Calling constructor overload when both overload have same signature

Question

Consider the following class,

class Foo
{
    public Foo(int count)
    {
        /* .. */
    }

    public Foo(int count)
    {
        /* .. */
    }
}
         


        

Answer 1

There is no ambiguity, because the compiler will choose the most specific overload of Foo(...) that matches. Since a method with a generic type parameter is considered less specific than a corresponding non-generic method, Foo(T) is therefore less specific than Foo(int) when T == int. Accordingly, you are invoking the Foo(int) overload.

Your first case (with two Foo(int) definitions) is an error because the compiler will allow only one definition of a method with precisely the same signature, and you have two.