Question

Given a method signature:

public bool AreTheSame<T>(Expression<Func<T, object>> exp1, Expression<Func<T, object>> exp2) 

What would be the most efficient way to say if the two expressions are the same? This only needs to work for simple expressions, by this I mean all that would be ‘supported’ would be simple MemberExpressions, eg c => c.ID.

An example call might be:

AreTheSame<User>(u1 => u1.ID, u2 => u2.ID); --> would return true 

Answer 1

Hmmm… I guess you’d have to parse the tree, checking the node-type and member of each. I’ll knock up an example…

using System; using System.Linq.Expressions; class Test {     public string Foo { get; set; }     public string Bar { get; set; }     static void Main()     {         bool test1 = FuncTest<Test>.FuncEqual(x => x.Bar, y => y.Bar),             test2 = FuncTest<Test>.FuncEqual(x => x.Foo, y => y.Bar);     }  } // this only exists to make it easier to call, i.e. so that I can use FuncTest<T> with // generic-type-inference; if you use the doubly-generic method, you need to specify // both arguments, which is a pain... static class FuncTest<TSource> {     public static bool FuncEqual<TValue>(         Expression<Func<TSource, TValue>> x,         Expression<Func<TSource, TValue>> y)     {         return FuncTest.FuncEqual<TSource, TValue>(x, y);     } } static class FuncTest {     public static bool FuncEqual<TSource, TValue>(         Expression<Func<TSource,TValue>> x,         Expression<Func<TSource,TValue>> y)     {         return ExpressionEqual(x, y);     }     private static bool ExpressionEqual(Expression x, Expression y)     {         // deal with the simple cases first...         if (ReferenceEquals(x, y)) return true;         if (x == null || y == null) return false;         if (   x.NodeType != y.NodeType             || x.Type != y.Type ) return false;          switch (x.NodeType)         {             case ExpressionType.Lambda:                 return ExpressionEqual(((LambdaExpression)x).Body, ((LambdaExpression)y).Body);             case ExpressionType.MemberAccess:                 MemberExpression mex = (MemberExpression)x, mey = (MemberExpression)y;                 return mex.Member == mey.Member; // should really test down-stream expression             default:                 throw new NotImplementedException(x.NodeType.ToString());         }     } }