标准是否保证以下代码可以工作(假设 st 不为空)?

Is the following code guaranteed by the standard to work(assuming st is not empty)?

#include <vector>
#include <stack>
int main()
{
   extern std::stack<int, std::vector<int> > st;
   int* end   = &st.top() + 1;
   int* begin = end - st.size();
   std::vector<int> stack_contents(begin, end);
}

推荐答案

是的.

std::stack 只是一个容器适配器.

std::stack is just a container adapter.

可以看到.top()其实是(§23.3.5.3.1)

You can see that .top() is actually (§23.3.5.3.1)

reference top() { return c.back(); }

其中 c 是容器，在本例中是 std::vector

Where c is the container, which in this case is a std::vector

也就是说你的代码基本翻译成:

Which means that your code is basically translated into:

   extern std::vector<int> st;
   int* end   = &st.back() + 1;
   int* begin = end - st.size();
   std::vector<int> stack_contents(begin, end);

并且由于 std::vector 保证是连续的，所以应该没有问题.

And as std::vector is guaranteed to be continuous there should be no problem.

但是，这并不意味着这是一个好主意.如果您需要像这样使用hacks"，这通常表明设计不佳.您可能想从一开始就使用 std::vector.

However, that does not mean that this is a good idea. If you need to use "hacks" like this it is generally an indicator of bad design. You probably want to use std::vector from the beginning.