Why does top()'s return value change after calling pop()?

Question

The const reference returned by priority_queue's top() changes after calling pop() (visual studio 2015)

priority_queue<int> queue;
queue.push(1);
queue.push(2);
queue.push(3);

const int & m = queue.top();
cout << m << endl; // 3
queue.pop();
cout << m << endl; // 2
queue.pop();
cout << m << endl; // 1

If get top value by auto & m = queue.top();, then output is also 3 2 1.

While if get top value by auto m = queue.top();, then output is 3 3 3.

What's mechanism behind this ?

Answer 1

If get top value by auto & m = queue.top();, then output is also 3 2 1.

Despite it is calling undefined behavior to use m after the 1st pop() call, it is likely that the next value is moved to that dangling reference (address). That is because the default underlying type of std::priority_queue is std::vector, which guarantees a contiguous array of elements.

But as mentioned that behavior is undefined, and there are no guarantees to reproduce that result with a different compiler.

While if get top value by auto m = queue.top();, then output is 3 3 3.

The value from top is stored into m once, and never changed afterwards.