1

I have a class template Z that I would like specialize when passed a type that is an any instantiation of a particular template N:

struct L {
    template <typename S> void foo(S &) {/*...*/}
};

template <class T>
struct M {
    template <typename S> void foo(S &) {/*...*/}
};

template <class T>
struct N {
    template <typename S> void foo(S &) {/*...*/}
};

// I'd like to specialize this for TY==N<anything>
template <typename TX, typename TY>
struct Z {
    void bar(TX &tx) { /*...*/ ty->foo(tx); /*...*/ }
    TY *ty;
};

Since Z<int, L> and Z<int, N<int>> and Z<int, M<int>> are all valid use cases, I can't do anything along the lines of turning Z into a template template, and there is a compelling complexity reduction possible in Z<TX, TY>::bar(TX &) when TY is a class built from N. Is there a way to achieve this?

Jeff
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1 Answers1

1

This should effect the specialization you desire:

template <typename TX, typename ANY>
struct Z< TX, N<ANY> > {
    // ...
};

Z gets specialized when the first parameter is TX, and the second parameter is N<ANY>. A quick illustration:

template <typename A> struct N { A a; };

template <typename TX, typename TY>
struct Z { Z () { std::cout << "not special" << std::endl; } };

template <typename TX, typename ANY>
struct Z< TX, N<ANY> > { Z () { std::cout << "special" << std::endl; } };

int main ()
{
    Z<int, int> z1;
    Z<int, N<int> > z2;
}

Results in the output:

not special
special
jxh
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  • Thanks, this is exactly what I was looking for. I had fiddled with the signature but was unable to deduce the right one. I think I actually get the parameter syntax of templates in general a bit better now. – Jeff Jun 14 '12 at 00:12
  • Glad to be of help. +1 on your question from me. – jxh Jun 14 '12 at 06:33