操作符<<函数的转储成员函数 [英] Generic implementation of operator<< function in terms of the dump member function

问题描述

我的所有类都实现了一个 dump 成员函数，例如：

  struct A {
 template< typename charT> 
 std :: basic_ostream< charT> & 
 dump（std :: basic_ostream< charT>& o）const {
 return（o<< x）; 
} 
 int x = 5; 
}; 
  

我想实现运算符<< 函数一次：

  template< typename charT，typename T> 
 std :: basic_ostream< charT> & 
 operator<< （std :: basic_ostream< charT>& o，const T& t）{
 return t.dump（o）; 
} 
  

问题是所有类型都被此模板捕获，包括标准类型。有没有办法解决这个问题？

解决方案

  typename charT> 
 auto operator<< （std :: basic_ostream< charT>& str，const T& t） - > decltype（t.dump（str））
 {
 static_assert（std :: is_same 
< decltype（t.dump（str）），
 std :: basic_ostream< charT> ;&> :: value，
.dump（ostream&）does not return ostream&！）; 
 
 return t.dump（str）; 
} 
  

这会重载运算符<< 仅适用于定义适当的转储成员的类型。

编辑：添加static_assert以获得更好的消息。 / p>

All my classes implement a dump member function, e.g.:

struct A {
    template <typename charT>
    std::basic_ostream<charT> &
    dump(std::basic_ostream<charT> &o) const {
        return (o << x);
    }
    int x = 5;
};

I would like to implement an operator<< function once for all such classes:

template<typename charT, typename T>
std::basic_ostream<charT> &
operator<< (std::basic_ostream<charT> &o, const T &t) {
    return t.dump(o);
}

The problem is that all types are caught by this template, including the standard types. Is there a way to get around this problem?

解决方案

template <typename T, typename charT>
auto operator<< (std::basic_ostream<charT> & str, const T & t) -> decltype(t.dump(str))
{
    static_assert(std::is_same
                   <decltype(t.dump(str)), 
                    std::basic_ostream<charT> &>::value, 
                  ".dump(ostream&) does not return ostream& !");

    return t.dump(str);
}

This overloads operator<< only for types that define an appropriate dump member.

Edit: added static_assert for better messages.

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