A FunctionObject type is the type of an object that can be used on the left of the function call operator.
[edit] RequirementsThe type T
satisfies FunctionObject if
T
satisfies std::is_object, andGiven
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 [edit] NotesFunctions 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 libraryDemonstrates 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)[edit] See also a type for which the invoke operation is defined
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