Partial specialisation of member function with non-type parameter

c++, partial-specialization, template-specialization, templates

Solution

You can wrap the function inside a class.

Only classes, not functions, may be partially specialized.

template<typename T, int R>
struct foo
{
    foo(const T& v) :
        value_(v)
    {}

    void bar()
    {
        return bar_impl< T, R >::bar( * this );
    }

    friend struct bar_impl< T, R >;

    T value_;
};

template< typename T, int R >
struct bar_impl {
    static void bar( foo< T, R > &t ) {
        std::cout << "Generic" << std::endl;
        for (int i = 0; i < R; ++i)
            std::cout << t.value_ << std::endl;
    }
};

template<>
struct bar_impl<float, 3> {
static void bar( foo< float, 3 > &t ) {
    std::cout << "Float" << std::endl;
    for (int i = 0; i < 3; ++i)
        std::cout << t.value_ << std::endl;
}
};

template<int R>
struct bar_impl<double, R> {
static void bar( foo< double, R > &t ) {
    std::cout << "Double" << std::endl;
    for (int i = 0; i < R; ++i)
        std::cout << t.value_ << std::endl;
}
};

Problem

I have a template class with both a type and a non-type template parameter. I want to specialize a member function, what I finding is, as in the example below, I can do a full specialization fine. ``` template<typename T, int R> struct foo { foo(const T& v) : value_(v) {} void bar() { std::cout << "Generic" << std::endl; for (int i = 0; i < R; ++i) std::cout << value_ << std::endl; } T value_; }; template<> void foo<float, 3>::bar() { std::cout << "Float" << std::endl; for (int i = 0; i < 3; ++i) std::cout << value_ << std::endl; } ``` However this partial specialization won't compile. ``` template<int R> void foo<double, R>::bar() { std::cout << "Double" << std::endl; for (int i = 0; i < R; ++i) std::cout << value_ << std::endl; } ``` Is there a way to achieve what I'm attempting would anyone know? I tried this in MSVC 2010.

Original source