从结构中的二维数组中释放动态分配的内存

Free dynamically allocated memory from 2D array within a struct

我有一个指向结构的指针,该结构中的对象之一是 int **。双指针用于为二维数组动态分配内存。我无法弄清楚如何释放该数组的内存。有什么想法吗?

struct time_data {
    int *week;
    int *sec;
    int **date;
};
typedef struct time_data time_data;

time_data *getTime(time_data *timeptr, int rows, int cols) {
    int i = 0;
    time_data time;

    // allocate memory for time.date field
    time.date = (int **)malloc(rows*(sizeof(int *))); // allocate rows
    if(time.date == NULL)
        printf("Out of memory\n");
    for(i=0; i<rows; i++) {
        time.date[i] = (int *)malloc(cols*sizeof(int));
        if(time.date[i] == NULL)
            printf("Out of memory\n");
    }
    timeptr = &time;
    return timeptr;
}

int main(int argc, const char * argv[]) {
    time_data *time = NULL;
    int rows = 43200, cols = 6;
    int i;
    time = getTime(time, rows, cols);

    for(i=0; i<rows; i++)
        free(time->date[i]); // problem here
    free(time->date);

}

修改版本(以防其他人有类似问题)

    struct time_data {
    int *week;
    int *sec;
    int **date;
};
typedef struct time_data time_data;

time_data *getTime(int rows, int cols) {
    int i = 0;
    time_data *time = malloc(sizeof(*time));

    // allocate memory for time.date field
    time->date = (int **)malloc(rows*(sizeof(int *))); // allocate rows
    if(time->date == NULL)
        printf("Out of memory\n");

    for(i=0; i<rows; i++) {
        time->date[i] = (int *)malloc(cols*sizeof(int));
        if(time->date[i] == NULL)
            printf("Out of memory\n");
    }
    return time;
}

int main(int argc, const char * argv[]) {
    time_data *time = NULL;
    int rows = 43200, cols = 6;
    int i;
    time = getTime(rows, cols);

    for(i=0; i<rows; i++)
        free(time->date[i]); // problem here
    free(time->date);
return 0;
}

你的释放没问题,但是你有一个严重的错误

timeptr = &time;
return timeptr;

您正在 return 获取局部变量的地址。

局部变量分配在函数的栈帧中,一旦函数returns,数据将不复存在

你也应该使用 malloc

timeptr = malloc(sizeof(*timeptr));

而且你必须 return 来自 main()

int