如何分配或数组免费仅部分？ [英] How allocate or free only parts of an array?

问题描述

看到这样的例子：

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

然后免费仅第3块？

Then free only the first 3 blocks?

或者等于做java的，有负面的指标，或者说不是始于索引0的数组。

Or make equal java, have an array with negative index, or an index that not began with 0.

非常感谢。

推荐答案

您不能直接释放第3个街区。您可以通过重新分配做类似的事情数组较小：

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;

至于你问题的第二部分，你可以增加阵列使其指向到你的内存块的中间。然后，您可以使用负的索引：

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);

同样的想法在相反方向工作的为好。您可以从阵列减去使起始指数大于0但要注意：作为@大卫索恩利指出，这在技术上是无效的根据ISO C标准和可能在所有平台上无法工作。

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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