--- Comment #7 from majbrock at dse dot nl 2010-07-29 13:30 ---
Thank you both for looking into it and explaining the behaviour.
I feel stupid and apologize, because I was certain that it was not read twice.
Yet now I can no longer reproduce that, so I guess I was wrong after all.
Thank
--- Comment #6 from rguenth at gcc dot gnu dot org 2010-07-29 13:27 ---
And with all other versions I tried (4.3 and 4.5)
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45126
--- Comment #5 from schwab at linux-m68k dot org 2010-07-29 12:55 ---
Works fine here with gcc 4.4.4.
movzwl vus, %eax
movzwl vus, %edx
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schwab at linux-m68k dot org changed:
What|Removed |Added
--- Comment #4 from majbrock at dse dot nl 2010-07-29 12:41 ---
Andreas said:
>> That does not change the fact that vus*vus can be assumed to be non-negative.
And then this bug was closed again.
So because one part of my report is dismissed you also dismiss the other part?
I already co
--- Comment #3 from schwab at linux-m68k dot org 2010-07-29 12:21 ---
That does not change the fact that vus*vus can be assumed to be non-negative.
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schwab at linux-m68k dot org changed:
What|Removed |Added
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--- Comment #2 from majbrock at dse dot nl 2010-07-29 12:00 ---
Ok, that is a choice.
But even then vus is read only once, where it appeared twice in the expression.
What about the possible side-effects of reading a volatile?
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majbrock at dse dot nl changed:
What|R
--- Comment #1 from rguenth at gcc dot gnu dot org 2010-07-29 11:25 ---
Possible values for vus are [0, 65535], volatileness does not change that.
Multiplying this as int is always positive (overflow is undefined), so
we can change the test to (vus * vus) != 0.
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rguenth at gcc dot
