c++ - 制作 shared_ptr 的拷贝时会发生什么？

Question

我想了解当将 shared_ptr 分配给另一个时，shared_ptr 中托管对象的引用计数会受到怎样的影响。

我在 C++ primer, 5th edition 中看到以下声明:

For example, the counter associated with a shared_ptr is incremented when ... we use it as the right-hand operand of an assignment... The counter is decremented when we assign a new value to the shared_ptr...

举个例子:

auto p = make_shared<int>(42); // object to which p points has one user

auto q(p); // p and q point to the same object
           // object to which p and q point has two users

auto r = make_shared<int>(42); // int to which r points has one user
r = q; // assign to r, making it point to a different address
       // increase the use count for the object to which q points
       // reduce the use count of the object to which r had pointed
       // the object r had pointed to has no users; that object is automatically freed

当我运行类似的代码时，以上不是我的观察:

代码:

#include<iostream>
#include<memory>

int main()
{
  std::shared_ptr<int> sh1 = std::make_shared<int>(1);
  std::shared_ptr<int> sh2 = std::make_shared<int>(2);

  sh2 = sh1;

  std::cout << "sh1 use count: " << sh1.use_count() << std::endl;
  std::cout << "sh2 use count: " << sh2.use_count() << std::endl;

  return 0;
}

Output:

sh1 use count: 2

sh2 use count: 2

sh2的use_count怎么也是2？根据上面提到的文字，它不应该是 0 吗？我在这里遗漏了什么吗？

Answer 1

起初你有 sh1.use_count=1 和 sh2.use_count=1。现在，当您使用 sh2=sh1 进行分配时，会发生以下情况:

1. sh2 计数器减一，因为 sh2(shared_ptr)将采用另一个指针
2. 从 sh2.use_count=0 开始，其指针下的对象 int(2) 被销毁。
3. 现在您将 sh2 分配给了一个属于 sh1 的新对象，因此它的计数器增加了 1，所以:sh2.use_count=2 ，当然还有 sh1.use_count=2，因为两个 shared_ptr 对象都指向同一个对象，即 int(1)。