[MLIR] Unconditionally take min of max lhs/rhs value in inferRemU

`arith.remu` cannot be larger than (rhs - 1) or lhs.

Author: goldsteinn

mlir/lib/Interfaces/Utils/InferIntRangeCommon.cpp

@@ -444,10 +444,10 @@ mlir::intrange::inferRemU(ArrayRef<ConstantIntRanges> argRanges) {
 
   unsigned width = rhsMin.getBitWidth();
   APInt umin = APInt::getZero(width);
-  APInt umax = APInt::getMaxValue(width);
+  // Remainder can't be larger than either of its arguments.
+  APInt umax = llvm::APIntOps::umin((rhsMax - 1), lhs.umax());
 
   if (!rhsMin.isZero()) {
-    umax = rhsMax - 1;
     // Special case: sweeping out a contiguous range in N/[modulus]
     if (rhsMin == rhsMax) {
       const APInt &lhsMin = lhs.umin(), &lhsMax = lhs.umax();

mlir/test/Dialect/Arith/int-range-interface.mlir

@@ -267,8 +267,8 @@ func.func @remui_base(%arg0 : index, %arg1 : index ) -> i1 {
 }
 
 // CHECK-LABEL: func @remui_base_maybe_zero
-// CHECK: %[[ret:.*]] = arith.cmpi ult
-// CHECK: return %[[ret]]
+// CHECK: %[[true:.*]] = arith.constant true
+// CHECK: return %[[true]]
 func.func @remui_base_maybe_zero(%arg0 : index, %arg1 : index ) -> i1 {
     %c4 = arith.constant 4 : index