std::reference_wrapper::operator() - cppreference.com (original) (raw)

| template< class... ArgTypes > typename std::result_of<T&(ArgTypes&&...)>::type operator() ( ArgTypes&&... args ) const; | | (since C++11) (until C++17) | | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | ------------------------------------- | | template< class... ArgTypes > std::invoke_result_t<T&, ArgTypes...> operator() ( ArgTypes&&... args ) const noexcept(/* see below */); | | (since C++17) (constexpr since C++20) |

Calls the Callable object, reference to which is stored, as if by INVOKE(get(), std::forward<ArgTypes>(args)...). This function is available only if the stored reference points to a Callable object.

T must be a complete type.

[edit] Parameters

args	-	arguments to pass to the called function

[edit] Return value

The return value of the called function.

[edit] Exceptions

[edit] Example

#include #include void f1() { std::cout << "reference to function called\n"; } void f2(int n) { std::cout << "bind expression called with " << n << " as the argument\n"; } int main() { std::reference_wrapper<void()> ref1 = std::ref(f1); ref1(); auto b = std::bind(f2, std::placeholders::_1); auto ref2 = std::ref(b); ref2(7); auto c = []{ std::cout << "lambda function called\n"; }; auto ref3 = std::ref(c); ref3(); }

Output:

reference to function called bind expression called with 7 as the argument lambda function called

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.