Subject: struct hashmap_entry packing

From: Jeff King <[email-masked]>
Date: Mon, 28 Jul 2014 13:17:44 -0400

Hi Karsten,

The hashmap_entry documentation claims:

  `struct hashmap_entry`::

	An opaque structure representing an entry in the hash table,
	which must be used as first member of user data structures.
	Ideally it should be followed by an int-sized member to prevent
	unused memory on 64-bit systems due to alignment.

I'm not sure if the statement about alignment is true. If I have a
struct like:

    struct magic {
	    struct hashmap_entry map;
	    int x;
    };

the statement above implies that I should be able to fit this into only
16 bytes on an LP64 system. But I can't convince gcc to do it. And I
think that makes sense, if you consider code like:

   memset(&magic.map, 0, sizeof(struct hashmap_entry));

The sizeof() has to be the same regardless of whether the hashmap_entry
is standalone or in another struct, and therefore must be padded up to
16 bytes. If we stored "x" in that padding in the combined struct, it
would be overwritten by our memset.

Am I missing anything? If this is the case, we should probably drop that
bit from the documentation. It's possible that we could get around it by
embedding the hashmap_entry elements directly into the parent struct,
but we would be counting on a reader dereferencing it as a hashmap_entry
seeing the members at the exact same offset. I'd imagine that's one of
those things that holds most of the time, but is violating the standard.
It's probably not worth it to save a few bytes.

-Peff