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[Bug c++/70495] false warning: comparison between signed and unsigned integer expressions
- From: "redi at gcc dot gnu.org" <gcc-bugzilla at gcc dot gnu dot org>
- To: gcc-bugs at gcc dot gnu dot org
- Date: Fri, 01 Apr 2016 12:21:14 +0000
- Subject: [Bug c++/70495] false warning: comparison between signed and unsigned integer expressions
- Auto-submitted: auto-generated
- References: <bug-70495-4 at http dot gcc dot gnu dot org/bugzilla/>
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=70495
--- Comment #4 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to AndrÃs from comment #3)
> > b+b has type int
> Note: this does not give the warning. You can see the linked example for
> more details.
I know, but b+b+b adds b to an int and still produces an int, which is
obviously enough to confuse the compiler. I'm not saying it's not a bug, just
explaining where the signed value comes from.