C ++ STL堆栈：什么时候可以安全地使用pop（）(C++ STL stack: When is it safe to pop())

C ++ STL堆栈：什么时候可以安全地使用pop（）(C++ STL stack: When is it safe to pop())

考虑：

#include <iostream> #include <stack> class Abc { int x = 5; public: void display() { std::cout << x << std::endl; } }; int main() { std::stack<Abc> S; S.emplace(); auto obj = S.top(); S.pop(); obj.display(); return 0; }

来自： http ： //www.cplusplus.com/reference/stack/stack/pop/ ，“这将调用已移除元素的析构函数”。 另外，从http://www.cplusplus.com/reference/stack/stack/top/，stack.top （）通过引用返回。

如果S.top()通过引用返回，并且如果S.pop()销毁该对象，为什么不obj.display()失败？

我知道栈调用底层容器的back()和pop_back()方法。 通过扩展，为什么没有失败？

Consider:

#include <iostream> #include <stack> class Abc { int x = 5; public: void display() { std::cout << x << std::endl; } }; int main() { std::stack<Abc> S; S.emplace(); auto obj = S.top(); S.pop(); obj.display(); return 0; }

From: http://www.cplusplus.com/reference/stack/stack/pop/, "This calls the removed element's destructor". Also, from http://www.cplusplus.com/reference/stack/stack/top/, stack.top() returns by reference.

If S.top() returned by reference and if S.pop() destructed the object, why doesn't obj.display() fail?

I know that stack calls the back() and pop_back() methods of the underlying container. By extension, why doesn't that fail?

最满意答案

auto obj = S.top(); 复制从S.top() 初始化 obj 。 它是随后移除的元素的副本。

你的代码片段具有明确定义的行为。

如果你需要一个引用，如auto& obj = S.top();

auto obj = S.top(); copy initializes obj from S.top(). It's a copy of the then removed element.

You code snippet has well defined behavior.

It will be undefined if you were to take a reference, like auto& obj = S.top();