On 10/06/2016 01:18 AM, Richard Biener wrote:
It ought to be easy to fold x ^ y to zero when x == y (famous last words).
I'm sure I'll regret saying that when I go to look at how to twiddle DOM
appropriately.
Interesting idea. Though it get's (theoretically) interesting for
ternary ops where
On Wed, 5 Oct 2016, Jeff Law wrote:
> On 07/08/2016 05:17 AM, Richard Biener wrote:
> > On Fri, 8 Jul 2016, Prathamesh Kulkarni wrote:
> >
> > > Hi Richard,
> > > For the following test-case:
> > >
> > > int f(int x, int y)
> > > {
> > >int ret;
> > >
> > >if (x == y)
> > > ret = x
On 07/20/2016 04:09 PM, Prathamesh Kulkarni wrote:
Oops wrong test-case, the dump below is of the following test-case:
int f(int x, int y)
{
int ret = 10;
if (x == y)
ret = x - y;
return ret;
}
DOM ought to be catching this as well and does with the hack patch I've
done for the firs
On 07/20/2016 04:07 PM, Prathamesh Kulkarni wrote:
Not record both equivalences. This might break the testcase it was
introduced for (obviously). Which is why I CCed Jeff for his opinion.
Well, folding happens for x - y, if x == y.
int f(int x, int y)
{
int ret;
if (x == y)
ret = x -
On 07/08/2016 05:17 AM, Richard Biener wrote:
On Fri, 8 Jul 2016, Prathamesh Kulkarni wrote:
Hi Richard,
For the following test-case:
int f(int x, int y)
{
int ret;
if (x == y)
ret = x ^ y;
else
ret = 1;
return ret;
}
I was wondering if x ^ y should be folded to 0 sinc
On Wed, 20 Jul 2016, Prathamesh Kulkarni wrote:
> On 20 July 2016 at 16:35, Richard Biener wrote:
> > On Wed, 20 Jul 2016, Prathamesh Kulkarni wrote:
> >
> >> On 8 July 2016 at 12:29, Richard Biener wrote:
> >> > On Fri, 8 Jul 2016, Richard Biener wrote:
> >> >
> >> >> On Fri, 8 Jul 2016, Pratha
On 20 July 2016 at 23:07, Prathamesh Kulkarni
wrote:
> On 20 July 2016 at 16:35, Richard Biener wrote:
>> On Wed, 20 Jul 2016, Prathamesh Kulkarni wrote:
>>
>>> On 8 July 2016 at 12:29, Richard Biener wrote:
>>> > On Fri, 8 Jul 2016, Richard Biener wrote:
>>> >
>>> >> On Fri, 8 Jul 2016, Pratham
On 20 July 2016 at 16:35, Richard Biener wrote:
> On Wed, 20 Jul 2016, Prathamesh Kulkarni wrote:
>
>> On 8 July 2016 at 12:29, Richard Biener wrote:
>> > On Fri, 8 Jul 2016, Richard Biener wrote:
>> >
>> >> On Fri, 8 Jul 2016, Prathamesh Kulkarni wrote:
>> >>
>> >> > Hi Richard,
>> >> > For the
On 07/20/2016 09:35 AM, Richard Biener wrote:
I have reported it as PR71947.
Could you help me point out how to fix this ?
Not record both equivalences. This might break the testcase it was
introduced for (obviously). Which is why I CCed Jeff for his opinion.
It's on my todo list. I'm still
On Wed, 20 Jul 2016, Prathamesh Kulkarni wrote:
> On 8 July 2016 at 12:29, Richard Biener wrote:
> > On Fri, 8 Jul 2016, Richard Biener wrote:
> >
> >> On Fri, 8 Jul 2016, Prathamesh Kulkarni wrote:
> >>
> >> > Hi Richard,
> >> > For the following test-case:
> >> >
> >> > int f(int x, int y)
> >>
On 8 July 2016 at 12:29, Richard Biener wrote:
> On Fri, 8 Jul 2016, Richard Biener wrote:
>
>> On Fri, 8 Jul 2016, Prathamesh Kulkarni wrote:
>>
>> > Hi Richard,
>> > For the following test-case:
>> >
>> > int f(int x, int y)
>> > {
>> >int ret;
>> >
>> >if (x == y)
>> > ret = x ^ y;
On Fri, 8 Jul 2016, Richard Biener wrote:
> On Fri, 8 Jul 2016, Prathamesh Kulkarni wrote:
>
> > Hi Richard,
> > For the following test-case:
> >
> > int f(int x, int y)
> > {
> >int ret;
> >
> >if (x == y)
> > ret = x ^ y;
> >else
> > ret = 1;
> >
> >return ret;
> >
On Fri, 8 Jul 2016, Prathamesh Kulkarni wrote:
> Hi Richard,
> For the following test-case:
>
> int f(int x, int y)
> {
>int ret;
>
>if (x == y)
> ret = x ^ y;
>else
> ret = 1;
>
>return ret;
> }
>
> I was wondering if x ^ y should be folded to 0 since
> it's guarded
Hi Richard,
For the following test-case:
int f(int x, int y)
{
int ret;
if (x == y)
ret = x ^ y;
else
ret = 1;
return ret;
}
I was wondering if x ^ y should be folded to 0 since
it's guarded by condition x == y ?
optimized dump shows:
f (int x, int y)
{
int iftmp.0_1;
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