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Home/ Questions/Q 6214839
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T06:56:50+00:00 2026-05-24T06:56:50+00:00

Why the following code throws an exception at runtime, whereas doing it in the

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Why the following code throws an exception at runtime, whereas doing it in the traditional way compiles without problem?

var left = Expression.Constant(25d);
var right = Expression.Constant(20);

// Throws an InvalidOperationException!
var multiplyExpression = Expression.Multiply(left, right); 

var multiply = 25d * 20;
Debug.WriteLine(multiply.ToString()); // Works normally!

I won’t use Expression.Convert since I can’t determine exactly which expression should be converted.

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  1. Editorial Team
    Editorial Team
    2026-05-24T06:56:51+00:00Added an answer on May 24, 2026 at 6:56 am

    Well, I figured out how to solve the problem using TypeCode enumeration to determine which node would have higher type precision, then convert the latter node’s type to the former’s type, and vice versa:

      private static void Visit(ref Expression left, ref Expression right)
      {
         var leftTypeCode = Type.GetTypeCode(left.Type);
         var rightTypeCode = Type.GetTypeCode(right.Type);
    
         if (leftTypeCode == rightTypeCode)
             return;
    
         if (leftTypeCode > rightTypeCode)
            right = Expression.Convert(right, left.Type);
         else
            left = Expression.Convert(left, right.Type);
      }
    
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