Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
801 views
in Technique[技术] by (71.8m points)

c# - Retrieving the name of the invoked method executed in a Func

I would like to get the name of the method that is being delegated as a Func.

Func<MyObject, object> func = x => x.DoSomeMethod();
string name = ExtractMethodName(func); // should equal "DoSomeMethod"

How can I achieve this?

-- For bragging rights --

Make ExtractMethodName also work with a property invocation, having it return the property name in that instance.

eg.

Func<MyObject, object> func = x => x.Property;
string name = ExtractMethodName(func); // should equal "Property"
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Look Ma! No expression trees!

Here's a quick, dirty and implementation-specific version that grabs the metadata token from the IL stream of the underlying lambda and resolves it.

private static string ExtractMethodName(Func<MyObject, object> func)
{
    var il = func.Method.GetMethodBody().GetILAsByteArray();

    // first byte is ldarg.0
    // second byte is callvirt
    // next four bytes are the MethodDef token
    var mdToken = (il[5] << 24) | (il[4] << 16) | (il[3] << 8) | il[2];
    var innerMethod = func.Method.Module.ResolveMethod(mdToken);

    // Check to see if this is a property getter and grab property if it is...
    if (innerMethod.IsSpecialName && innerMethod.Name.StartsWith("get_"))
    {
        var prop = (from p in innerMethod.DeclaringType.GetProperties()
                    where p.GetGetMethod() == innerMethod
                    select p).FirstOrDefault();
        if (prop != null)
            return prop.Name;
    }

    return innerMethod.Name;
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...