Most efficient way to test equality of lambda expressions

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());
        }
    }
}