mi_malloc_aligned does not return aligned pointers for small-sized allocations #206
Description
Hi,
After the upgrade to version 1.6.0, mi_malloc_aligned does not always return aligned pointers for small-sized allocations.
Here is a program for test:
#include <stdio.h>
#include "mimalloc.h"
int main() {
int i;
for (i = 0; i < 4; i++) {
void* ptr = mi_malloc_aligned(8, 16);
printf("%p %d\n", ptr, (size_t)ptr % 16);
}
return 0;
}Output:
0x53bda001100 0
0x53bda001108 8
0x53bda001110 0
0x53bda001118 8
I expect all pointers are aligned by 16-bytes, but some of them are 8-bytes aligned.
After some debugging, I found the problem may be caused by commit 4531367 :
static void* mi_heap_malloc_zero_aligned_at(mi_heap_t* const heap, const size_t size, const size_t alignment, const size_t offset, const bool zero) mi_attr_noexcept {
// note: we don't require `size > offset`, we just guarantee that
// the address at offset is aligned regardless of the allocated size.
mi_assert(alignment > 0 && alignment % sizeof(void*) == 0);
if (alignment <= MI_MAX_ALIGN_SIZE && offset==0) return _mi_heap_malloc_zero(heap, size, zero);If alignment <= 16 bytes, the function _mi_heap_malloc_zero is called without alignment requirement. But It seems _mi_heap_malloc_zero does not always return a pointer aligned by MI_MAX_ALIGN_SIZE.
I'm not sure if I understand correctly. Is this a bug?
Thanks.