[llvm] [AMDGPU][AggressiveInstCombine] Narrow 64 bit math to 32 bit if profitable (PR #130577)

[via llvm-commits]

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@@ -1224,6 +1224,68 @@ static bool foldLibCalls(Instruction &I, TargetTransformInfo &TTI,
   return false;
 }
 
+static bool tryNarrowMathIfNoOverflow(Instruction &I, TargetTransformInfo &TTI,
+                                      const DataLayout &DL) {
+  unsigned opc = I.getOpcode();
+  Type *OldType = I.getType();
+  if (opc != Instruction::Add && opc != Instruction::Mul &&
+      !OldType->isIntOrIntVectorTy()) {
+    return false;
+  }
+  unsigned OrigBit = OldType->getScalarSizeInBits();
+  unsigned MaxBitsNeed = OrigBit;
+  switch (opc) {
+  case Instruction::Add:
+    MaxBitsNeed = KnownBits::add(computeKnownBits(I.getOperand(0), DL),
+                                 computeKnownBits(I.getOperand(1), DL))
+                      .countMaxActiveBits();
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Shoreshen wrote:

Hi @dtcxzyw , what kind of generalized proof??

Roughly if `a` is n bit integer (0~(n-1) bit) and and the highest bit that is possible to be 1 is m<n-1, then a must be positive. Then we have $a \leq \sum_{i=0}^m2^i,b \leq \sum_{i=0}^m2^i$. In conclusion $a+b\leq 2*\sum_{i=0}^m2^i=\sum_{i=1}^{m+1}2^i$ and $a\times b\leq \left(\sum_{i=0}^m2^i\right)\cdot\left(\sum_{i=0}^m2^i\right)=2^{2m+2}-2^{m+2}+1$




https://github.com/llvm/llvm-project/pull/130577