Re: Expected Behaviour?
On Mar 21, 12:51 pm, Jiang <goo.mai...@yahoo.com> wrote:
On Mar 19, 12:48 am, rkld...@gmail.com wrote:
I would like to know the expected behavior of the code below?
struct A{
void fn(){cout << "1";}
};
template<class T>
struct B{
void fn(){cout << "2";}
};
template<class T>
struct C : A, B<T>{
C(){ fn(); } // fn is a non dependent name.
};
int main(){
C<int> c;
}
'fn()' is a non dependent name and hence will be bound at the time of
parsing the template. Since dependent bases are not searched, it will
be bound to A::fn. That means that when the template is instantiated,
the call 'fn()' has already been bound to A::fn.
Sorry I don't see why A::fn should be selected here.
According to ?14.6.2/3, the compiler will not find B<T>::fn() when
looking up the call to "fn":
"In the definition of a class template or a member of a class
template, if a base class of the class template depends on a template-
parameter, the base class scope is not examined during unqualified
name lookup either at the point of definition of the class template or
member or during an instantiation of the class template or member."
And in fact, the example program will not compile if A::fn() is
commented out, gcc reports: "there are no arguments to 'fn' that
depend on a template parameter, so a declaration of 'fn' must be
available".
Now, clearly the compiler must first find B<T>::fn(), before it can
diagnose the call to fn() as ambiguous.So at what point does the
compiler find B<T>::fn (if ever)? I for one cannott find a clear
answer to that question in the current Standard.
Greg
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