我有一个指向结构的指针,结构中的一个对象是一个int ** . 双指针用于为2d数组动态分配内存 . 我无法弄清楚如何释放这个数组的内存 . 有任何想法吗?
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(如果其他人有类似的问题)
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;
}
1 回答
你的解放是好的,但你有一个严重的错误
您正在返回局部变量的地址 .
局部变量在函数的堆栈帧中分配,一旦函数返回,数据将不再存在 .
您也应该使用
malloc
而且你必须从
main()
返回int