Question

I’m trying to use partial template specialization on a (non-member) function, and I’m tripping up on the syntax. I’ve searched StackOverflow for other partial template specialization questions, but those deal with partial specialization of a class or member function template.

For a starting point, I have:

struct RGBA {
    RGBA(uint8 red, uint8 green, uint8 blue, uint8 alpha = 255) :
        r(red), g(green), b(blue), a(alpha)
    {}

    uint8 r, g, b, a;
};

struct Grayscale {
    Grayscale(uint8 intensity) : value(intensity) {}

    uint8 value;
};

inline uint8 IntensityFromRGB(uint8 r, uint8 g, uint8 b) {
    return static_cast<uint8>(0.30*r + 0.59*g + 0.11*b);
}

// Generic pixel conversion.  Must specialize this template for specific
// conversions.
template <typename InType, typename OutType>
OutType ConvertPixel(InType source);

I can do a complete specialization of ConvertPixel to make an RGBA to Grayscale conversion function like this:

template <>
Grayscale ConvertPixel<RGBA, Grayscale>(RGBA source) {
    return Grayscale(IntensityFromRGB(source.r, source.g, source.b));
}

I’ll conceivably have more pixel types that offer red, green, and blue, but perhaps in a different format, so what I’d really like to do is a partial specialization by specifying Grayscale for OutType and still allow for a variety of InTypes. I’ve tried a variety of approaches like this:

template <typename InType>
Grayscale ConvertPixel<InType, Grayscale>(InType source) {
    return Grayscale(IntensityFromRGB(source.r, source.g, source.b));
}

But the (Microsoft VS 2008 C++) compiler rejects it.

Is what I’m attempting possible? If so, what’s the right syntax?

Answer 1

It is possible using class partitial specialization:

template<class A, class B>
struct Functor {
    static A convert(B source);
};

template<class B>
struct Functor<GrayScale, B> {
    static GrayScale convert(B source) {
         return Grayscale(IntensityFromRGB(source.r, source.g, source.b));
    }
};

// Common function
template<class A, class B>
A Convert(B source) {
   return typename Functor<A,B>::convert(source);
}