[Bug tree-optimization/110852] [14 Regression] ICE: in get_predictor_value, at predict.cc:2695 with -O -fno-tree-fre and __builtin_expect() since r14-2219-geab57b825bcc35

["jakub at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=110852

--- Comment #7 from Jakub Jelinek <jakub at gcc dot gnu.org> ---
So, what about following patch (which also fixes the ICE, would of course need
to add the testcase) and doesn't regress any predict-*.c tests)?

--- gcc/predict.cc.jj   2024-01-03 11:51:32.000000000 +0100
+++ gcc/predict.cc      2024-01-04 16:28:55.041507010 +0100
@@ -2583,44 +2583,36 @@ expr_expected_value_1 (tree type, tree o
   if (get_gimple_rhs_class (code) == GIMPLE_BINARY_RHS)
     {
       tree res;
-      tree nop0 = op0;
-      tree nop1 = op1;
-      if (TREE_CODE (op0) != INTEGER_CST)
-       {
-         /* See if expected value of op0 is good enough to determine the
result.  */
-         nop0 = expr_expected_value (op0, visited, predictor, probability);
-         if (nop0
-             && (res = fold_build2 (code, type, nop0, op1)) != NULL
-             && TREE_CODE (res) == INTEGER_CST)
-           return res;
-         if (!nop0)
-           nop0 = op0;
-        }
       enum br_predictor predictor2;
       HOST_WIDE_INT probability2;
-      if (TREE_CODE (op1) != INTEGER_CST)
+      tree nop0 = expr_expected_value (op0, visited, predictor, probability);
+      if (!nop0)
+       {
+         nop0 = op0;
+         *predictor = PRED_UNCONDITIONAL;
+         *probability = -1;
+       }
+      tree nop1 = expr_expected_value (op1, visited, &predictor2,
&probability2);
+      if (!nop1)
+       {
+         nop1 = op1;
+         predictor2 = PRED_UNCONDITIONAL;
+         probability2 = -1;
+       }
+      /* Finally see if we have two known values.  */
+      res = fold_build2 (code, type, nop0, nop1);
+      if (TREE_CODE (res) == INTEGER_CST)
        {
-         /* See if expected value of op1 is good enough to determine the
result.  */
-         nop1 = expr_expected_value (op1, visited, &predictor2,
&probability2);
-         if (nop1
-             && (res = fold_build2 (code, type, op0, nop1)) != NULL
-             && TREE_CODE (res) == INTEGER_CST)
+         /* If one operand is PRED_UNCONDITIONAL, aka directly or indirectly
+            constant, prefer the other predictor.  */
+         if (predictor2 == PRED_UNCONDITIONAL)
+           return res;
+         if (*predictor == PRED_UNCONDITIONAL)
            {
              *predictor = predictor2;
              *probability = probability2;
              return res;
            }
-         if (!nop1)
-           nop1 = op1;
-        }
-      if (nop0 == op0 || nop1 == op1)
-       return NULL;
-      /* Finally see if we have two known values.  */
-      res = fold_build2 (code, type, nop0, nop1);
-      if (TREE_CODE (res) == INTEGER_CST
-         && TREE_CODE (nop0) == INTEGER_CST
-         && TREE_CODE (nop1) == INTEGER_CST)
-       {
          /* Combine binary predictions.  */
          if (*probability != -1 || probability2 != -1)
            {
@@ -2631,6 +2623,9 @@ expr_expected_value_1 (tree type, tree o

          if (predictor2 < *predictor)
            *predictor = predictor2;
+         if (*predictor != PRED_BUILTIN_EXPECT
+             && *predictor != PRED_BUILTIN_EXPECT_WITH_PROBABILITY)
+           *probability = -1;

          return res;
        }