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Home/ Questions/Q 6167327
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T22:29:33+00:00 2026-05-23T22:29:33+00:00

Consider the following code snippet namespace ConsoleApplication1 { public delegate TResult Function<in T, out

  • 0

Consider the following code snippet

namespace ConsoleApplication1
{

public delegate TResult Function<in T, out TResult>(T args);

 class Program
 {
      static void Main(string[] args)
    {
        Program pg =new Program();
        Function<Object, DerivedClass> fn1 = null;
        Function<String, BaseClass> fn2 = null;
        fn1 = new Function<object, DerivedClass>(pg.myCheckFuntion)
        fn2=fn1;
        fn2("");// calls myCheckFuntion(Object a)
        pg.myCheckFuntion("Hello"); //calls myCheckFuntion(String a)
     }

     public DerivedClass myCheckFuntion(Object a)
    {
        return  new DerivedClass();
    }
    public DerivedClass myCheckFuntion(String a)
    { 
        return new DerivedClass();
    }
 }

why the delegate call and normal method invocation call different methods.

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  1. Editorial Team
    Editorial Team
    2026-05-23T22:29:34+00:00Added an answer on May 23, 2026 at 10:29 pm

    The delegate is bound to myCheckFuntion(Object) at compile time – you’re telling it to find a method which accepts an Object. That binding is just to a single method – it doesn’t perform overload resolution at execution time based on the actual argument type.

    When you call pg.myCheckFuntion("Hello") that will bind to myCheckFuntion(String) at compile-time because "Hello" is a string, and the conversion from string to string is preferred over the conversion from string to object in overload resolution.

    Note that if you write:

    object text = "Hello";
    pg.myCheckFuntion(text);
    

    then that will call myCheckFuntion(Object).

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