通过C ++切换在运行时选择模板实例化 [英] Choosing a template instantiation at runtime though switch in C++

问题描述

我有一组由整数类型 Index 和类类型 T 组成的函数，我以下面的方式部分专门化：

I have a set of functions that are templated both by an integer type Index and a class type T, that I "partially specialize" in the following manner:

// Integer type
enum Index{One,Two,Three,Four};

// Default implementation
template<int I>
struct Foo{
  template<typename T> static void bar(const T& x){ std::cout <<"default" << endl; }
};

// Template specializations
template<> 
struct Foo<One>{
  template<typename T> static void bar(const T& x){ std::cout << "one" << endl; }
};

这是我用来在程序运行时使用切换语句选择一个特定索引（应该会产生一个有效的查询表）。开关独立于 T ：

This I use to select a particular index at the runtime of the program using a switch-statement (which should result in an efficient look-up table). The switch is independent of T:

template<typename T>
void barSwitch(int k, const T& x){
  switch(k){
    case ONE: Foo<ONE>::bar(x); break;
    case TWO: Foo<TWO>::bar(x); break;
    case THREE: Foo<THREE>::bar(x); break;
  }
}

这当然很好，但是该类 Foo 不是我要为其应用切换的唯一类。实际上，我有很多类都由相同的整数类型作为模板。因此，我也想使用功能 Foo对上面的类 barSwitch 进行模板化，以便可以插入其他类或功能。我唯一想到的方法是使用宏：

This works fine, of course, but the class Foo is not the only class for which I would like to apply the switch. In fact, I have a lot of classes that are all templated by the same integer type. So I would like to "template" the class barSwitch above with the function "Foo" as well, so that I can plug in a different class or a different function. The only way I can think of to achieve this is to use a macro:

#define createBarSwitch(f,b) \
template<typename T> \
void barSwitch(int k, const T& x){ \
  switch(k){ \
    case ONE: f<ONE>::b(x); break; \
    case TWO: f<TWO>::b(x); break; \
    case THREE: f<THREE>::b(x); break; \
  }\
}

是否有更好的，更多的C ++

Is there some better, more C++ style way of doing this?

推荐答案

模板模板参数是关键：

enum Index { One, Two, Three, Four };

template <template <Index> class Switcher, typename T>
void barSwitch(int k, const T & x)
{
    switch (k)
    {
        case 1: Switcher<One>::template bar<T>(x); break;
        case 2: Switcher<Two>::template bar<T>(x); break;
        default: assert(false);
    }
}

用法：

template <Index I> struct Foo
{
    template <typename T> static void bar(const T & x);
};

barSwitch<Foo>(1, Blue);

（您有责任确保用每个可能的模板代替当然，Switcher 的成员模板为 bar 。如果没有，则会出现编译错误。）

(It is your responsibility to ensure that every possible template that you substitute for Switcher has a member template bar, of course. If not, you'll get a compile error.)

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