http://gcc.gnu.org/bugzilla/show_bug.cgi?id=59999
--- Comment #16 from Paulo J. Matos <pa...@matos-sorge.com> ---
(In reply to rguent...@suse.de from comment #15)
> Exactly the same problem. C integral type promotion rules make
> that i = (short)((int)i + 1) again. Note that (int)i + 1
> does not overflow, (short) ((int)i + 1) invokes implementation-defined
> behavior which in our case is modulo-2 reduction.
>
> Nothing guarantees that (short)i + 1 does not overflow.
OK, that makes sense. But in GCC 4.8 that doesn't seem to be what happens.
It seems to be i = (short) ((unsigned short) i + 1)
Later i is cast to int for comparison.
Before ivopts this is the end of the loop body:
i.7_19 = (unsigned short) i_26;
_20 = i.7_19 + 1;
i_21 = (short intD.8) _20;
_10 = (intD.1) i_21;
if (_10 < _25)
goto <bb 7>;
else
goto <bb 6>;
i is initially a short, then moved to unsigned short. The addition is performed
and returned to short. Then cast to int for the comparison.
For GCC 4.5.4 the end of loop body is:
iD.2767_18 = iD.2767_26 + 1;
D.5046_9 = (intD.0) iD.2767_18;
if (D.5046_9 < D.5047_25)
goto <bb 5>;
else
goto <bb 6>;
Here the addition is made in short int and then there's only one cast to int.
