shared_ptr 和 use_count [英] shared_ptr and use_count

问题描述

在以下代码片段中:

shared_ptrp;{p = shared_ptr(new int);cout<<p.use_count()<<endl;}cout<<p.use_count()<<endl;

输出结果为

<前>11

我不明白为什么第一个输出是 1 -- 不应该是 2 吗?

解决方案

临时对象的生命周期不够长，第一个 p.use_count() 返回 2.临时对象被销毁首先，放弃对其拥有的任何事物的所有权.

此外，由于临时值是一个右值，对 p 的赋值将导致移动赋值，这意味着无论如何使用计数永远不会是 2(假设一个质量实现).所有权只是从临时转移到 p，永远不会超过 1.

In the following code snippet:

shared_ptr<int> p;

{
    p = shared_ptr<int>(new int);
    cout<<p.use_count()<<endl;
}

cout<<p.use_count()<<endl;

The output comes out to be

1
1

I don't understand why the 1st output is 1 -- shouldn't it be 2?

解决方案

The temporary object's lifetime does not last long enough for the first p.use_count() to return 2. The temporary object is destroyed first, relinquishing its ownership on anything it owned.

Furthermore, since the temporary is an rvalue, the assignment to p will result in a move-assignment, which means the use-count will never be 2 anyway (assuming a quality implementation). Ownership is simply transferred from the temporary to p, never exceeding 1.

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