views:

54

answers:

2

The following code doesn't compile (error CS0123: No overload for 'System.Convert.ToString(object)' matches delegate 'System.Converter<T,string>'):

class A<T> {
    void Method(T obj) {
        Converter<T, string> toString = Convert.ToString;

        // this doesn't work either (on .NET 4):
        Converter<object, string> toString2 = Convert.ToString;
        Converter<T, string> toString3 = toString2;            
    }
}

however, this does:

class A<T> {
    void Method(T obj) {
        // o is a T, and Convert.ToString(o) is using
        // string Convert.ToString(object o)

        Converter<T, string> toString = o => Convert.ToString(o);
    }
}

In c# 4, co/contra-variant delegates can be assigned to each other, and delegates can be created from co/contra-variant methods, so the ToString(object) method can be used as a Converter<T, string>, as T is always guarenteed to be convertable to an object.

So, the first example (method group overload resolution) should be finding the only applicable method string Convert.ToString(object o), the same as the method call overload resolution. Why is the method group & method call overload resolution producing different results?

A: 

The second code compiles because o derives from object, so obviously you can invoke a method that takes an object as input with any type of parameter.

The delegate types, however, are not equal. Unless T is object the method signatures don't match. If, say, T is int you would have a Converter<int, string> which is not the same as Converter<object, string. They are two entirely different types.

You are hitting the issues around C# 3.0's lack of co/contra-variance. Should be better in C# 4.

Mark Seemann
This is in .NET 4. I've updated my question
thecoop
+2  A: 

This has to do with the fact that variance is not applicable to value types, so if you restrict T like where T : class you get variance on T and the first snippet of code will compile.

From the Covariance and Contravariance FAQ:

Variance is supported only if a type parameter is a reference type. Variance is not supported for value types.

João Angelo
More specifically, it's doesn't work for an unconstrained `T` because it _might be_ a value type.
thecoop