how much extra memory does posix_memalign() take?

Question

I am trying to decide if I should use memalign() over malloc() because aligned memory would make my job easier. I read the GNU documentation here (http://www.gnu.org/software/libc/manual/html_node/Aligned-Memory-Blocks.html) which mentions that The function memalign works by allocating a somewhat larger block. I want to know the exact value for that "somewhat larger block".

Logically thinking the extra memory required should be equal to the the value of alignment required. But I am not sure if there is an optimization over that.

Answer 1

Posix is a standard, not a specific set of code, but we can look at libc for an example.

Here's what posix_memalign() initially allocates in that implementation.

mem = malloc (size + 2 * alignment);

With this beautiful ASCII illustration.

    /*
      ______________________ TOTAL _________________________
     /                                                      \
    +---------------+-------------------------+--------------+
    |               |                         |              |
    +---------------+-------------------------+--------------+
    \____ INIT ____/ \______ RETURNED _______/ \____ END ___/

   */

It then returns to the heap the unused storage on either end of the allocation.

This means that fragmentation may get worse, though the heap memory used is the same amount.