Question

I have a generic class which takes two type parameters, Generic<A, B>. This class has methods with signatures that are distinct so long and A and B are distinct. However, if A == B the signatures match exactly and overload resolution cannot be performed. Is it possible to somehow specify a specialisation of the method for this case? Or force the compiler to arbitrarily choose one of the matching overloads?

using System;  namespace Test {     class Generic<A, B>     {         public string Method(A a, B b)         {             return a.ToString() + b.ToString();         }          public string Method(B b, A a)         {             return b.ToString() + a.ToString();         }     }      class Program     {         static void Main(string[] args)         {             Generic<int, double> t1 = new Generic<int, double>();             Console.WriteLine(t1.Method(1.23, 1));              Generic<int, int> t2 = new Generic<int, int>(); // Following line gives: //     The call is ambiguous between the following methods //     or properties: 'Test.Generic<A,B>.Method(A, B)' and //     'Test.Generic<A,B>.Method(B, A)'             Console.WriteLine(t2.Method(1, 2));            }     } } 

Answer 1

Thanks for the good answers, they prompted me into this solution:

using System;  namespace Test {     class Generic<A, B>     {         public string Method(A a, B b)         {             return this.DefaultMethod(a, b);         }          protected string DefaultMethod(A a, B b)         {             return a.ToString() + b.ToString();         }          public string Method(B b, A a)         {             return b.ToString() + a.ToString();         }     }      class Generic<A> : Generic<A, A>     {         public new string Method(A a, A b)         {             return base.DefaultMethod(a, b);         }     }      class Program     {         static void Main(string[] args)         {             Generic<int, double> t1 = new Generic<int, double>();             Console.WriteLine(t1.Method(1.23, 1));              Generic<int> t2 = new Generic<int>();             Console.WriteLine(t2.Method(1, 2));         }     } }