memcpy to an unaligned address

[Shaun Jackman]

On 8/3/05, Richard Henderson <rth@redhat.com> wrote:
> It is nevertheless correct.  Examine all of the parts of the expression.
> 
> In particular, "&s->b".  What type does it have?  In an ideal world, it
> would be "pointer to unaligned integer".  But we have no such type in
> our type system, so it is "pointer to integer".  This expression is ONLY
> THEN passed to memcpy.  At which point we query the argument for its
> alignment, and get the non-intuitive result.
> 
> If you instead pass "s" to memcpy, you should get the correct unaligned
> copy.  If that isn't happening, that's a bug.

I'm not sure I understood the last line. s is a structure, and its
address is aligned. How would you pass it to memcpy, and why would it
generate an unaligned copy?

Cheers,
Shaun