Better C++ syntax for template base class typedefs and functions?

Question

I have code that compiles fine with VC9 (Microsoft Visual C++ 2008 SP1) but not with GCC 4.2 (on Mac, if that matters). If I pile on enough qualifiers and keywords I can force it to work in GCC but this doesn't seem right.

Here's a minimal code sample exhibiting my problems:

template< typename N >
struct B {
    typedef N n_type;                     // can derived class access typedef?
    void foo() {}                         // can derived class access function?
};

template< typename N >
struct D : public B<N> {

    typedef B<N> b_type;
    typedef typename b_type::n_type bn_type;

    void f1( n_type ) {}                  // ERROR: 'n_type' has not been
                                          // declared

    void f2( typename B<N>::n_type ) {}   // OK, verbose

    void f3( b_type::n_type ) {}          // ERROR: 'struct B<N>::n_type' is 
                                          // not a type

    void f4( typename b_type::n_type ) {} // OK, verbose

    void f5( bn_type ) {}                 // OK, verbose typedefs

    void f6() { foo(); }                  // ERROR: there are no arguments to
                                          // 'foo' that depend on a template
                                          // parameter, so a declaration of
                                          // 'foo' must be available

    void f7() { b_type::foo(); }          // OK, verbose

};

Am I wrong to expect a template class derived from another template class to be able to use inherited typedefs and functions directly? Is there a better way to do this than what I've come up with so far?

Answer 1

Am I wrong to expect a template class derived from another template class to be able to use inherited typedefs and functions directly?

Yes, this will not generally work as you expect it. The C++ name lookup rules specify that a name is only searched in a templated base classes if it depends on a template parameter (if it is a "dependent name"). If a name does not depend on a template parameter it isn't searched there. (Also see this C++ FAQ Lite entry)

To call functions from a dependent base class the most easy way is to use this->, since this is always implicitly a dependent name:

void f6() { this->foo(); }