Question

In C++, I am trying to specialize a templated function for a object that is templated itself.

Here is a basic example:
test.h:

template <class T>
class myC {
    T x;
};

template <class U>
void f(U y) {
}

template <>
template <class T>
void f<myC<T> >(myC<T> y) {
}

test.cpp

#include "test.h"
int main() {
    myC<double> m;
    f(m);
}

GCC 4.6.1 gives me the following error message:

In file included from test.cpp:1:0:
test.h:13:25: error: too many template parameter lists in declaration of ‘void f(myC<T>)’
test.h:13:6: error: template-id ‘f<myC<T> >’ for ‘void f(myC<T>)’ does not match any template declaration
test.h:13:25: note: saw 2 ‘template<>’, need 1 for specializing a member function template

Is this at all possible? Or is there another way to accomplish the same goal?

Answer 1

You cannot specialize a template function; only template classes can be specialized. Edit: Nawaz’s answer is correct: it’s partial specialization that is not allowed for template functions, only for classes. A full specialization is possible:

template <class U> void f(U y) {}
template<> void f<double>(double y) {} // specialization for double

Note that the template argument need not be explicitly specified if it can be deduced from the context:

template<> void f<>(int y) {} // specialization for int

In your case, full specialization is not possible because the function argument is a template class. However, a template function, like any function, can be overloaded. In your case, it will be like this:

template <class T>
class myC {
    T x;
};

template <class U>
void f(U y) {
}

template <class T>
void f(myC<T> y) {
}

int main() {
    myC<double> m;
    f(m);
    return 0;
}