C中的内存对齐：返回地址中的偏移量是如何考虑的？

Question

我遇到了以下代码here。

/* Allocate aligned memory in a portable way.
 *
 * Memory allocated with aligned alloc *MUST* be freed using aligned_free.
 *
 * @param alignment The number of bytes to which memory must be aligned. This
 *  value *must* be <= 255.
 * @param bytes The number of bytes to allocate.
 * @param zero If true, the returned memory will be zeroed. If false, the
 *  contents of the returned memory are undefined.
 * @returns A pointer to `size` bytes of memory, aligned to an `alignment`-byte
 *  boundary.
 */
void *aligned_alloc(size_t alignment, size_t size, bool zero) {
    size_t request_size = size + alignment;
    char* buf = (char*)(zero ? calloc(1, request_size) : malloc(request_size));

    size_t remainder = ((size_t)buf) % alignment;
    size_t offset = alignment - remainder;
    char* ret = buf + (unsigned char)offset;

    // store how many extra bytes we allocated in the byte just before the
    // pointer we return
    *(unsigned char*)(ret - 1) = offset;

    return (void*)ret;
}

/* Free memory allocated with aligned_alloc */
void aligned_free(void* aligned_ptr) {
    int offset = *(((char*)aligned_ptr) - 1);
    free(((char*)aligned_ptr) - offset);
}

解释：

char *ret = buf + (unsigned char)offset; 在这里，我们设置了一个新指针，它比 buf 的基地址提前了偏移字节。

例如我们想在一个 16 位对齐的内存中分配 68 个字节，它看起来像这样：

1. requested_size = 68+16 = 84 假设 buf 的基地址是 buf = 0x112223341 然后
2. remainder = sizeof(buf)%16 = (84%16) = 4
3. offset = 16 - 4 = 12 (i.e. 0x0C)
4. ret = &buf + offset = 0x11223341+0x0C = 0x1122334D

问题：

以下行是做什么的？ 我在理解这种语法和实现结果时遇到了一些麻烦。

*(unsigned char*)(ret - 1) = offset

当我们返回 ret 时，那些被分配但绝不是 ret 基地址的一部分的额外字节会发生什么？也就是说，如果我们分配了 16 个额外的字节，但只需要 12 个字节进行对齐，那么剩下的字节会发生什么？

=======更新问题中的代码=======

感谢@ThomasMailund 和他的见解，我认为我可以安全地修改上述代码以简化一些类型转换，如下所示：

/* Allocate aligned memory in a portable way.
 *
 * Memory allocated with aligned alloc *MUST* be freed using aligned_free.
 *
 * @param alignment The number of bytes to which memory must be aligned. This
 *  value *must* be <= 255.
 * @param bytes The number of bytes to allocate.
 * @param zero If true, the returned memory will be zeroed. If false, the
 *  contents of the returned memory are undefined.
 * @returns A pointer to `size` bytes of memory, aligned to an `alignment`-byte
 *  boundary.
 */
void *aligned_alloc(size_t alignment, size_t size, bool zero) {
    size_t request_size = size + alignment;
    unsigned char *buf = zero ? calloc(1, request_size) : malloc(request_size);

    size_t remainder = ((size_t)buf) % alignment;
    size_t offset = alignment - remainder;
    unsigned char *ret = buf + (unsigned char)offset;

    // store how many extra bytes we allocated in the byte just before the
    // pointer we return
    *(ret - 1) = offset;

    return ret;
}

/* Free memory allocated with aligned_alloc */
void aligned_free(void* aligned_ptr) {
    int offset = *(((char*)aligned_ptr) - 1);
    free(((char*)aligned_ptr) - offset);
}

Answer 1

对于将来正在查看此内容的任何人：

当我查看void aligned_free 函数时，我想我现在明白了。 我们只是保存在 (ret-1) 的地址处计算的偏移字节数，然后在准备释放内存时将其读回，以释放这些额外的字节。 即从(ret-1)读取偏移量，从基地址中减去那些我们要求释放的许多#字节以获得原始buf基地址，然后释放buf，这样我们就不会遇到内存泄漏！

注意：那些额外的偏移字节，它们仍然分配在内存中，只是没有使用！