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Date: Sat, 22 Sep 2018 23:45:03 -0400
From: Rich Felker <dalias@...c.org>
To: musl@...ts.openwall.com
Subject: Re: un-UBify-strings

On Sat, Sep 22, 2018 at 11:15:02PM -0400, Rich Felker wrote:
> On Sun, Sep 23, 2018 at 03:10:14AM +0000, Pascal Cuoq wrote:
> > 
> > On 23 Sep 2018, at 04:45, Rich Felker <dalias@...c.org<mailto:dalias@...c.org>> wrote:
> > I'm also trying to fix the UB in
> > address range checks for implementing memmove as memcpy, etc. Is this
> > correct:
> > 
> > if ((uintptr_t)s-(uintptr_t)d-n <= -2*n) return memcpy(d, s, n);
> > 
> > ?
> > 
> > It looks okay to me. You want to test whether
> > (uintptr_t)s-(uintptr_t)d, computed as a mathematical integer, is
> > between -n and n, and since uintptr_t is unsigned, you are using the
> > well-known trick of aligning one of the bounds with 0 so that both
> > inequalities can be tested in one instruction.
> 
> Right.
> 
> > It would seen more natural to me to work on the right-hand side of
> > zero, that it, to compute (uintptr_t)s-(uintptr_t)d+n and to check
> > whether that is <= 2*n (overlap) or > 2*n (no overlap). The
> > generated code may even be one instruction shorter. Apart from that,
> > as long as we have the hypothesis that n <= UINTPTR_MAX/2, I cannot
> > immediately see any reason why it would not work.
> 
> dist(s,d)==n is a no-overlap case. Otherwise I think this is correct
> and we can use:
> 
> 	if ((uintptr_t)s-(uintptr_t)d+n >= 2*n) return memcpy(d, s, n);
> 
> Yes?

BTW just below there's a conditional if (d<s) that, as far as I can
tell, does not need any fixing. If we reach that point (if we don't
just call memcpy for the non-overlapping case) then, assuming n is
valid, d and s necessarily point into the same array, and therefore
d<s is well-defined.

Rich

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