如何将模板化函数重载转换为部分专用模板化类静态方法? [英] How To Convert Templated Function Overloads to Partial-Specialized Templated Class Static Methods?

问题描述

我有几个要根据类型质量来专门研究的功能，例如字符，有符号整数，无符号整数，浮点数，指针"；使用type_traits似乎是 的方式，并且具有类似于以下内容的代码:

I have several functions that I want to specialize based on type qualities, such as "character, signed-integer, unsigned-integer, floating-point, pointer"; using type_traits seems like the way to do this, and have code similar to the to the following:

#include <tr1/type_traits>
#include <iostream>

template<bool, typename _Tp = void>
struct enable_if 
{ };

template<typename _Tp>
struct enable_if<true, _Tp>
{
    typedef _Tp type;
};


template< typename T >
inline void
foo_impl( typename enable_if< std::tr1::is_integral< T >::value, T >::type const& )
{
    std::cout << "This is the function-overloaded integral implementation.\n";
}

template< typename T >
inline void
foo_impl( typename enable_if< std::tr1::is_floating_point< T >::value, T >::type const& )
{
    std::cout << "This is the function-overloaded floating-point implementation.\n";
}

template< typename T >
inline void
function_overloads_foo( T const& arg )
{
    foo_impl< T >( arg ); // vital to specify the template-type
}

void function_overloads_example()
{
    function_overloads_foo( int() );
    function_overloads_foo( float() );
}

除了我的真实代码外，我还有bar，baz等，以及foo.

except in my real code, I also have bar,baz, etc., along with foo.

但是，我想按质量将所有这些功能分组为一个模板类，作为static方法.如何做到最好?这是我天真，尝试使用Tag，SFINAE和部分专业化的尝试:

However, I would like to group all of these functions per quality into one templated class as static methods. How is this best done? Here's my naive, and broken attempt to use Tags, SFINAE, and partial-specialization:

struct IntegralTypeTag;
struct FloatingPointTypeTag;

template< typename T, typename U = void >
class Foo
{
};

template< typename T >
class Foo< T, typename enable_if< std::tr1::is_integral< T >::value, IntegralTypeTag >::type >
{
    static void foo( T const& )
    {
        std::cout << "This is the integral partial-specialization class implementation.\n";
    }
};

template< typename T >
class Foo< T, typename enable_if< std::tr1::is_floating_point< T >::value, FloatingPointTypeTag >::type >
{
    static void foo( T const& )
    {
        std::cout << "This is the floating-point partial-specialization class implementation.\n";
    }
};

template< typename T >
inline void
partial_specialization_class_foo( T const& arg )
{
    Foo< T >::foo( arg );
}

void partial_specialization_class_example()
{
    partial_specialization_class_foo( int() );
    partial_specialization_class_foo( float() );
}

注意:在我的真实代码中，我将具有bar，baz等，以及foo静态方法.

Note: in my real code, I'd have bar,baz, etc., along with foo static-methods.

仅供参考，这是C ++ 03.

FYI, this is C++03.

顺便说一句，我是否以常规方式进行模板化函数重载?

As an aside, am I doing the templated function overloading in the conventional way?

推荐答案

这是一种方法:

#include <tr1/type_traits>
#include <iostream>

struct IntegralTypeTag;
struct FloatingPointTypeTag;

template <
  typename T,
  bool is_integral = std::tr1::is_integral<T>::value,
  bool is_floating_point = std::tr1::is_floating_point<T>::value
> struct TypeTag;

template <typename T>
struct TypeTag<T,true,false> {
  typedef IntegralTypeTag Type;
};

template <typename T>
struct TypeTag<T,false,true> {
  typedef FloatingPointTypeTag Type;
};

template <typename T,typename TypeTag = typename TypeTag<T>::Type> struct Foo;


template <typename T>
struct Foo<T,IntegralTypeTag> {
  static void foo( T const& )
  {
    std::cout << "This is the integral partial-specialization class implementation.\n";
  }
};

template <typename T>
struct Foo<T,FloatingPointTypeTag> {
  static void foo( T const& )
  {
    std::cout << "This is the floating-point partial-specialization class implementation.\n";
  }
};

template< typename T >
inline void
partial_specialization_class_foo( T const& arg )
{
      Foo< T >::foo( arg );
}

int main(int,char**)
{
  partial_specialization_class_foo(int());
  partial_specialization_class_foo(float());
  return 0;
}

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