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# std::reference_wrapper::operator()

From cppreference.com
< cpp‎ | utility‎ | functional‎ | reference wrapper

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; |  | (since C++17)

Calls the `Callable` object, reference to which is stored. This function is available only if the stored reference points to a `Callable` object.

### [edit] Parameters

args | - | arguments to pass to the called function

### [edit] Return value

The return value of the called function.

### [edit] Exceptions

(none)

### [edit] Example

Run this code

    #include <iostream>
    #include <functional>

    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] See also

get operator T& | accesses the stored reference (public member function) [edit]

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