如何分配与页面大小对齐的内存?

问题描述

我需要分配应该与页面大小对齐的内存.我需要将此内存传递给ASM代码，该代码计算所有数据块的异或.我需要使用malloc()来完成此操作.

I need to allocate memory which should be page size aligned. I need to pass this memory to an ASM code which calculates xor of all data blocks. I need to do this with malloc().

推荐答案

您应该使用此功能.

如果由于某种原因不能这样做，那么通常的方法是将块大小添加到分配大小中，然后使用整数算术欺骗指针.

If you can't, for whatever reason, then the way this is generally done is by adding the block size to the allocation size, then using integer-math trickery to round the pointer.

类似这样的东西:

/* Note that alignment must be a power of two. */
void * allocate_aligned(size_t size, size_t alignment)
{
  const size_t mask = alignment - 1;
  const uintptr_t mem = (uintptr_t) malloc(size + alignment);
  return (void *) ((mem + mask) & ~mask);
}

这还没有经过非常深入的测试，但是您知道了.

This has not been very deeply tested but you get the idea.

请注意，以后不可能找到指向free()内存的正确指针.要解决此问题，我们必须添加一些其他机制:

Note that it becomes impossible to figure out the proper pointer to free() the memory later. To fix that, we would have to add some additional machinery:

typedef struct {
  void *aligned;
} AlignedMemory;

AlignedMemory * allocate_aligned2(size_t size, size_t alignment)
{
  const size_t mask = alignment - 1;
  AlignedMemory *am = malloc(sizeof *am + size + alignment);
  am->aligned = (void *) ((((uintptr_t) (am + 1)) + mask) & ~mask);
  return am;
}

这可以稍微复杂点指针，并为您提供一个可以使用free()的指针，但是您需要取消引用aligned指针才能获得正确对齐的指针.

This wraps the pointer trickery a bit, and gives you a pointer you can free(), but you need to dereference into the aligned pointer to get the properly aligned pointer.

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