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, …

可能重复:
通过隐式转换为字符串来流对象时,重载决策失败

我知道这样做并不是一个好主意,但我真的想知道下面的代码无法编译的原因(即为什么"没有可接受的转换"):

#include <iostream>
#include <string>


class Test
{
public:
    operator std::string () const;
};

Test::operator std::string () const
{
    return std::string("Test!");
}

int main ()
{
    std::string str = "Blah!";
    std::cout << str << std::endl;

    Test test;

    str = test;//implicitly calls operator std::string without complaining

    std::cout << str << std::endl;

    std::cout << test;//refuses to implicitly cast test to std::string

    return 0;
}

在Visual Studio 2010上,我收到此错误:" error C2679: binary '<<' : no operator found which takes a right-hand operand …