我可以为（非成员）函数使用部分模板专门化吗？

问题描述

我试图对（非成员）函数使用部分模板专门化，我跳过了语法。我已经搜索StackOverflow的其他部分模板专门化问题，但那些处理类或成员函数模板的部分专门化。

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.

对于一个起点，我有： / p>

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);

我可以做一个完整的ConvertPixel专业化，使RGBA到灰度转换功能像这样： p>

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));
}

我可以想象有更多的像素类型提供红，绿，蓝，但也许以不同的格式，所以我真正想做的是通过指定 Grayscale OutType 并仍然允许各种 InType 。我尝试了这样的各种方法：

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));
}

但是Microsoft VS 2008 C ++编译器拒绝它。

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

我试图可能吗？

推荐答案

可以使用类别partitial specialization：

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);
}

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