字符串隐式转换为对象时,重载解析失败怎么办?

内容来源于 Stack Overflow,并遵循CC BY-SA 3.0许可协议进行翻译与使用

  • 回答 (2)
  • 关注 (0)
  • 查看 (47)

下面的代码:

#include <string>
#include <ostream>
#include <iostream>

struct NameType {
   operator std::string() { return "wobble"; }
};

struct Person {
   NameType name;
};

int main() {
   std::cout << std::string("bobble");
   std::cout << "wibble";

   Person p;
   std::cout << p.name;
}

在GCC 4.3.4上输出以下结果

prog.cpp: In function ‘int main()’:
prog.cpp:18: error: no match for ‘operator<<’ in ‘std::cout << p.Person::name’
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:112: note: candidates are: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>& (*)(std::basic_ostream<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:121: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ios<_CharT, _Traits>& (*)(std::basic_ios<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:131: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::ios_base& (*)(std::ios_base&)) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:169: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:173: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long unsigned int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:177: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(bool) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/bits/ostream.tcc:97: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(short int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:184: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(short unsigned int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/bits/ostream.tcc:111: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:195: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(unsigned int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:204: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long long int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:208: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long long unsigned int) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:213: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(double) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:217: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(float) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:225: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long double) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:229: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(const void*) [with _CharT = char, _Traits = std::char_traits<char>]
/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/bits/ostream.tcc:125: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_streambuf<_CharT, _Traits>*) [with _CharT = char, _Traits = std::char_traits<char>]

op<<(ostream&, string const&)如何进入超载设置?

提问于
用户回答回答于

C ++ 98中的14.8.1 / 4

如果参数类型不包含参与模板参数推导的模板参数,则将对函数参数执行隐式转换(第4节),以将其转换为相应函数参数的类型。

在这里你想要一个实例化

template <class charT, class traits, class Allocator>
  basic_ostream<charT, traits>&
    operator<<(basic_ostream<charT, traits>&,
               const basic_string<charT, traits, Allocator>&);

在不明确提供任何模板参数的情况下被推导出来。因此,所有的参数包含一个模板参数是参与了模板参数推导,因此没有人可以从隐式转换得到其数值。

用户回答回答于

这是因为它是一个模板。

为此,需要首先实例化模板,然后使用转换操作符。这是错误的顺序,所以它不起作用。

如果在程序的早期使用了特定的操作符,则无关紧要。

但如果我们添加一个非模板运算符?

#include <string> 
#include <ostream> 
#include <iostream>  

struct NameType {
   operator std::string() { return "wobble"; } 
};  

struct Person {
    NameType name;
};  

void operator<<(std::ostream& os, const std::string& s)   // ** added **
{ std::operator<<(os, s); }

int main() 
{    
    std::cout << std::string("bobble");
    std::cout << "wibble";

     Person p;
     std::cout << p.name; 
}  

现在它工作,并输出

 bobblewibblewobble

所属标签

可能回答问题的人

  • HKC

    红客学院 · 创始人 (已认证)

    26 粉丝7 提问5 回答
  • Dingda

    Dingda · 站长 (已认证)

    4 粉丝0 提问3 回答
  • 西风

    renzha.net · 站长 (已认证)

    9 粉丝1 提问3 回答
  • 螃蟹居

    1 粉丝0 提问2 回答

扫码关注云+社区

领取腾讯云代金券