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);
}

Hi,

In the case where remainder works out to be 0 and since the allocation somehow came back aligned, wouldn't the writing of the offset be out of bounds?

Thanks

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);

Not really sure who still uses non-power of two aligned memory and why alignment is part of the memory (counter to what alignment typically is in C and C++).

With for example 16-byte alignment this simplifies to void * ptr = (void *)(((uintptr_t)mem+15) & ~ (uintptr_t)0x0F); and generally one can do bit shifts instead of more expensive modulo.

However, this one is much simpler. Feel free to reuse/adjust for storing the alignment:

#include <assert.h>
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static void memset_16aligned(void * ptr, char byte, size_t size_bytes, uint16_t alignment) {
    assert((size_bytes & (alignment-1)) == 0); // Size aligned
    assert(((uintptr_t)ptr & (alignment-1)) == 0); // Pointer aligned
    memset(ptr, byte, size_bytes);
}

// 1. Careful with segmented address spaces: lookup uintptr_t semantics
// 2. Careful with long standing existing optimization compiler bugs pointer to
// integer and back optimizations in for example clang and gcc
// 3. Careful with LTO potentially creating problem 2.
// 4. Consider C11 aligned_alloc or posix_memalign
void ptrtointtoptr() {
  const uint16_t alignment = 16;
  const uint16_t align_min_1 = alignment - 1;
  void * mem = malloc(1024+align_min_1);
  // C89: void *ptr = (void *)(((INT_WITH_PTR_SIZE)mem+align_min_1) & ~(INT_WITH_PTR_SIZE)align_min_1);
  // ie void *ptr = (void *)(((uint64_t)mem+align_min_1) & ~(uint64_t)align_min_1);
  // offset ptr to next alignment byte boundary
  void * ptr = (void *)(((uintptr_t)mem+align_min_1) & ~(uintptr_t)align_min_1);
  printf("0x%08" PRIXPTR ", 0x%08" PRIXPTR "\n", (uintptr_t)mem, (uintptr_t)ptr);
  memset_16aligned(ptr, 0, 1024, alignment);
  free(mem);
}

other nit: size_t requires C99.