指针被删除后，指针变化时保存的地址

问题描述

限时删除！！

在下面的代码中，为什么指针 x 保存的地址在 delete 后改变？据我理解， delete 调用应该释放从堆分配的内存，但不应该更改指针地址。

In the following code, why is the address held by pointer x changing after the delete? As I understand, the deletecall should free up allocated memory from heap, but it shouldn't change the pointer address.

using namespace std;
#include <iostream>
#include <cstdlib>

int main()
{
    int* x = new int;
    *x = 2;

    cout << x << endl << *x << endl ;

    delete x;

    cout << x << endl;

    system("Pause");
    return 0;
}

OUTPUT:
01103ED8
2
00008123

观察：我使用Visual Studio 2013和Windows 8.据报告，这不工作在其他编译器相同。此外，我明白这是坏的做法，我应该重新分配指针为NULL，删除后，我只是想了解是什么驱动这种奇怪的行为。

Observations: I'm using Visual Studio 2013 and Windows 8. Reportedly this doesn't work the same in other compilers. Also, I understand this is bad practice and that I should just reassign the pointer to NULL after it's deletion, I'm simply trying to understand what is driving this weird behaviour.

推荐答案

好吧，为什么不呢？这是完全合法的输出 - 读取指针后删除它导致未定义的行为。这包括指针的值改变。 （实际上，这甚至不需要UB; delete d指针可以指向任何地方。）

Well, why not? It's perfectly legal output -- reading a pointer after having deleted it leads to undefined behavior. And that includes the pointer's value changing. (In fact, that doesn't even need UB; a deleted pointer can really point anywhere.)

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