指向矢量元素的指针崩溃

问题描述

    vector<int> v;
    v.push_back(1);
    int * p = &v[0];
    for (int i = 2; i <= 100; ++i)
    {
        v.push_back(i);
    }
    *p = 5;

我知道vector重新分配了一块新的内存以增加容量，但是p只是指向某个内存地址的指针，而p本身并没有改变.即使向量重新分配后，p所指向的内存也位于同一进程的地址空间中.为什么会崩溃?

I know vector reallocated new piece of memory to increase capacity, but p is just a pointer to some memory address and p itself didn't change. Also memory pointed to by p is in the address space of the same process even after the vector reallocates. Why would it crash?

推荐答案

如果将代码更改为以下内容:

If you change your code to the following:

#include <stdio.h>
#include <vector>

int
main(int argc, char* argv[])
{
    std::vector<int> v;
    v.push_back(1);
    int * p = &v[0];

    printf( "Old: %08X\n", &v[0] );
    for (int i = 2; i <= 100; ++i)
    {
        v.push_back(i);
    }
    printf( "New: %08X\n", &v[0] );

    getchar();

    return 0;
}

您将看到&v[0]的内存地址几乎总是与重新分配之前的地址不同.这意味着您创建的指针现在指向(潜在地)无效的内存.
现在，您只有一个指针p指向某个内存块.无法保证该内存块中的内容，即使它是有效的.

You will see that the memory address of &v[0] is almost always different to what it was before the reallocation. That means the pointer you created is now pointing to (potentially) invalid memory.
You now just have a pointer p pointing to some block of memory. There are no guarantees as to what is in that block of memory, or even if it is valid.

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