Allocating Aligned Memory Blocks

The address of a block returned by malloc or realloc in the GNU system is always a multiple of eight (or sixteen on 64-bit systems). If you need a block whose address is a multiple of a higher power of two than that, use memalign, posix_memalign, or valloc. These functions are declared in `stdlib.h'.

With the GNU library, you can use free to free the blocks that memalign, posix_memalign, and valloc return. That does not work in BSD, however--BSD does not provide any way to free such blocks.

Function: void * memalign (size_t boundary, size_t size)

The memalign function allocates a block of size bytes whose address is a multiple of boundary. The boundary must be a power of two! The function memalign works by allocating a somewhat larger block, and then returning an address within the block that is on the specified boundary.

Function: int posix_memalign (void **memptr, size_t alignment, size_t size)

The posix_memalign function is similar to the memalign function in that it returns a buffer of size bytes aligned to a multiple of alignment. But it adds one requirement to the parameter alignment: the value must be a power of two multiple of sizeof (void *).

If the function succeeds in allocation memory a pointer to the allocated memory is returned in *memptr and the return value is zero. Otherwise the function returns an error value indicating the problem.

This function was introduced in POSIX 1003.1d.

Function: void * valloc (size_t size)

Using valloc is like using memalign and passing the page size as the value of the second argument. It is implemented like this:

void *
valloc (size_t size)
{
  return memalign (getpagesize (), size);
}

section How to get information about the memory subsystem? for more information about the memory subsystem.

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