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This noncompliant code example aligns ptr to a 4096-byte boundary, whereas the realloc() function aligns the memory to a different alignmentsuitable, but likely different, alignment:
| Code Block | ||||
|---|---|---|---|---|
| ||||
#include <stdlib.h>
void func(void) {
size_t resize = 1024;
size_t alignment = 1 << 12;
int *ptr;
int *ptr1;
if ((ptr = aligned_alloc(alignment , sizeof(int))) == NULL) {
/* Handle error */
}
if ((ptr1 = realloc(ptr, resize)) == NULL) {
/* Handle error */
}
} |
The resulting program has undefined behavior because when the alignment that realloc() enforces is different from that of aligned_alloc().
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| Code Block |
|---|
memory aligned to 4096 bytes ptr = 0x1621b000 After realloc(): ptr1 = 0x1621a010 |
Unfortunately, ptr1 is no longer aligned to 4096 bytes.
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This compliant solution implements an aligned realloc() function. It allocates resize bytes of new memory with the same alignment as the old memory and then moves the old memory there, consequently freeing up then frees the old memory:
| Code Block | ||||
|---|---|---|---|---|
| ||||
#include <stdlib.h>
void func(void) {
size_t resize = 1024;
size_t alignment = 1 << 12;
int *ptr;
int *ptr1;
if ((ptr = aligned_alloc(alignment, sizeof(int))) == NULL) {
/* Handle error */
}
if ((ptr1 = aligned_alloc(alignment, resize)) == NULL) {
/* Handle error */
}
if ((memcpy(ptr1, ptr, sizeof(int)) == NULL) {
/* Handle error */
}
free(ptr);
} |
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| Code Block | ||||
|---|---|---|---|---|
| ||||
#include <malloc.h>
void func(void) {
size_t alignment = 1 << 12;
int *ptr;
int *ptr1;
//* Original allocation */
if ((ptr = _aligned_malloc(sizeof(int), alignment))
== NULL) {
/* Handle error */
}
//* Reallocation */
if ((ptr1 = _aligned_realloc(ptr, 1024, alignment))
== NULL) {
_aligned_free(ptr);
/* Handle error */
}
_aligned_free(ptr1);
} |
Note that on Windows, _aligned_malloc() takes the size and alignment arguments in reverse order from C's _aligned_alloc().
Risk Assessment
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