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Date:	Sat, 9 Feb 2002 07:34:30 -0500
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From:	DJ Delorie <dj AT delorie DOT com>
To:	djgpp-workers AT delorie DOT com
In-reply-to:	<9003-Sat09Feb2002090616+0200-eliz@is.elta.co.il>
Subject:	Re: Alignment problem
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> I'm not sure I understand this: what is the ``required'' alignment
> for a buffer that is 8 bytes large or larger?  Isn't the required
> alignment for these buffers 8-byte alignment?  That is, if I malloc a
> double, shouldn't I expect to get a double aligned on 8-byte boundary?

What it means is that the pointer returned will be sufficiently
aligned so that it can be used as a pointer to any possible data type
the CPU supports without causing a misalignment exception.  For
example, on a sparc you must have at least 4-byte alignment or you
can't use the pointer to store an "int".  I don't know if the x86 has
any memory access opcodes that *require* an aligned address.  If it
does, then the returned pointer must be at least that aligned.

Malloc doesn't know the difference between allocating 8 bytes for a
double, or for 8 chars.  And since you can always realloc a buffer, I
don't think we can rely on the size of the allocation to determine
what kinds of data types will be used with it.

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