如何仅分配或释放数组的一部分? [英] How allocate or free only parts of an array?

问题描述

看这个例子:

int *array = malloc (10 * sizeof(int))

有没有办法只释放前 3 个块?

Is there a way to free only the first 3 blocks?

或者有一个带有负索引的数组，或不以 0 开头的索引?

Or to have an array with negative indexes, or indexes that don't begin with 0?

推荐答案

您不能直接释放前 3 个块.您可以通过将数组重新分配更小来做类似的事情:

You can't directly free the first 3 blocks. You can do something similar by reallocating the array smaller:

/* Shift array entries to the left 3 spaces. Note the use of memmove
 * and not memcpy since the areas overlap.
 */
memmove(array, array + 3, 7);

/* Reallocate memory. realloc will "probably" just shrink the previously
 * allocated memory block, but it's allowed to allocate a new block of
 * memory and free the old one if it so desires.
 */
int *new_array = realloc(array, 7 * sizeof(int));

if (new_array == NULL) {
    perror("realloc");
    exit(1);
}

/* Now array has only 7 items. */
array = new_array;

至于问题的第二部分，您可以增加 array 使其指向内存块的中间.然后您可以使用负索引:

As to the second part of your question, you can increment array so it points into the middle of your memory block. You could then use negative indices:

array += 3;
int first_int = array[-3];

/* When finished remember to decrement and free. */
free(array - 3);

同样的想法也适用于相反的方向.您可以从 array 中减去以使起始索引大于 0.但要小心:正如@David Thornley 指出的那样，根据 ISO C 标准，这在技术上是无效的，并且可能不适用于所有平台.

The same idea works in the opposite direction as well. You can subtract from array to make the starting index greater than 0. But be careful: as @David Thornley points out, this is technically invalid according to the ISO C standard and may not work on all platforms.

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