使用 realloc 缩小分配的内存 [英] Using realloc to shrink the allocated memory

问题描述

关于C中realloc函数的简单问题:如果我使用 realloc 来缩小指针指向的内存块，是否会释放额外"内存?还是需要手动释放?

Simple question about the realloc function in C: If I use realloc to shrink the memory block that a pointer is pointing to, does the "extra" memory get freed? Or does it need to be freed manually somehow?

例如，如果我这样做

int *myPointer = malloc(100*sizeof(int));
myPointer = realloc(myPointer,50*sizeof(int));
free(myPointer);

我会出现内存泄漏吗?

推荐答案

不，您不会有内存泄漏.realloc 将简单地将其余的标记为可用"以供将来的 malloc 操作使用.

No, you won't have a memory leak. realloc will simply mark the rest "available" for future malloc operations.

但是你以后仍然需要free myPointer.顺便说一句，如果您在 realloc 中使用 0 作为大小，它将与 free 在某些实现中具有相同的效果.正如 Steve Jessop 和 R.. 在评论中所说，你不应该依赖它.

But you still have to free myPointer later on. As an aside, if you use 0 as the size in realloc, it will have the same effect as free on some implementations. As Steve Jessop and R.. said in the comments, you shouldn't rely on it.

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