static member function make_shared of shared_ptr

Question

Using libc++ I find out std::shared_ptr::make_shared() static member function in public section. It is very handy when I have already defined type alias to std::shared_ptr's specialization:

using T = int;
using P = std::shared_ptr< T >;
auto p = P::make_shared(123); // <=> std::make_shared< T >(123)
static_assert(std::is_same< decltype(p), P >::value);

I'm worry about standard compliance, because articles (1, 2) from trusted source mentions nothing about static member function make_shared of std::shared_ptr.

Is it bad parctice to use the function currently? Why?

Isn't the fact that it doesn't exist in the standard a good enough reason? — T.C., Jan 28 '16 at 23:27
You seem to already know that this function is non-standard. Isn't this just an opinion-based question? — Brian Bi, Jan 28 '16 at 23:27
I just checked the standard specification and it's not described there. This looks like implementation specific. This does not work with VS2015 c++ standard library. — Guillaume Gris, Jan 28 '16 at 23:28
@Brian I can suppose, that my source is outdated or incomplete or inaccurate, because it is public wiki. — Tomilov Anatoliy, Jan 28 '16 at 23:29

score 5 · Accepted Answer · answered Jan 28 '16 at 23:42

When using this static make_shared member function you depend on an implementation-specific extension of g++ / its standard library, as you already know. For the little gain you get from it, I would rather keep my code portable and use std::make_shared.

For the case you mentioned, i.e. constructing a new object using copy or move constructor from an existing one, I can suggest an alternative (untested; for C++11-compatibility you would have to add a trailing return type):

template <typename T>
auto share_ptr_to_new(T&& v) {
    using T2 = typename std::remove_reference<T>::type;
    return std::make_shared<T2>(std::forward<T>(v));
}

In the example above, you could then just write auto p = share_ptr_to_new(123).

Really close alternative for me is [overloading](http://coliru.stacked-crooked.com/a/f87c5e9131d5eda7) of `make_shared` for `std::shared_ptr` itself somewhere in user namespace. — Tomilov Anatoliy, Jan 28 '16 at 23:55

score 3 · Answer 2 · answered Jan 28 '16 at 23:33

3

FWIW, there is

template<class T, class... Args> shared_ptr<T> make_shared(Args&&... args);

as a non-member function.

You should be able to use std::make_shared instead of std::shared_ptr::make_shared.

answered Jan 28 '16 at 23:33

R Sahu

204,454
14
159
270

It is copy-book maxim for me and many others, who deal with C++14's ``. – Tomilov Anatoliy Jan 28 '16 at 23:35
@Orient, I didn't understand that. – R Sahu Jan 28 '16 at 23:37
1

Copy-book maxim synonims: truism, common truth. – Tomilov Anatoliy Jan 28 '16 at 23:39

static member function make_shared of shared_ptr

2 Answers2