问题
How do you allocate memory that's aligned to a specific boundary in C (e.g., cache line boundary)? I'm looking for malloc/free like implementation that ideally would be as portable as possible --- at least between 32 and 64 bit architectures.
Edit to add: In other words, I'm looking for something that would behave like (the now obsolete?) memalign function, which can be freed using free.
回答1:
Here is a solution, which encapsulates the call to malloc, allocates a bigger buffer for alignment purpose, and stores the original allocated address just before the aligned buffer for a later call to free.
// cache line
#define ALIGN 64
void *aligned_malloc(int size) {
void *mem = malloc(size+ALIGN+sizeof(void*));
void **ptr = (void**)((uintptr_t)(mem+ALIGN+sizeof(void*)) & ~(ALIGN-1));
ptr[-1] = mem;
return ptr;
}
void aligned_free(void *ptr) {
free(((void**)ptr)[-1]);
}
回答2:
Use posix_memalign
/free
.
int posix_memalign(void **memptr, size_t alignment, size_t size);
void* ptr;
int rc = posix_memalign(&ptr, alignment, size);
...
free(ptr)
posix_memalign
is a standard replacement for memalign
which, as you mention is obsolete.
回答3:
What compiler are you using? If you're on MSVC, you can try _aligned_malloc() and _aligned_free()
.
来源:https://stackoverflow.com/questions/1919183/how-to-allocate-and-free-aligned-memory-in-c