Date: Mon, 15 Mar 1999 11:03:09 +1000 From: Greg Black <gjb@comkey.com.au> To: Mark Turpin <mturpin@saturn.spel.com> Cc: questions@FreeBSD.ORG Subject: Re: Programming Question Message-ID: <19990315010309.16892.qmail@alpha.comkey.com.au> In-Reply-To: <Pine.BSF.4.05.9903141646550.6443-100000@saturn.spel.com> of Sun, 14 Mar 1999 16:50:32 EST References: <Pine.BSF.4.05.9903141646550.6443-100000@saturn.spel.com>
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> Why does my sizeof(mystruct) come out as 4 instead of 3 ( short + char ) ?
>
> struct {
> short shortvariable;
> char charvariable;
> } mystruct;
Because that's the size the compiler allocated for it. It could
legitimately be 8 (and other sizes, too). Read clause 6.5.2.1
"Semantics" of ISO/IEC 9899: 1990 (the ISO Standard for C) for
more info. In particular, note the paragraph that says:
Each non-bit-field member of a structure or union object is
aligned in an implementation-defined manner appropriate to
its type.
Then see the later paragraph that says:
There may also be unnamed padding at the end of a structure
or union, as necessary to achieve the appropriate alignment
were the structure or union to be an element of an array.
Because it's implementation-defined, compiler authors are at
liberty to offer normal default behaviour with padding for
alignment needs (as you have seen here) and optional packed
behaviour, provided they document it.
Final note: this has nothing to do with FreeBSD. Try a news
group like comp.lang.c if you have Usenet access.
--
Greg Black <gjb@acm.org>
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