make_unique and perfect forwarding

Question

Why is there no std::make_unique function template in the standard C++11 library? I find

std::unique_ptr<SomeUserDefinedType> p(new SomeUserDefinedType(1, 2, 3));

a bit verbose. Wouldn't the following be much nicer?

auto p = std::make_unique<SomeUserDefinedType>(1, 2, 3);

This hides the new nicely and only mentions the type once.

Anyway, here is my attempt at an implementation of make_unique:

template<typename T, typename... Args>
std::unique_ptr<T> make_unique(Args&&... args)
{
    return std::unique_ptr<T>(new T(std::forward<Args>(args)...));
}

It took me quite a while to get the std::forward stuff to compile, but I'm not sure if it's correct. Is it? What exactly does std::forward<Args>(args)... mean? What does the compiler make of that?

Answer 1

Herb Sutter, chair of the C++ standardization committee, writes on his blog:

That C++11 doesn’t include make_unique is partly an oversight, and it will almost certainly be added in the future.

He also gives an implementation that is identical with the one given by the OP.

Edit: std::make_unique now is part of C++14.