你可以分配一个相当于make_shared的数组吗？ [英] Can you allocate an array with something equivalent to make_shared?

问题描述

  buffer = new char [64]; 
 buffer = std :: make_shared< char>（char [64]）; ???可以使用 make_shared<>（）为数组分配内存。    / code>？
 
 
 
我可以做： buffer = std :: make_shared< char>（new char [64]）; / code> 
 
 
 
但是这还涉及调用new，这是因为我理解 make_shared 更安全，更多效率。
解决方案
 make_shared的意义是将托管对象合并到共享指针的控制块中，
 
 
 
因为你在处理C ++ 11，或许使用C ++ 11数组才能满足你的目标？
  #include< memory> 
 #include< array> 
 int main（）
 {
 auto buffer = std :: make_shared< std :: array< char，64& 
} 
  
注意，你不能像指针一样使用共享指针你会从新的[]，因为 std :: shared_ptr （不同于例如 std :: unique_ptr 不提供 operator [] 。您必须取消引用它：（* buffer）[n] ='a';  
 
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.
解决方案
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';

                        
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