[llvm] [InstCombine] Reduce multiplicands of even numbers when a shift is involved (PR #92475)

[via llvm-commits]

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@@ -1504,17 +1504,32 @@ Instruction *InstCombinerImpl::visitLShr(BinaryOperator &I) {
       // able to invert the transform and perf may suffer with an extra mul
       // instruction.
       if (Op0->hasOneUse()) {
-        APInt NewMulC = MulC->lshr(ShAmtC);
-        // if c is divisible by (1 << ShAmtC):
-        // lshr (mul nuw x, MulC), ShAmtC -> mul nuw nsw x, (MulC >> ShAmtC)
-        if (MulC->eq(NewMulC.shl(ShAmtC))) {
-          auto *NewMul =
-              BinaryOperator::CreateNUWMul(X, ConstantInt::get(Ty, NewMulC));
-          assert(ShAmtC != 0 &&
-                 "lshr X, 0 should be handled by simplifyLShrInst.");
-          NewMul->setHasNoSignedWrap(true);
-          return NewMul;
+        unsigned CommonZeros = std::min(MulC->countr_zero(), ShAmtC);
+        if (CommonZeros != 0) {
+          APInt NewMulC = MulC->lshr(CommonZeros);
+          unsigned NewShAmtC = ShAmtC - CommonZeros;
+          // if c is divisible by (1 << ShAmtC):
+          // lshr (mul nuw x, MulC), ShAmtC -> mul nuw nsw x, (MulC >> ShAmtC)
+          if (NewShAmtC == 0) {
+            auto *NewMul =
+                BinaryOperator::CreateNUWMul(X, ConstantInt::get(Ty, NewMulC));
+            NewMul->setHasNoSignedWrap(true);
+            return NewMul;
+          }
+
+          // We can reduce expressions such as like lshr (mul nuw x, 6), 2 ->
+          // lshr (mul nuw nsw x, 3), 1
+          // TODO: What about if ALL uses can be simplified in this way? Is that
+          // likely enough to happen to justify even caring?
+          auto *NewMul = Builder.CreateMul(X, ConstantInt::get(Ty, NewMulC), "",
+                                           /*NUW*/ true, /*NSW*/ true);
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AtariDreams wrote:

Okay, so pigeonhole theory:

If an expression is (x * c) >> c1, and said expression is nuw, that means it does not overflow, which means there is no set bit being cutoff/truncated.

This means that if the expression is reduced, then that means one less bit is shifted out, which means, when reduced and evaluated, the high bit of the expression MUST be 0, and therefore, nsw.

https://github.com/llvm/llvm-project/pull/92475