被释放的malloc错误指针没有被分配

Question

我试图从函数返回字符串数组，然后释放它使用的内存。代码如下:

int main(int argc, const char * argv[])
{
    for (int m = 0; m < 10000; m++) {
        char **data = dataTest();
        int i = 0;
        while(data[i]) {
            printf("%p ",data[i]);
            free(data[i]);
            i++;
        }
        printf(" address= %p.n",data);
        free(data);
    }
    return 0;
}

函数如下:

char **dataTest()
{
    char *row[] = {"this", "is", "a", "data", "string", NULL};
    char **str = row;
    char **dataReturn = (char **) malloc(sizeof(char *) * 6);
    int i = 0;
    while (*str) {
        dataReturn[i] = malloc(sizeof(char) * strlen(*str));
        strcpy(dataReturn[i++], *str);
        str++;
    }
    return dataReturn;
}

一开始运行得很好，但很快就出现了错误。结果如下。地址出了问题，malloc错误发生了。有人遇到过同样的问题吗?

<>之前0x100300030 0x100300040 0x100300050 0x100300060 0x100300070地址= 0x1003000000x100300030 0x100300040 0x100300050 0x100300060 0x100300070地址= 0x1003000000x100400030 0x100300030 0x100300040 0x100300050 0x100300060地址= 0x100400000testC(562,0x7fff73e71310) malloc: ***错误的对象0x3000000000000;未分配被释放的指针***在malloc_error_break中设置一个断点进行调试0x100300060 0x100300070 0x100300030 0x100300040 0x100300050 0x3000000000000程序以退出码:9结束

Answer 1

您需要将此添加到dataTest函数的return dataReturn;之前:

dataReturn[i] = NULL ;

否则您的while (data[i]) {}将继续超出所需的范围。

而不是:

dataReturn[i] = malloc( sizeof(char) * (strlen(*str)) );

写:

dataReturn[i] = malloc(strlen(*str) + 1);

，以便为结束零分配空间。

BTW sizeof (char)总是1.