如何释放C中的2D malloc数组?

I'm creating a 2D Array in C; am I freeing it correctly?

我用C创建一个二维数组;我是否正确地释放了它?

// create
int n = 3;
int (*X)[n] = malloc(sizeof(int[n][n]));

// set to 0
for(int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
        X[i][j] = 0;
    }
}   

// is this all I need?
free(X);

2 个解决方案

#1

You must call free once for each malloc. Your code has one malloc and one free (with the same address) so it is correct.

你必须为每一个malloc打一次免费电话。您的代码有一个malloc和一个free(具有相同的地址)，所以它是正确的。

#2

For every malloc(), there must be a matching free(). So if you call malloc() in a loop, there must be a similar loop later on that calls free() just as many times. Your code has one malloc() and one corresponding free(), so yes, you have freed everything you need to.

对于每个malloc()，必须有一个匹配的free()。因此，如果您在循环中调用malloc()，那么在以后调用free()的时候一定会有一个类似的循环。您的代码有一个malloc()和一个相应的free()，因此，是的，您已经释放了需要的所有内容。

The most common way I've seen to allocate a two-dimensional array is to do something like:

我所见过的分配二维数组的最常见方法是:

int **arr = malloc(sizeof (int *) * num_rows);

for (i = 0; i < num_rows; ++i)
        arr[i] = malloc(sizeof (int) * num_cols);
/*
  * Using `sizeof *arr` and `sizeof **arr`
  * respectively would have been more safe,
  * but for the purpose of this example, I think
  * using `int *` and `int` is simpler to
  * understand.
  */

then later to free:

然后*:

for (i = 0; i < num_rows; ++i)
        free(arr[i]);

free(arr);

This makes the outer dimension hold num_row pointers to pointers to int (each of the num_row pointers points to the starting address of the 'column'). So each element in the outer dimension points to a row, and in each row there are num_cols elements in the inner dimension (the 'columns'), which is just a group of num_cols integers. Does this make sense to you? So you have to allocate num_rows integer pointers, and each one of these points to the first 'column' of that row--for each row you have to allocate space for num_cols integers, so you can see the loop will make a total of num_rows * num cols integers, and they way they are allocated allows you to use array indexing notation (two-dimensional in this case--each element in the outer dimension points to the start of a 'row', which contains a pointer to the start of a 'column', hence the double pointer) to access the elements. I know this is probably confusing, that is why I tried to describe it so many different times/ways, but please just ask any questions, especially about what you don't understand in particular, and I will be more than happy to work with you in helping understand it.

这使得外部维度保存指向int的num_row指针(每个num_row指针指向'column'的起始地址)。因此，外部维度中的每个元素指向一行，而在每一行中，内部维度(“列”)中都有num_cols元素，这是一组num_cols整数。这对你有意义吗?所以你必须分配num_rows整数指针,和每一个点到第一列的这一行,每一行需要分配空间num_cols整数,所以你可以看到循环将总共num_rows * num整数关口,他们的方式分配允许您使用数组索引符号(二维在这种情况下,每个元素在外层维度指向“行”的开始,它包含一个指针指向“列”的开始,因此有双指针)来访问元素。我知道这可能会让人困惑，这就是为什么我试着用不同的方式描述它，但是请提出任何问题，特别是关于你不理解的东西，我将非常高兴与你一起帮助你理解它。

I'm not saying you have to create your 2-D array this way; your way works fine too, I just wanted to show you this way since you are likely to run into it since it's very common.

我不是说你必须用这种方法创建二维数组;你的方法也很有效，我只是想给你演示一下，因为你很可能会遇到它，因为它很常见。

Also, look into Valgrind. You can use that tool to see if you have forgotten any free()s / have an unmatched malloc(), and lets you know if there is any allocated memory unreachable/leaked (maybe the pointer was changed before the call to free() so the call doesn't properly free the memory and you get a leak).

另外,看着Valgrind。您可以使用这个工具来看看你是否忘记了任何*()s /有一个无与伦比的malloc(),并让你知道如果有任何分配的内存访问/泄露(可能调用前的指针改变了*()的调用并不妥善释放内存,你得到一个泄漏)。

#1