Problem with functions accepting inner classes of template classes

Question

I have a problem with inner classes in class templates. I have a template class (say: Matrix), and a subclass (say: Matrix::Row).

Answer 1

It would need to deduce the type T for the call to template function x, and template argument deduction is only allowed in a specific set of circumstances:

http://publib.boulder.ibm.com/infocenter/compbgpl/v9v111/index.jsp?topic=/com.ibm.xlcpp9.bg.doc/language_ref/template_argument_deduction.htm

A<T>::B

does not seem to be one of them :/

Answer 2

How should the compiler be able to deduce this? Imagine the following setup:

struct A { typedef int T; };
struct B { typedef int T; };

template <typename S> void foo(typename S::T);

Now when you say int x; foo(x);, there's no way to match this unambiguously.

The point is that you are not deducing a template parameter from a given class template, but rather just an arbitrary, free-standing type. The fact that that type was defined inside another class is not relevant for that.

Answer 3

That is non-deducible context. That is why the template argument cannot be deduced by the compiler.

Just imagine, you might have specialized A as follows:

template <>
struct A<SomeType>
{
    typedef std::map <double, double> B;
};

Now this specialization has a nested type called B which is a typedef of std::map<double,double>.

So how would the compiler deduce the type SomeType, given that A<SomeType>::B is std::map<double, double>?

And in fact, there can be many such specializations, as such:

template <>
struct A<SomeOtherType>
{
    typedef std::map <double, double> B;
};

Even this specialization has B as nested type.

Now if I say A<T>::B is std::map<double,double>, then can you say what T is? Is it SomeType? or SomeOtherType?