Getting root of expression

Question

Let me show you some code right away:

class MyClass {

    public MyClass(Customer c) {
        ExtractRoot(() => c.FirstName);
    }
}

ExtractRoot is accepting a Expression<Func<object>> as its parameter.

I am trying to extract the customer object 'c' from the expression tree. I am passing it to the ExtractRoot method, and I know how to get the ConstantExpression which represents the customer object. But, when I call ToString() on that object I get this printed:
MyAssemblyName.CustomNamespace.MyClass+<>c__DisplayClass2

Can someone please explain what this means and why I am not getting my customer's ToString() method called, it is as if the object I am getting is not the customer, why and what am I getting back?

Finally, how can I get the actual customer object, or the root of my expression?

I'm not sure I understand the question, do you mean `constantExpression.Value`? — Joachim Isaksson, Jul 28 '13 at 19:18
() => c.FirstName is a typeless delegate. To get to the root you need a delegate in the form of c => c.FirstName, i think — Rob van der Veer, Jul 28 '13 at 19:19
@JoachimIsaksson correct, i keep descending the tree until i arrive at the `ConstantExpression` then i get its value like you said but i get what i showed you ? — Ibrahim Najjar, Jul 28 '13 at 19:20
possible duplicate of [What does "DisplayClass" name mean when calling lambda?](http://stackoverflow.com/questions/16401860/what-does-displayclass-name-mean-when-calling-lambda) — Erik Schierboom, Jul 28 '13 at 19:35
Why the down vote !! and this isn't a complete duplicate @ErikSchierboom because the related answer covers part of my question but what about getting the actual root object. — Ibrahim Najjar, Jul 28 '13 at 21:52
@RobvanderVeer it turned out that you can but you need some reflection, i found the answer in another stackoverflow question and put the link in my answer. Thank You. — Ibrahim Najjar, Jul 29 '13 at 11:48

score 0 · Accepted Answer · edited May 23 '17 at 10:31

0

I have found the complete solution to my answer which has two parts:

Why the DisplayClass and what it means can be found in the answer to this question: What does “DisplayClass” name mean when calling lambda?
How to get the actual root of the expression (which is possible by the way) can be found in this questions answer: Getting the object out of a MemberExpression.

edited May 23 '17 at 10:31

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answered Jul 29 '13 at 11:46

Ibrahim Najjar

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Getting root of expression

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