On Wed, 20 Jul 2016, Prathamesh Kulkarni wrote:
> On 8 July 2016 at 12:29, Richard Biener <rguent...@suse.de> 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;
> >> > 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;
> >> > int iftmp.0_4;
> >> >
> >> > <bb 2>:
> >> > if (x_2(D) == y_3(D))
> >> > goto <bb 3>;
> >> > else
> >> > goto <bb 4>;
> >> >
> >> > <bb 3>:
> >> > iftmp.0_4 = x_2(D) ^ y_3(D);
> >> >
> >> > <bb 4>:
> >> > # iftmp.0_1 = PHI <iftmp.0_4(3), 1(2)>
> >> > return iftmp.0_1;
> >> >
> >> > }
> >> >
> >> > The attached patch tries to fold for above case.
> >> > I am checking if op0 and op1 are equal using:
> >> > if (bitmap_intersect_p (vr1->equiv, vr2->equiv)
> >> > && operand_equal_p (vr1->min, vr1->max)
> >> > && operand_equal_p (vr2->min, vr2->max))
> >> > { /* equal /* }
> >> >
> >> > I suppose intersection would check if op0 and op1 have equivalent ranges,
> >> > and added operand_equal_p check to ensure that there is only one
> >> > element within the range. Does that look correct ?
> >> > Bootstrap+test in progress on x86_64-unknown-linux-gnu.
> >>
> >> I think VRP is the wrong place to catch this and DOM should have but it
> >> does
> >>
> >> Optimizing block #3
> >>
> >> 1>>> STMT 1 = x_2(D) le_expr y_3(D)
> >> 1>>> STMT 1 = x_2(D) ge_expr y_3(D)
> >> 1>>> STMT 1 = x_2(D) eq_expr y_3(D)
> >> 1>>> STMT 0 = x_2(D) ne_expr y_3(D)
> >> 0>>> COPY x_2(D) = y_3(D)
> >> 0>>> COPY y_3(D) = x_2(D)
> >> Optimizing statement ret_4 = x_2(D) ^ y_3(D);
> >> Replaced 'x_2(D)' with variable 'y_3(D)'
> >> Replaced 'y_3(D)' with variable 'x_2(D)'
> >> Folded to: ret_4 = x_2(D) ^ y_3(D);
> >> LKUP STMT ret_4 = x_2(D) bit_xor_expr y_3(D)
> >>
> >> heh, registering both reqivalencies is obviously not going to help...
> >>
> >> The 2nd equivalence is from doing
> >>
> >> /* We already recorded that LHS = RHS, with canonicalization,
> >> value chain following, etc.
> >>
> >> We also want to record RHS = LHS, but without any
> >> canonicalization
> >> or value chain following. */
> >> if (TREE_CODE (rhs) == SSA_NAME)
> >> const_and_copies->record_const_or_copy_raw (rhs, lhs,
> >> SSA_NAME_VALUE (rhs));
> >>
> >> generally recording both is not helpful. Jeff? This seems to be
> >> r233207 (fix for PR65917) which must have regressed this testcase.
> >
> > Just verified it works fine on the GCC 5 branch:
> >
> > Optimizing block #3
> >
> > 0>>> COPY y_3(D) = x_2(D)
> > 1>>> STMT 1 = x_2(D) le_expr y_3(D)
> > 1>>> STMT 1 = x_2(D) ge_expr y_3(D)
> > 1>>> STMT 1 = x_2(D) eq_expr y_3(D)
> > 1>>> STMT 0 = x_2(D) ne_expr y_3(D)
> > Optimizing statement ret_4 = x_2(D) ^ y_3(D);
> > Replaced 'y_3(D)' with variable 'x_2(D)'
> > Applying pattern match.pd:240, gimple-match.c:11346
> > gimple_simplified to ret_4 = 0;
> > Folded to: ret_4 = 0;
> 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.
Richard.
--
Richard Biener <rguent...@suse.de>
SUSE LINUX GmbH, GF: Felix Imendoerffer, Jane Smithard, Graham Norton, HRB
21284 (AG Nuernberg)
