Fix 2-bit rotation normalization

srem treats both operands as signed. In 2-bit representation
2 is interpretted as -2. Since the rotation amount is 0 or 1
in 2-bit, we can determine by checking if the amount is odd or even

Author: chaitanyav

clang/lib/AST/ExprConstant.cpp

@@ -19761,9 +19761,13 @@ APSInt NormalizeRotateAmount(const APSInt &Value, const APSInt &Amount) {
   if (BitWidth == 1) {
     // Rotating a 1-bit value is always a no-op
     NormAmt = APSInt(APInt(AmtBitWidth, 0), NormAmt.isUnsigned());
+  } else if (BitWidth == 2) {
+    // For 2-bit values: rotation amount is 0 or 1 based on
+    // whether the amount is even or odd. We can't use srem here because
+    // the divisor (2) would be misinterpreted as -2 in 2-bit signed arithmetic.
+    NormAmt =
+        APSInt(APInt(AmtBitWidth, NormAmt[0] ? 1 : 0), NormAmt.isUnsigned());
   } else {
-    // Divisor is always unsigned to avoid misinterpreting BitWidth as
-    // negative in small bit widths (e.g., BitWidth=2 would be -2 if signed).
     APInt Divisor;
     if (AmtBitWidth > BitWidth) {
       Divisor = llvm::APInt(AmtBitWidth, BitWidth);

clang/lib/CodeGen/CGBuiltin.cpp

@@ -2497,6 +2497,14 @@ RValue CodeGenFunction::emitRotate(const CallExpr *E, bool IsRotateRight) {
   if (BitWidth == 1) {
     // Rotating a 1-bit value is always a no-op
     ShiftAmt = ConstantInt::get(ShiftTy, 0);
+  } else if (BitWidth == 2) {
+    // For 2-bit values: rotation amount is 0 or 1 based on
+    // whether the amount is even or odd. We can't use srem here because
+    // the divisor (2) would be misinterpreted as -2 in 2-bit signed arithmetic.
+    llvm::Value *One = ConstantInt::get(ShiftTy, 1);
+    ShiftAmt = Builder.CreateAnd(ShiftAmt, One);
+    if (ShiftTy != Ty)
+      ShiftAmt = Builder.CreateIntCast(ShiftAmt, Ty, false);
   } else {
     unsigned ShiftAmtBitWidth = ShiftTy->getIntegerBitWidth();
     bool ShiftAmtIsSigned = E->getArg(1)->getType()->isSignedIntegerType();