c++ - 元编程，试图避免许多特化

Question

我正在为旧程序添加新内容而工作，旧程序到处都使用元编程。我仍在使用 C++ 03 和 boost。所以这是一个问题:我制作了模板函数，但我不想对其进行专门研究，因为获取特定值的四个函数调用的唯一区别是:

template < typename Message >
void function(const Message & message)
{
    ....
    int value = getHelper...getValue();
    ....
}

有许多不同的消息类型:

1. MessageA: public BaseForA< MessageA >
2. MessageB: public BaseForB< MessageB >
3. template < typename Appendage >
MessageAWithAppendage < Appendage >: public BaseForA< MessageA < Appendage > >
4. template < typename Appendage >
MessageB: public BaseForB< MessageB >: public BaseForB< MessageB < Appendage > >

还有两种附肢类型:
SmallAppendage
BigAppendage

每条消息的header中都有条件变量，依赖于它 getValue()应该从消息中获取字段或返回零。如果类型没有附加，则该字段可以在消息本身中，或者在附件中，或者如果消息带有附件，则同时在消息本身。

我需要像 Base class 这样的东西来处理没有附件的消息，而 Extended 来处理有附件的消息像这样的东西:

template < typename Message >
 class Helper
 {
 public:
    virtual int getValue(const Message & msg)
    {
        if(..)
        {
            return msg.value;
        }
        ...
    }
};

template< template < class > class Message, typename Appendage >
class ExtendedHelper : public Helper < Message < Appendage > >
{
public:
    virtual int getValue(const Message<Appendage> & msg)
    {
        int value = Helper::getValue(msg);
        if(value)
        {
            return value;
        }
        return msg.appendage.getValue();
    }
};

在那之后，我认为这样的事情会奏效，但事实并非如此:

template < class Type >
struct AppendageTraits
{
    enum { appendageIncluded = false };
};

template < class Appendage >
struct AppendageTraits < MessageAWithAppendage < Appendage > >
{
    enum { appendageIncluded = true };
};

template < class Appendage >
struct AppendageTraits < MessageBWithAppendage < Appendage  > >
{
    enum { appendageIncluded = true };
};

template< typename Message , bool >
struct GetHelper
{
    Helper< Message > * operator()( )
    {
       static Helper< Message > helper;
       return &helper;
    }
};

编辑:我的特征现在可以编译了。是否有可能使这个工作:

template < typename Appendage >
struct GetHelper<MessageAWithAppendage <Appendage>, true>
{
    Helper< MessageAWithAppendage <Appendage> > * operator()( )
    {
        static Helper< MessageAWithAppendage <Appendage>, Appendage > helper;
        return &helper;
    }
};

template < typename Appendage >
struct GetHelper<MessageBWithAppendage <Appendage>, true>
{
    Helper< MessageBWithAppendage <Appendage> > * operator()( )
    {
        static ExtendedHelper< MessageBWithAppendage <Appendage>, Appendage > helper;
        return &helper;
    }
};

编辑:现在参数 1 的类型/值不匹配

static ExtendedHelper< MessageAWithAppendage <Appendage>, Appendage > helper;

期望类模板得到..类模板!，但不知何故未被识别。

编辑:我解决了这个错误，这是因为:

Like normal (non-template) classes, class templates have an injected-class-name (clause 9). The injected-class-name can be used with or without a template-argument-list. When it is used without a template- argument-list, it is equivalent to the injected-class-name followed by the template-parameters of the class template enclosed in <>. When it is used with a template-argument-list, it refers to the specified class template specialization, which could be the current specialization or another specialization.

正确的代码:

template < typename Appendage >
struct GetHelper<MessageAWithAppendage <Appendage>, true>
{
    Helper< MessageAWithAppendage <Appendage> > * operator()( )
    {
        static Helper< MessageAWithAppendage, Appendage > helper;
        return &helper;
    }
};

template < typename Appendage >
struct GetHelper<MessageBWithAppendage <Appendage>, true>
{
    Helper< MessageBWithAppendage <Appendage> > * operator()( )
    {
        static ExtendedHelper< MessageBWithAppendage, Appendage > helper;
        return &helper;
    }
};

Answer 1

读完所有内容后我的头很痛，但如果您只想专注于一个操作，为什么不制作一个具有(可能是模板化的)重载(本质上是静态访问者)的仿函数:

struct ValueGetter
{
  int operator()(const MessageA& ma) const {
    return ma.whatever_we_need_to_do();
  } 
  int operator()(const MessageB& mb) const {
    return mb.whatever_we_need_to_do();
  } 
};

// this is now completely generic
template < typename Message >
void function(const Message & message)
{
    ....
    ValueGetter vg;
    int value = vg(message);
    ....
}