函数的转储成员函数

函数的转储成员函数

本文介绍了操作符<<函数的转储成员函数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

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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08-16 01:23