Do std::make_shared and std::make_unique have a "nothrow" version?

Question

For the new operator, we have the std::nothrow version:

std::unique_ptr<T> p = new(std::nothrow) T();

Do we have something like this for the std::make_shared or std::make_unique?

Answer 1

No, we don't. Looking through the cppreference pages for make_unique and make_shared, we see that every version uses the default new overload.

It is not difficult to implement one, though, something like this:

template <class T, class... Args>
std::unique_ptr<T> make_unique_nothrow(Args&&... args)
    noexcept(noexcept(T(std::forward<Args>(args)...)))
{
    return std::unique_ptr<T>(new (std::nothrow) T(std::forward<Args>(args)...));
}

template <class T, class... Args>
std::shared_ptr<T> make_shared_nothrow(Args&&... args)
    noexcept(noexcept(T(std::forward<Args>(args)...)))
{
    return std::shared_ptr<T>(new (std::nothrow) T(std::forward<Args>(args)...));
}

(Note that this version of make_shared_nothrow does not avoid double allocation as make_shared does.) C++20 added many new overloads for make_unique, but they can be implemented in a similar way. Also, per comment,

Don't forget to check the pointer before using it, when using this version. — Superlokkus Jul 18 '19 at 10:46