内存错误重挫 [英] Memory Clobbering Error

问题描述

我有一小块code的。我与 -lmcheck 编译它，因为我尝试调试code，其中我也有同样类似的错误。

当我运行这个code我得到这个错误：

 内存分配惨败前挡
 

有人能解释为什么免费（PTR）将抛出我这个错误？

怎么我还能释放指针？

感谢。

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆string.h中GT;
＃包括LT＆;＆stdlib.h中GT;
＃包括LT＆;＆unistd.h中GT;
的#define LEN 5
INT主（INT ARGC，CHAR *的argv []）{    字符* PTR = NULL;    PTR =（字符*）malloc的（LEN + 1）; // +1字符串
    的strcpy（PTR，你好）;    INT I = 0;
    对于（i = 0; I＆LT; LEN，我++）
    {
        的printf（PTR [％d个=％C \\ N，我，PTR [I]）;
        PTR ++;
    }
    免费（PTR）;
    返回0;
}
 

解决方案

正在递增 PTR ，因此改变它指向的地址。你不能做到这一点。

在你的情况，有一个单独的指针，假设的char * p = PTR 和 P 离开 PTR 不变，以便您可以免费（PTR）之后。

修改以你的code第二次看，我发现你正在做 PTR ++ 时，你不应该。您正在访问像 PTR数组中的字符[I] ，如果乱用 PTR 的指针，你是改变基址和 PTR访问字符[I] 可能会导致（并会导致）意外的结果。

如果你只是删除了这一行（ PTR ++ ）的code会奇迹般地工作。
如果你想探索指针的概念，尝试另一种解决方案，您的code可能是这个样子：

  INT主（INT ARGC，CHAR *的argv []）{    字符* PTR = NULL;
    字符* P;    PTR =（字符*）malloc的（LEN + 1）; // +1字符串（请检查NULL）
    P = PTR;    的strcpy（PTR，你好）;    INT I = 0;
    而（* P）//注意我是如何变成一个while循环，C字符串是NULL终止，所以这将打破，一旦我们得到的字符串的结尾。我们得到的是，这适用于任何字符串大小工作。
    {
        的printf（PTR [％d个=％C \\ N，我++，* P）; //这里我解引用指针，访问其个人CHAR
        p ++;
    }
    免费（PTR）;
    返回0;
}
 

I have a small piece of code. I compiled it with -lmcheck as I am trying to debug a code where I have the same similar error.

I get this error when I run this code:

memory clobbered before allocated block

Can someone explain the reason why free(ptr) will throw me this error?

How else can I free the pointer?

Thanks.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>
#define LEN 5


int main(int argc, char *argv[]){

    char *ptr = NULL;

    ptr = (char *) malloc(LEN+1);// +1 for string
    strcpy(ptr, "hello");

    int i = 0;
    for(i = 0; i<LEN; i++)
    {
        printf("ptr[%d] = %c\n", i, ptr[i]);
        ptr++;
    }
    free(ptr);


    return 0;
}

解决方案

You are incrementing ptr, therefore changing the address that it points to. You can't do that.

In your case, have a separate pointer, let's say char * p = ptr and do your operations with p leaving ptr intact so you can free(ptr) later.

EDIT Taking a second look at your code, I found that you are doing ptr++ when you shouldn't. You are accessing the characters in the array like ptr[i], if you mess with the ptr pointer, you are changing the base address and accessing the characters with ptr[i] can lead (and will lead) to unexpected results.

If you simply remove that line (ptr++) your code will magically work. If you want to explore the pointer concept and try another solution, your code could look something like this:

int main(int argc, char *argv[]){

    char *ptr = NULL;
    char * p; 

    ptr = (char *) malloc(LEN+1);// +1 for string (please check for NULL)
    p = ptr;

    strcpy(ptr, "hello");

    int i = 0;
    while (*p) // note how I changed it to a while loop, C strings are NULL terminated, so this will break once we get to the end of the string. What we gain is that this will work for ANY string size.
    {
        printf("ptr[%d] = %c\n", i++, *p); // here i dereference the pointer, accessing its individual char
        p++;
    }
    free(ptr);


    return 0;
}

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