C / C++ portable aligned memory allocation

align.c

/* 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);
}

I think in that case, we need to add a check that sets offset to length of alignment and that should fix it.

In the case that remainder == 0, we get offset = alignment, not offset = 0. So it doesn't end up out of bounds.

Thank you for this implementation.
But in C++ I stored offset at the end of the array.
I modified the code like char* buf = new char[alignment + type_size * size + sizeof(size_t)]; and the offset stored by *((size_t*)((char*)buf+ type_size * size)) = (size_t)alignment - remainder;.
And to free the pointer:

size_t size = mSize;
size_t type_size = sizeof(double);
size_t offset = *((size_t*)((char*)data+ type_size * size));
delete[]((char*)mImagData - offset);