Re: Koenig lookup and template arguments

Mark Van Peteghem wrote:
:: Consider the following code:
::
:: namespace A
:: {
:: struct P {};
:: struct Q {};
::
:: template <class T, class U>
:: U operator&(T, U) { return U(); }
:: }
::
:: namespace B
:: {
:: template <class T>
:: struct X {};
:: struct Y {};
::
:: template <class T, class U>
:: U operator&(T, U) { return U(); }
:: }
::
:: B::Y y1 = B::X<int>() & B::Y();
:: B::Y y2 = B::X<A::P>() & B::Y();
::
:: The line that declares y1 compiles fine, as expected. But the line
:: that declares y2 gives an error, saying that operator& is
:: ambiguous, it could be the one in namespace A or the one in
:: namespace B (with MSVC 2005, gcc 3.4.2 Cygwin and Comeau online).
::
:: I would have thought that a B::X object only lives in the
:: namespace B since it doesn't derive from a class in another
:: namespace. But apparently its template argument can make Koenig
:: lookup check the namespace A. Even the following line gives the
:: same error:
::
:: B::Y y3 = B::X<B::X<A::P> >() & B::Y();
::
:: This is counterintuitive to me. In my code, I have a similar
:: situation where the classes in namespace A all have a method
:: void call(), that is called by operator&. Therefore it is pointless
:: to use A::operator& with objects from namespace B.

The template parameters also affect the namespaces of the lookup. The
standard explicitly says so.

I think the problem can be with your definition of operator&. Isn't it
overly broad to define it for just ANY two types T and U? That kind of
"highjacks" the operator for any expression where one of your types is
involved. Having several of these operators doesn't make it any
better.

What about defining it like

template<class T, class U>
U operator&(X<T>, U);

That would probably limit it to the types you actually want it to
operate on.

Bo Persson

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