Date: Sun, 14 Mar 1999 20:31:44 -0500 (EST) From: Thomas David Rivers <rivers@dignus.com> To: ian@bulinfo.net, mturpin@saturn.spel.com Cc: freebsd-questions@FreeBSD.ORG Subject: Re: Programming Question Message-ID: <199903150131.UAA10509@lakes.dignus.com> In-Reply-To: <36EC34C0.13D218B4@bulinfo.net>
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> Mark Turpin wrote:
> >
> > I feel really silly having to ask this.
> >
> > Why does my sizeof(mystruct) come out as 4 instead of 3 ( short + char ) ?
> >
> > struct {
> > short shortvariable;
> > char charvariable;
> > } mystruct;
> >
> > sizeof(mystruct) == 4
>
> Try finding the addresses of the structure members. As I know the compiler
> aligns every variable at even address. Maybe that's the reason.
Not exactly... the size of the structure depends on the maximum alignment
requirements of all of its members.
In this case; on most architectures, the alignment of the `shortvariable'
is 2, the alignment of `charvariable' is 1.
So, the alignment of the entire structure is 2. The size of the
structure is then padded - so that if you had an array of these structures,
each array element would be properly aligned so that the structure
alignment is preserved. Thus, the size of the structure is padded to
reach a two-byte alignment... adding one byte... making the size 4.
That's why; when dealing with structures - the whole is typically greater
than the sum of the parts... :-)
- Dave Rivers -
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