openssl/crypto/aligned_alloc.c

69 lines
2.2 KiB
C

/*
* Copyright 2024-2025 The OpenSSL Project Authors. All Rights Reserved.
*
* Licensed under the Apache License 2.0 (the "License"). You may not use
* this file except in compliance with the License. You can obtain a copy
* in the file LICENSE in the source distribution or at
* https://www.openssl.org/source/license.html
*/
#include "internal/e_os.h"
#include "internal/cryptlib.h"
#include "internal/mem_alloc_utils.h"
#include "crypto/cryptlib.h"
#include <stdlib.h>
void *ossl_malloc_align(size_t num, size_t alignment, void **freeptr,
const char *file, int line)
{
size_t alloc_bytes;
void *ret;
*freeptr = NULL;
/* Ensure that alignment is a power of two no larger than 65536 */
if (alignment == 0 || (alignment & (alignment - 1)) != 0
|| alignment > 65536) {
ossl_report_alloc_err_inv(file, line);
return NULL;
}
/*
* Note: Windows supports an _aligned_malloc call, but we choose
* not to use it here, because allocations from that function
* require that they be freed via _aligned_free. Given that
* we can't differentiate plain malloc blocks from blocks obtained
* via _aligned_malloc, just avoid its use entirely
*/
if (ossl_unlikely(!ossl_size_add(num, alignment, &alloc_bytes, file, line)))
return NULL;
/*
* Step 1: Allocate an amount of memory that is <alignment>
* bytes bigger than requested
*/
*freeptr = CRYPTO_malloc(alloc_bytes, file, line);
if (*freeptr == NULL)
return NULL;
/*
* Step 2: Add <alignment - 1> bytes to the pointer
* This will cross the alignment boundary that is
* requested
*/
ret = (void *)((char *)*freeptr + (alignment - 1));
/*
* Step 3: Use the alignment as a mask to translate the
* least significant bits of the allocation at the alignment
* boundary to 0. ret now holds a pointer to the memory
* buffer at the requested alignment
* NOTE: It is a documented requirement that alignment be a
* power of 2, which is what allows this to work
*/
ret = (void *)((uintptr_t)ret & (uintptr_t)(~(alignment - 1)));
return ret;
}