c++ - 了解 use_count 与 shared_ptr

Question

我想出了下面的例子

std::shared_ptr<foo> a(new foo());
{
    std::shared_ptr<foo> b = a;
    std::cout << "before" << b.use_count() << "\n"; //returns 2
    b.reset();
    std::cout << "after" << b.use_count() << "\n";  //returns 0
} 
std::cout << "Finished\n";

现在在上面的代码中，第二个 use_count 语句返回零。在那种情况下，析构函数不应该在打印出 "Finished" 之前被调用。为什么第二条语句中的 use_count 打印为 0 ？我读到 use_count 的定义是:

Returns the number of shared_ptr objects that share ownership over the same pointer as this object (including it).

如果我在使用计数之前执行了一个reset()，这仅仅意味着它的引用计数减少了 1。如果我错了请纠正我。

以上是我的理解，如有错误请指正

std::shared_ptr<foo> a(new foo());   //reference count is 1
{
    std::shared_ptr<foo> b = a;      //reference count becomes 2
    std::cout << "before" << b.use_count() << "\n"; //returns 2 //OK this I understand
    b.reset(); //b smart pointer gives up its reference count so now it should be 1.
    std::cout << "after" << b.use_count() << "\n";  //This should be 1 why is it 0 ?
} 
std::cout << "Finished\n";

所以我的问题是为什么 b.use_count() 返回 0 ？

Answer 1

b.reset(); 之后，b 为空(即不指向任何对象)。

根据标准(引用自N4527 §20.8.2.2.5[util.smartptr.shared.obs])

long use_count() const noexcept;

7 Returns: the number of shared_ptr objects, *this included, that share ownership with *this, or 0 when *this is empty.