我以为是内存入侵问题,对吗?

问题描述 投票:0回答:1

如下,我的代码是:

int     *text(char *str)
{
    int     *cond;
    int     *temp;
    int     cond_size;
    int     num;
    int     i;

    cond_size = -1;
    cond = malloc(sizeof(int) * 1);
    *cond = 0;
    while (*str != '\0')
    {
        if (*str == ' ')
            str++;
        num = 0;
        while (*str >= '0' && *str <= '9')
            num = num * 10 + *(str++) - '0';
        temp = cond;
        cond = malloc(sizeof(int) * (++cond_size));
        i = -1;
        while (++i < cond_size)
            cond[i] = temp[i];
        cond[i] = num;
        free(temp);
    }
    g_size = (i + 1) / 4;
    return (cond);
}

我的主要功能是:

int     *text(char *str);

#include <stdio.h>
#include <stdlib.h>

int     g_size = 0;

int     main(void)
{
    int     *test;
    int     i;

    i = 0;
    test = text(" 4 3 2 1 1 2 2 2 4 3 2 1 1 2 2 2");
    while(i < g_size)
    {
        printf("\n%d\n", test[i]);
        i++;
    }
}

使用输入字符串4 3 2 1 1 2 2 2 4 3 2 1 ...,将输出以下输出:

==============
|  4 3 2 1   |
|4        '1'| <==
|3         2 |  
|2         2 |
|1         2 |
|  1 2 2 2   |
==============

但是,当我检查(<==),

'1'被打印为'4'

而且这不符合我的预期。

使用malloc或其他错误时代码中是否存在任何内存入侵?

c malloc heap-memory
1个回答
0
投票

我可以看到一些问题:

  • cond_size为-1:

    cond_size = -1; // Here is problem
    cond = malloc(sizeof(int) * 1);
    *cond = 0;
    while (*str != '\0')
    {
        if (*str == ' ')
            str++;
        num = 0;
        while (*str >= '0' && *str <= '9')
           num = num * 10 + *(str++) - '0';
        temp = cond;
        cond = malloc(sizeof(int) * (++cond_size)); // you are trying allocate memory with size 0
        i = -1;
        while (++i < cond_size)
            cond[i] = temp[i];
        cond[i] = num; // you are writing to not allocated memory
    

    您可以找到有关malloc的更多信息。

  • 温度超出范围:

    temp = cond;
    cond = malloc(sizeof(int) * (++cond_size));
    i = -1;
    while (++i < cond_size)
        cond[i] = temp[i]; // temp size is cond_size-1
    

    所以您应该通过cond_size-1限制循环。

请看fixed version.

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