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Date: Fri, 17 Feb 2006 20:14:00 -0000
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From: Eli Zaretskii <eliz@gnu.org>
To: Paul Koning <pkoning@equallogic.com>
CC: ghost@cs.msu.su, gdb@sources.redhat.com
In-reply-to: <17398.11182.747232.774924@gargle.gargle.HOWL> (message from Paul 	Koning on Fri, 17 Feb 2006 15:01:50 -0500)
Subject: Re: MI: reporting of multiple breakpoints
Reply-to: Eli Zaretskii <eliz@gnu.org>
References: <dt4ofp$60k$1@sea.gmane.org> 	<20060217153211.GA21402@nevyn.them.org> 	<dt4r7g$h69$1@sea.gmane.org> 	<ulkw9hj1y.fsf@gnu.org> 	<20060217194426.GA28988@nevyn.them.org> 	<ufymhhi81.fsf@gnu.org> <17398.11182.747232.774924@gargle.gargle.HOWL>
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> Date: Fri, 17 Feb 2006 15:01:50 -0500
> From: Paul Koning <pkoning@equallogic.com>
> Cc: ghost@cs.msu.su, gdb@sources.redhat.com
> 
> It is a bad thing for the break at x to fail due to the bad luck of
> having a watch exception at the preceding instruction.

I agree.  But the current code was written for a reason, I think, so
I'm curious to know what was that reason.

> If the two stops happened to be the SAME instruction, then you have
> plausible deniability.

Yes; except that if they happen on the same instruction, they are
still two different events: the breakpoint breaks _before_ the
instruction, the watchpoint _after_ it.  At least on x86.  So we can
announce both even in this case.