使用 SFINAE 检查全局操作符<<?

2021-12-13 00:00:00 templates c++ sfinae

我想要几个重载的全局 to_string() 函数，它们采用某种类型的 T 并将其转换为它的字符串表示形式.对于一般情况，我希望能够写:

I want to have several overloaded, global to_string() functions that take some type T and convert it to its string representation. For the general case, I want to be able to write:

template<typename T,class OutputStringType> inline
typename enable_if<!std::is_pointer<T>::value
                && has_insertion_operator<T>::value,
                   void>::type
to_string( T const &t, OutputStringType *out ) {
  std::ostringstream o;
  o << t;
  *out = o.str();
}

到目前为止，我对 has_insertion_operator 的实现是:

My implementation of has_insertion_operator so far is:

struct sfinae_base {
  typedef char yes[1];
  typedef char no[2];
};

template<typename T>
struct has_insertion_operator : sfinae_base {
  template<typename U> static yes& test( U& );
  template<typename U> static no& test(...);

  static std::ostream &s;
  static T const &t;

  static bool const value = sizeof( test( s << t ) ) == sizeof( yes ); // line 48
};

(它借鉴了this和这个.)这似乎有效.但是现在我想要一个 to_string 的重载版本，用于 not 有 operator<<< 但do 的类型> 有自己的to_string() member 函数，即:

(It borrows from this and this.) That seems to work. But now I want to have an overloaded version of to_string for types that do not have operator<< but do have their own to_string() member function, i.e.:

template<class T,class OutputStringType> inline
typename enable_if<!has_insertion_operator<T>::value
                && has_to_string<T,std::string (T::*)() const>::value,
                   void>::type
to_string( T const &t, OutputStringType *out ) {
  *out = t.to_string();
}

has_to_string 的实现是:

#define DECL_HAS_MEM_FN(FN_NAME)                                      
  template<typename T,typename S>                                     
  struct has_##FN_NAME : sfinae_base {                                
    template<typename SignatureType,SignatureType> struct type_check; 
    template<class U> static yes& test(type_check<S,&U::FN_NAME>*);   
    template<class U> static no& test(...);                           
    static bool const value = sizeof( test<T>(0) ) == sizeof( yes );  
  }

DECL_HAS_MEM_FN( to_string );

(这部分似乎工作正常.它改编自 this.)但是，当我有:

(This part seems to work fine. It's adapted from this.) However, when I have:

struct S {
  string to_string() const {
    return "42";
  }
};

int main() {
  string buf;
  S s;
  to_string( s, &buf ); // line 104
}

我明白了:

foo.cpp: In instantiation of ‘const bool has_insertion_operator<S>::value’:
foo.cpp:104:   instantiated from here
foo.cpp:48: error: no match for ‘operator<<’ in ‘has_insertion_operator<S>::s << has_insertion_operator<S>::t’

似乎 SFINAE 没有发生.如何正确编写 has_insertion_operator 以便确定全局 operator<< 是否可用?

It seems like SFINAE is not happening. How do I write has_insertion_operator correctly such that it determines whether a global operator<< is available?

仅供参考:我使用的是 g++ 4.2.1(在 Mac OS X 上作为 Xcode 的一部分提供).另外，我希望代码只是标准的 C++03，没有 3rd 方库，例如 Boost.

FYI: I'm using g++ 4.2.1 (that which ships as part of Xcode on Mac OS X). Also, I'd like the code to be only standard C++03 without 3rd-party libraries, e.g., Boost.

谢谢！

推荐答案

我应该更忠实于这个的答案.一个有效的实现是:

I should have simply been more faithful to this answer. A working implementation is:

namespace has_insertion_operator_impl {
  typedef char no;
  typedef char yes[2];

  struct any_t {
    template<typename T> any_t( T const& );
  };

  no operator<<( std::ostream const&, any_t const& );

  yes& test( std::ostream& );
  no test( no );

  template<typename T>
  struct has_insertion_operator {
    static std::ostream &s;
    static T const &t;
    static bool const value = sizeof( test(s << t) ) == sizeof( yes );
  };
}

template<typename T>
struct has_insertion_operator :
  has_insertion_operator_impl::has_insertion_operator<T> {
};

我相信它不实际上依赖于 SFINAE.

I believe that it does not actually rely on SFINAE.

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