如果malloc()在同一位置分配内存，则访问释放的指针可能会导致数据损坏，除非将释放的指针设置为NULL.

问题描述

由于用户(GeorgSchölly116K声望)在其问题中的陈述，出现了此Question陈述.应该真正设置指针吗?释放它们后将其设置为"NULL"?

This Question statement is came in picture due to statement made by user (Georg Schölly 116K Reputation) in his Question Should one really set pointers to `NULL` after freeing them?

那我怎么弄不到数据将如何损坏?

代码

#include<iostream>
int main()
{
    int count_1=1, count_2=11, i;
    
    
    int *p=(int*)malloc(4*sizeof(int));
    
    std::cout<<p<<"\n";
    
    for(i=0;i<=3;i++)
    {
        *(p+i)=count_1++;
    }
    
    for(i=0;i<=3;i++)
    {
        std::cout<<*(p+i)<<" ";
    }
    
    std::cout<<"\n";
    free(p);
    
    
    
    p=(int*)malloc(6*sizeof(int));
    
    std::cout<<p<<"\n";
    
    for(i=0;i<=5;i++)
    {
        *(p+i)=count_2++;
    }
    
    for(i=0;i<=3;i++)
    {
        std::cout<<*(p+i)<<" ";
    }
    
}

输出

0xb91a50
1 2 3 4
0xb91a50
11 12 13 14

再次在释放(0xb91a50)后分配相同的内存位置，但是它工作正常，不是吗?

推荐答案

您不要在代码中重用旧指针.在 p =(int *)malloc(6 * sizeof(int)); 之后， p 指向一个不错的新分配数组，您可以毫无问题地使用它.Georg引用的数据损坏问题将在类似于以下内容的代码中发生:

You do not reuse the old pointer in your code. After p=(int*)malloc(6*sizeof(int));, p point to a nice new allocated array and you can use it without any problem. The data corruption problem quoted by Georg would occur in code similar to that:

int *p=(int*)malloc(4*sizeof(int));
...
free(p);

// use a different pointer but will get same address because of previous free
int *pp=(int*)malloc(6*sizeof(int));

std::cout<<p<<"\n";

for(i=0;i<=5;i++)
{
    *(pp+i)=count_2++;
}

p[2] = 23;             //erroneouly using the old pointer will corrupt the new array
for(i=0;i<=3;i++)
{
    std::cout<<*(pp+i)<<" ";
}

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