std::make_shared and std::forward — what's the point?

Question

I'm trying to understand why std::make_shared is declared/implemented the way it is:

template<class _Tp, class ..._Args>
inline _LIBCPP_INLINE_VISIBILITY
typename enable_if
<
    !is_array<_Tp>::value,
    shared_ptr<_Tp>
>::type
make_shared(_Args&& ...__args)
{
    return shared_ptr<_Tp>::make_shared(_VSTD::forward<_Args>(__args)...);
}

What I'm curious about is:

Why does make_shared accept only rvalue-references? Why there's no overload for reference-to-constant (i.e., const _Args &)?
What's the point of _VSTD::forward call?

Good explanation or good hyperlink will both be highly appreciated.

It's not an rvalue reference. It's a forwarding reference. You need to learn about perfect forwarding. — Simple, Dec 10 '15 at 12:41
Possible duplicate of [Advantages of using forward](http://stackoverflow.com/questions/3582001/advantages-of-using-forward) — Simple, Dec 10 '15 at 12:44
There are some implementation details here, not sure that duplicate is a good idea. But since the OP is happy with the links, just add that to the answer. — Karoly Horvath, Dec 10 '15 at 12:45

Maxim Egorushkin · Accepted Answer · 2015-12-10T13:31:48.863

Why does make_shared accept only rvalue-references? Why there's no overload for reference-to-constant (i.e., const _Args &)?

This is a universal reference that binds both to r-values and l-values. For r-values the deduced type is T, for l-values it is T&.

What's the point of _VSTD::forward call?

When a temporary is bound to an r-value reference, that reference (the name) is not r-value any more. You need to be able to restore the original value category of the value and this is what std::forward<> is for: to forward the r-value arguments as r-values and l-value arguments as l-values.

Essentially std::forward applies && to T or T&. When && is applied to

T it becomes T&& - the r-value reference, and
T& becomes T&&&, which is again T&.

std::make_shared and std::forward — what's the point?

1 Answers1