How to invoke lambda method that is non-static using MethodInfo (uses <>c__DisplayClass1)

Question

Consider the following code:

bool result;

Func<int, bool> lambda1 = i => i == 9000;
MethodInfo lambda1Method = lambda1.Method;
result = (bool)lambda1Method.Invoke(null, new object[] { 9000 }); // this works, result = true

int myLocalVariable = 9000;
Func<int, bool> lambda2 = i => i == myLocalVariable;
MethodInfo lambda2Method = lambda2.Method;
result = (bool)lambda2Method.Invoke(null, new object[] { 9000 }); // error

Invoking lambda2Method results in a System.Reflection.TargetException:

Non-static method requires a target.

This question here explains why the lambda1Method is static, while lambda2Method is non-static. Basically if lambdas contain local variables, a class is dynamically created that interprets each local variable as a field. lambda2Method becomes an instance method of that new class. I know this because lambda2Method.DeclaringType is <>c__DisplayClass1, and lambda2Method.IsStatic is false.

My question is, how can I make this work? I understand that because lambda2Method is non-static, I need to supply a value for the object obj parameter of MethodBase.Invoke(), and it needs to be an instance of <>c__DisplayClass1, but how do I obtain this instance?

What is wrong with `result = lambda2.Invoke(9000);` ? Why converting to `MethodInfo` ? — Zein Makki, Aug 08 '16 at 05:19
@user3185569 Absolutely nothing, but the project I am working on involves using the `MethodInfo Delegate.Method` property. — Mr Anderson, Aug 08 '16 at 05:21
@MrAnderson Ok, then check the edited answer on how to create an instance using reflection. — Zein Makki, Aug 08 '16 at 05:24

Zein Makki · Accepted Answer · 2016-08-08T05:53:58.883

2

The main issue that you need to address in you question is How to create an instance of a type generated by the compiler ?

So, if you really have to use MethodInfo, then you can create the required instance by using Reflection:

var instance = Activator.CreateInstance(lambda2Method.DeclaringType);

result = lambda2Method.Invoke(instance, new object[] { 9000 });

Summary:

The declaring type for your method lambda2 is a hidden class generated by the compiler. MethodInfo.Invoke requires a target instance of that type to invoke a non-static method.

Edit:

To get the captured value of myVariable correct, you can make use of the Target property :

result = lambda2Method.Invoke(lambda2.Target, new object[] { 9000 });

edited Aug 08 '16 at 05:53

answered Aug 08 '16 at 05:22

Zein Makki

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It does let me create the instance using `Activator`, but the value of the field `myLocalVariable` is 0, not 9000, so `result` is false. – Mr Anderson Aug 08 '16 at 05:28
@MrAnderson That's because we created a new instance of that type. – Zein Makki Aug 08 '16 at 05:32
@MrAnderson You can use `lambda2.Target`, check the edited answer. – Zein Makki Aug 08 '16 at 05:35

How to invoke lambda method that is non-static using MethodInfo (uses <>c__DisplayClass1)

1 Answers1