Can you allocate an array with something equivalent to make_shared?

问题

buffer = new char[64];
buffer = std::make_shared<char>(char[64]); ???

Can you allocate memory to an array using make_shared<>()?

I could do: buffer = std::make_shared<char>( new char[64] );

But that still involves calling new, it's to my understanding make_shared is safer and more efficient.

回答1:

The point of make_shared is to incorporate the managed object into the control block of the shared pointer,

Since you're dealing with C++11, perhaps using a C++11 array would satisfy your goals?

#include <memory>
#include <array>
int main()
{
    auto buffer = std::make_shared<std::array<char, 64>>();
}

Note that you can't use a shared pointer the same way as a pointer you'd get from new[], because std::shared_ptr (unlike std::unique_ptr, for example) does not provide operator[]. You'd have to dereference it: (*buffer)[n] = 'a';

回答2:

Do you need the allocated memory to be shared? You can use a std::unique_ptr instead and the std::make_unique available on C++14:

auto buffer = std::make_unique<char[]>(64);

There will be a std::make_shared version available in C++20:

auto buffer = std::make_shared<char[]>(64);

回答3:

How about this?

template<typename T>
inline std::shared_ptr<T> MakeArray(int size)
{
    return std::shared_ptr<T>( new T[size], []( T *p ){ delete [] p; } );
}

auto  buffer = new char[64];
auto  buffer = MakeArray<char>(64);

来源：https://stackoverflow.com/questions/13794447/can-you-allocate-an-array-with-something-equivalent-to-make-shared