When should I std::forward a function call?

问题:

A code snippet I saw in Effective Modern C++ has a clever implementation of the instrumentation rationale to create a function timer :

auto timeFuncInvocation = 
    [](auto&& func, auto&&... params)
    {
        start timer; 
        std::forward<decltype(func)>(func)(
            std::forward<decltype(params)>(params)...); 
        stop timer and record elapsed time; 
    };

My question is about std::forward<decltype(func)>(func)(...

• To my understanding, we are actually casting the function to its original type, but why is this needed? It looks like a simple call would do the trick.
• Are there any other cases where we use perfect forwarding to make a function call ?

---

^{This looks like a good use case for the use of familiar template syntax in lambda expressions in case we wanted to make the timer type a compile time constant.}

回答1:

A better description of what std::forward<decltype(func)>(func)(...) is doing would be preserving the value category of the argument passed to the lambda.

Consider the following functor with ref-qualified operator() overloads.

struct foo
{
    void operator()() const &&
    { std::cout << __PRETTY_FUNCTION__ << '\n'; }

    void operator()() const &
    { std::cout << __PRETTY_FUNCTION__ << '\n'; }
};

Remember that within the body of the lambda func is an lvalue (because it has a name). If you didn't forward the function argument the && qualified overload can never be invoked. Moreover, if the & qualified overload were absent, then even if the caller passed you an rvalue foo instance, your code would fail to compile.

Live demo