相关疑难解决方法(0)

免责声明:我知道应该避免隐式转换为字符串,并且正确的方法是op<<过载Person.

请考虑以下代码:

#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;
}

它在GCC 4.3.4上产生以下结果:

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, …

这个工作,打印1:

#include <iostream>

struct Int {
    int i;
    operator int() const noexcept {return i;}
};

int main() {
    Int i;
    i.i = 1;
    std::cout << i;
}

但是,这无法在GCC 4.8.1上编译:

#include <iostream>
#include <string>

struct String {
    std::string s;
    operator std::string() const {return s;}
};

int main() {
    String s;
    s.s = "hi";
    std::cout << s;
}

以下是错误的相关部分:

错误:'operator <<'不匹配(操作数类型是'std :: ostream {aka std :: basic_ostream}'和'String')
std :: cout << s;

SNIP

template std :: basic_ostream <_CharT,_Traits>&std :: …