C++ named requirements:FunctionObject
From cppreference.com
A FunctionObject type is the type of an object that can be used on the left of the function call operator.
Contents |
[edit]Requirements
The type T satisfies FunctionObject if
- The type
Tsatisfies std::is_object, and
Given
f, a value of typeTorconst T,args, suitable argument list, which may be empty.
The following expressions must be valid:
| Expression | Requirements |
|---|---|
| f(args) | performs a function call |
[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 <functional>.
- Some return types of functions of <functional>.
[edit]Example
Demonstrates different types of function objects.
Run this code
#include <functional>#include <iostream> 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 <functional> 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 pointerstatic_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) |
