------- Comment #2 from pinskia at gmail dot com 2008-09-24 17:44 ------- Subject: Re: New: GCC applies signed strict-overflow rules to unsigned short type
When doing addition unsigned short is promoted to an signed int. So this is not a bug. That is unsigned short + 1 is a signed int since 1 is a signed int. Sent from my iPhone On Sep 24, 2008, at 10:29 AM, "alexandre dot nunes at gmail dot com" <[EMAIL PROTECTED] > wrote: > I'll submit a testcase that apparently demonstrates that gcc is > trying to apply > signed strict overflow rules to an unsigned short type, at least on > 32 bit > machines when short is 16 bit. > > Here is the output: > arm-elf-gcc -O2 -W -Wall -Wstrict-overflow=5 -c testcase.c > testcase.c: In function âincr_counterâ: > testcase.c:13: warning: assuming signed overflow does not occur when > assuming > that (X + c) < X is always false > > this is from gcc 4.3.1; my native build also has the same semantics, > and I've > tested with 4.2.4 also. > > The thing is that the type is unsigned (even if it is smaller than > the target > machine register), so that it can overflow and it can be detected > (costly > perhaps, but can). > > > -- > Summary: GCC applies signed strict-overflow rules to > unsigned > short type > Product: gcc > Version: 4.2.4 > Status: UNCONFIRMED > Severity: normal > Priority: P3 > Component: c > AssignedTo: unassigned at gcc dot gnu dot org > ReportedBy: alexandre dot nunes at gmail dot com > > > http://gcc.gnu.org/bugzilla/show_bug.cgi?id=37642 > -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=37642
