C ++指针 - 删除对数组指针的操作

While I was using delete operator on pointers, I saw something that I could not understand. I think it is best to show it on an example:

当我在指针上使用delete运算符时,我看到了一些我无法理解的东西。我认为最好在一个例子中显示它:

#include <iostream>
using namespace std;

int main() {

    int* p = new int[5]{3, 4, 5, 6, 7};

    cout << p[0] << p[1] << p[2] << p[3] << p[4] << '\n';

    delete[] p;

    cout << p[0] << p[1] << p[2] << p[3] << p[4] << '\n';

    return 0;
}

The result:

34567
00567

After the delete operation why first two elements turned to zeros? Thanks for your attention. I’m looking forward to your replies.

删除操作后为什么前两个元素变为零?感谢您的关注。我很期待你的回复。

4 个解决方案

#1

 delete[] p;

 cout << p[0] << p[1] << p[2] << p[3] << p[4] << endl;

The moment you use p after calling delete[] on it, it's undefined behavior. As the wording says the behavior is undefined so questioning why x or y happens is pointless as this is uncertain.

在调用delete []之后使用p的那一刻,它是未定义的行为。正如措辞所说,行为是不确定的,所以质疑为什么x或y发生是没有意义的,因为这是不确定的。

#2

You don't want to pay for what you don't need in C++.

你不想为C ++中不需要的东西买单。

Why would you expect released memory to be zeroed? delete just marks the memory as not used anymore, there's no need to zero it since you can't use it anyway without invoking undefined behavior.

为什么你期望释放的内存归零?删除只是将内存标记为不再使用,不需要将其归零,因为无论如何都不能使用它而不调用未定义的行为。

#3

When you 'delete' a pointer or variable, there's no reason for it to be set to '0'.
By deleting the array you simply tell the computer that it can use the given space freely. The interesting question is not why the rest of the values persist, but actually why are the 2 first ones are set to '0', and the answer could really be anything

当你'删除'指针或变量时,没有理由将它设置为'0'。通过删除数组,您只需告诉计算机它可以*使用给定的空间。有趣的问题不是为什么剩下的值仍然存在,但实际上为什么2个第一个被设置为'0',答案可能真的是什么

#4

-1

System do not have to clear this memory with zeroes.

系统不必用零清除此内存。

delete[] just means you do not not need this memory anymore and you return this memory to the system. System marked this memory as free and will overwrite it with new data sometime in future.

delete []只是意味着您不再需要此内存并将此内存返回给系统。系统将此内存标记为空闲,并在将来某个时间用新数据覆盖它。

#1

 delete[] p;

 cout << p[0] << p[1] << p[2] << p[3] << p[4] << endl;