A FunctionObject type is the type of an object that can be used on the left of the function call operator.
The type T satisfies FunctionObject if.
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:
| Expression | Requirements |
|---|---|
f(args) | performs a function call |
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.
<functional> <functional> demonstrates different types of function objects.
#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)
| a type for which the invoke operation is defined (named requirement) |
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https://en.cppreference.com/w/cpp/named_req/FunctionObject