Method invocation will consist of the following steps:

1. Member lookup (14.3) - evaluate method group set (Base.f() and

Derived.f()) .Standart say:

"The compile-time processing of a method invocation of the form M(A), where

M is a method group and A is

an optional argument-list, consists of the following steps:

The set of candidate methods for the method invocation is constructed.

Starting with the set of methods

associated with M, which were found by a previous member lookup (14.3),

...."

2. Remove non-applicable methods and methods of base classes. In this

example both Base.f() and Derived.f() are applicable. Standart say:

"... the set is reduced to those

methods that are applicable with respect to the argument list A. The set

reduction consists of applying

the following rules to each method T.N in the set, where T is the type in

which the method N is declared:

If N is not applicable with respect to A (14.4.2.1), then N is removed from

the set.

If N is applicable with respect to A (14.4.2.1), then all methods declared

in a base type of T are removed

from the set."

But what mean words "then all methods declared in a base type of T are

removed from the set"? If all methods must be removed than test() method in

example must invoke Derived.f(), but MSVC# invoke Base.f(). Why? Any

suggestions.

class Source

{

static public implicit operator Target ( Source source )

{ return new Target(); }

}

class Target

{}

class Base

{

public void f ( Source source )

{}

}

class Derived : Base

{

public void f ( Target target ) // this function is applicable

(user-defined implicit conversion used)

{}

public void test()

{

Source source = new Source();

f ( source ); // invoke Base.f() - why?

}

}

Alex.