隐式转换为std :: string [英] Implicit conversion to std::string

问题描述

可能重复:
通过隐式转换流对象时，重载分辨率失败字符串

Possible Duplicate:
Overload resolution failure when streaming object via implicit conversion to string

我知道这样做不是一个好主意，但是我真的很想知道以下代码为何无法编译的原因(即为什么没有可接受的转换"):

I know it's not such a good idea to do this, but I really want to know the reason why the code below does not compile (i.e. why there is "no acceptable conversion"):

#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 of type 'Test' (or there is no acceptable conversion)"

On Visual Studio 2010 I get this error: "error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'Test' (or there is no acceptable conversion)"

<<运算符是否可以将std::string隐式转换为其他内容以便使用?如果是，我需要在类中重载什么运算符才能使此功能正常工作?我拒绝相信我实际上需要使用operator char *.

Does the << operator implicitly cast std::string to something else in order to make use of it? If yes, what operator do I need to overload in my class to make such a thing work? I refuse to believe that I would actually need to use operator char *.

推荐答案

operator<<(std::basic_ostream&, std::basic_string)是函数模板，在模板自变量推导过程中不考虑用户定义的转换.您需要在课程中重载operator<<.

operator<<(std::basic_ostream&, std::basic_string) is a function template and user defined conversions are not considered during template argument deduction. You need to overload operator<< for your class.

当然，另一个选择是强制转换

Another option, of course, is a cast

std::cout << static_cast<std::string>(test);

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