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.
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 |
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>.
- Some return types of functions of <functional>.
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 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)
See also
| a type for which the invoke operation is defined (named requirement) |