> I believe it's intended.

> > Appreciate if any one can shed some lights for me.

> > See following:

> > func.h: In function 'void boo::rfunc(const boo::R<T>&) [with T =
> > hpc::B]':
> > main.cpp:16:   instantiated from here
> > func.h:13: error: call of overloaded 'work(const boo::R<hpc::B>&)' is
> > ambiguous
> > func.h:5: note: candidates are: void boo::work(T) [with T =
> > boo::R<hpc::B>]
> > main.cpp:23: note:                 void hpc::work(T) [with T =
> > boo::R<hpc::B>]

> I think you're running into a case of ADL (Argument-Dependent name
> Lookup).  To figure out which function to call, the compiler is allowed
> to look in the namespaces of arguments (and of template arguments).  In
> your case, since when instantiating 'boo::rfunc<T>', 'R' is from 'boo'
> and 'T' (deduced as 'hpc::B') is from 'hpc' namespace.  Both namespaces
> are considered making 'work' symbol ambiguous.

> > func.h:
> > namespace boo {
> > template <typename T>
> > void work(T n) { std::cout << "good bye work" << std::endl; }
> > template <typename T> class R { };
> > template <typename T>
> > void rfunc(const R<T>& a) { work(a); }
> > } // namespace boo

> > main.cpp:
> > #include <iostream>
> > #include "func.h"

> A side note: here you made inclusion of 'func.h' dependent on the
> previous inclusion of <iostream>.  Better to avoid this.  Consider
> including <iostream> in 'func.h', and still keep it here because
> this module uses 'cout' and 'endl' as well.

> > using std::cout;
> > using std::endl;
> > namespace hpc {
> > class B { };
> > class A {
> > public:
> >        void bfunc(void) { rfunc(n); }
> > protected:
> >        boo::R<B> n;
> > };

> > template <typename T>
> > void work(T n) { cout << "hello world work" << endl; }

> > } // namespace hpc

> > using hpc::A;
> > int main(int argc, char* argv[])
> > {
> >        A a;
> >        a.bfunc();
> >        return 0;
> > }

> --
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