How to find all methods that are declared as virtual in c# using reflection?

Question

For example

public class UserInfo {}

public interface IUser
{
  UserInfo Get(int id);
  string GetData(int id);
}

public class User : IUser
{
    public UserInfo Get(int id)
    {
        return null;
    }
    public virtual string GetData(int id)
    {
        return null;
    }
}

I tried the following but it returns true for both methods.

MethodInfo[] methods=typeof(User).GetMethods();
foreach(MethodInfo method in methods)
{
   if(method.IsVirtual)
   {
      Console.WriteLine(method.Name);
   }
}

Expected Output

GetData

Actual Output

Get
GetData

This link is some what similar but not working in my case: use-reflection-to-find-all-public-virtual-methods-and-provide-an-override

[Works for me](http://ideone.com/Gv0taZ). Can you provide a real, minimal, compilable, runnable example? — Blorgbeard, Jul 30 '14 at 10:52
Indeed, I had to do multiple things to test this from the original question: 1) Add a UserInfo class. 2) Add an IUser class. 3) Fix the "P" being public. 4) use `typeof(User)` instead of `typeof(Base)`. None of these are things that someone answering the question should have to do. Please provide short but *complete* code examples in future. — Jon Skeet, Jul 30 '14 at 11:01

score 4 · Accepted Answer · answered Jul 30 '14 at 10:54

4

The problem is Get is a virtual method in that it implements IUser.Get and so is called using execution-time dispatch - I suspect, anyway. (You haven't given us a complete program, so I've had to guess.)

If you actually want "methods that can be overridden" you also need to check MethodBase.IsFinal. So just change your loop to:

foreach (MethodInfo method in methods)
{
   if (method.IsVirtual && !method.IsFinal)
   {
      Console.WriteLine(method.Name);
   }
}

answered Jul 30 '14 at 10:54

Jon Skeet

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This is correct, if you (OP) remove the IUser interface from your User class, you will see your expected result. Adding IsFinal resolves this. – Troels Larsen Jul 30 '14 at 10:59
@TroelsLarsen: Yes, that's the test I did :) – Jon Skeet Jul 30 '14 at 10:59
this condition (method.IsVirtual && !method.IsFinal) satisfy for properties too (getter and setter methods like get_PropertyName, set_PropertyName) – Elanchezhian Narayanasamy Jul 31 '14 at 06:28
@ElanchezhianNarayanasamy: Are you asking that, or stating it? I'm not sure what the point of your comment it. – Jon Skeet Jul 31 '14 at 07:11

Troels Larsen · Answer 2 · 2014-07-30T11:15:52.047

Full, working sample:

class Program
{
    static void Main(string[] args)
    {
        MethodInfo[] methods = typeof(User).GetMethods();
        foreach (var method in methods)
        {
            if (method.IsVirtual && !method.IsFinal)
            {
                Console.WriteLine(method.Name);
            }
        }
        Console.ReadLine();
    }
}

public class User : IUser
{
    public string Get(int id)
    {
        // some codes
        return null;
    }
    public virtual string GetData(int id)
    {
        // I want to find this method declared as virtual
        return null;
    }
}

public interface IUser
{
    string Get(int id);
    string GetData(int id);
}

Output:

GetData
ToString
Equals
GetHashCode

Please mark the other answer if this works for you. I merely pasted it for OP to run. — Troels Larsen, Jul 30 '14 at 11:16

How to find all methods that are declared as virtual in c# using reflection?

2 Answers2