = void (*)(int); operator FP() const { return bar; } } t; t(6); // t is converted to a function pointer static_cast<void (*)(int)>(t)(6); // the same as t(6) t.operator T::FP()(6); // the same as t(6) }

Output:

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

[edit] See also

| | a type for which the invoke operation is defined(named requirement) | | ---------------------------------------------------------------------- |