I have a few related questions about managing aligned memory blocks. Cross-platform answers would be ideal. However, as I\'m pretty sure a cross-platform solution does not
Experiment on your system. On many systems (especially 64-bit ones), you get 16-byte aligned memory out of malloc() anyway. If not, you will have to allocate the extra space and move the pointer (by at most 8 bytes on almost every machine).
For example, 64-bit Linux on x86/64 has a 16-byte long double, which is 16-byte aligned - so all memory allocations are 16-byte aligned anyway. However, with a 32-bit program, sizeof(long double) is 8 and memory allocations are only 8-byte aligned.
Yes - you can only free() the pointer returned by malloc(). Anything else is a recipe for disaster.
If your system does 16-byte aligned allocations, there isn't a problem. If it doesn't, then you'll need your own reallocator, which does a 16-byte aligned allocation and then copies the data - or that uses the system realloc() and adjusts the realigned data when necessary.
Double check the manual page for your malloc(); there may be options and mechanisms to tweak it so it behaves as you want.
On MacOS X, there is posix_memalign() and valloc() (which gives a page-aligned allocation), and there is a whole series of 'zoned malloc' functions identified by man malloc_zoned_malloc and the header is .