Re: struct hashmap_entry packing

> 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.
>