Purpose of perfect forwarding for Callable argument in invocation expression?

Question

In Scott Meyer's book Effective Modern C++ on page 167 (of the print version), he gives the following example:

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

I completely understand the perfect forwarding of params, but it is unclear to me when perfect forwarding of func would ever be relevant. In other words, what are the advantages of the above over the following:

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

When `func` has a ref-qualified function call operator. – T.C. Apr 28 '16 at 16:31 — T.C., Apr 28 '16 at 16:31
Ohhhhhh.... makes perfect sense. – Daisy Sophia Hollman Apr 28 '16 at 16:37 — Daisy Sophia Hollman, Apr 28 '16 at 16:37

Jarod42 · Accepted Answer · 2016-04-28T16:44:59.817

13

For the same purpose as for arguments: so when Func::operator() is a ref-qualified:

struct Functor
{
    void operator ()() const &  { std::cout << "lvalue functor\n"; }
    void operator ()() const && { std::cout << "rvalue functor\n"; }
};

Demo

edited Apr 28 '16 at 16:44

answered Apr 28 '16 at 16:41

Jarod42

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4

Remove the `const` part, then it makes *better* sense. – Nawaz Apr 28 '16 at 16:44
no, as `const` rvalue reference will prefer `const &` overload to `&&` overload: `const Functor f() { return Functor{}; }; f()();` [example](http://coliru.stacked-crooked.com/a/3e1b8f476f9e3f05) – Dev Null Aug 10 '17 at 11:53

Purpose of perfect forwarding for Callable argument in invocation expression?

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