std :: shared_ptr初始化：make_shared< Foo>（）vs shared_ptr< T>（new Foo） [英] std::shared_ptr initialization: make_shared<Foo>() vs shared_ptr<T>(new Foo)

问题描述

之间有什么区别：

std::shared_ptr<int> p = std::shared_ptr<int>( new int );

和

std::shared_ptr<int> p = std::make_shared< int >();

？

为什么？

P。非常确定这一定已经回答了，但我找不到类似的问题。

P. S. Pretty sure this must have been answered already, but I can't find a similar question.

推荐答案

比必要的更详细：

std::shared_ptr<int> p(new int);  // or '=shared_ptr<int>(new int)' if you insist
auto p = std::make_shared<int>(); // or 'std::shared_ptr<int> p' if you insist

有什么区别？

What's the difference?

主要区别是第一个需要两个内存分配：一个用于托管对象（ new int ），一个用于引用计数。 make_shared 应该分配一块内存，并在其中创建两个。

The main difference is that the first requires two memory allocations: one for the managed object (new int), and one for the reference count. make_shared should allocate a single block of memory, and create both in that.

Which one should I prefer and why?

您通常应使用 make_shared 它更高效。正如在另一个答案中所指出的，它还避免了任何内存泄漏的可能性，因为你从来没有一个指向被管对象的原始指针。

You should usually use make_shared as it's more efficient. As noted in another answer, it also avoids any possibility of a memory leak, since you never have a raw pointer to the managed object.

但是， ，它有一个潜在的缺点，即当对象被破坏时，如果仍然有弱指针阻止共享计数被删除，则不会释放内存。

However, as noted in the comments, it has a potential disadvantage that the memory won't be released when the object is destroyed, if there are still weak pointers preventing the shared count from being deleted.

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