c++ - 自定义 std::ostream 子类中 operator<< 重载的歧义错误

Question

在尝试开发一个继承自 std::stringbuf 的类时和 std::ostream并给它一个自定义 operator<<()重载(应该适用于此类，但一般不适用于 std::ostream)，我遇到了一些我不理解的歧义错误。

问题出在下面代码的下半部分。如果我删除 operator<<()过载，我没有收到任何错误(g++ --std=c++17 -o foo.exe foo.cpp 使用 g++ 9.2.0，MSYS2 在 Windows 上构建)。

#include <iostream>
#include <ostream>
#include <sstream>

class MyStream:
    private std::stringbuf,
    public std::ostream
{
public:
    MyStream() noexcept:
        std::stringbuf(),
        std::ostream(this)
    {}

    ~MyStream()
    {
        if (pptr() - pbase() < 2)
        { sync(); }
    }

protected:
    virtual int sync() override
    {
        // make sure there actually is something to sync
        if (pptr() - pbase() < 2)
        { return 0; }

        const std::string message{str()};
        const std::size_t length{message.length()};

        // update the pointers for the next sync
        setp(pbase() + length, epptr());

        std::cout << message;

        return std::stringbuf::sync();
    }
};

MyStream& operator<<(MyStream& out, bool boolean) noexcept
{ return out << (boolean ? "yes" : "no"); }
/*           ↑
    more than one operator "<<" matches these operands:
        -- function template "std::basic_ostream<_Elem, _Traits>
            &std::operator<<(std::basic_ostream<_Elem, _Traits> &_Ostr, const char *_Val)"
        -- function template "std::basic_ostream<char, _Traits>
            &std::operator<<(std::basic_ostream<char, _Traits> &_Ostr, const char *_Val)"
        -- function template "std::basic_ostream<_Elem, _Traits>
            &std::operator<<(std::basic_ostream<_Elem, _Traits> &_Ostr, const _Elem *_Val)"
        -- function "operator<<(MyStream &out, bool boolean) noexcept"
        -- operand types are: MyStream << const char *
*/

int main()
{
    MyStream stream;

    stream << "Hello World" << std::endl;
    /*     ↑
        more than one operator "<<" matches these operands:
            -- function template "std::basic_ostream<_Elem, _Traits>
                &std::operator<<(std::basic_ostream<_Elem, _Traits> &_Ostr, const char *_Val)"
            -- function template "std::basic_ostream<char, _Traits>
                &std::operator<<(std::basic_ostream<char, _Traits> &_Ostr, const char *_Val)"
            -- function template "std::basic_ostream<_Elem, _Traits>
                &std::operator<<(std::basic_ostream<_Elem, _Traits> &_Ostr, const _Elem *_Val)"
            -- function "operator<<(MyStream &out, bool boolean) noexcept"
            -- operand types are: MyStream << const char [12]
    */
}

为什么我的 bool 过载了？ char* 的有效候选人争论？我怎样才能正确地做到这一点？

Answer 1

This似乎在 linux (g++ (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0) 下对我有用:

MyStream& operator<<(MyStream& out, bool boolean)
{
    out << (boolean ? "yes" : "no");
    return out;
}

我只更改了返回调用。那是因为 out << (boolean ? "yes" : "no")将使用基础 ostream operator<<对于 char*类型，因此返回 std::basic_ostream<char>输入而不是 MyStream返回值类型。

然后在main()使用 bool输入 MyStream operator<<作品:

int main()
{
    MyStream stream;
    bool b=true;

    stream << b << "Hello World" << b << std::endl;  

    //by contrast, this should output "yesHello Worldyes"
    //stream << b << "Hello World";
    //stream << b << std::endl;

    //and so would this kind of ugliness! lol
    //static_cast<MyStream&>(stream << b << "Hello World") << b << std::endl;
}

..但是一旦基类ostream operator<<为 "Hello World" 调用声明的一部分，它当然会返回一个 std::basic_ostream<char>输入 :)

..然后在那之后，因为不存在 ostream operator<<对于 bool类型，bool b类型将隐式提升为 int所以输出是:

yesHello World1 !

---

Windows 构建

但是您似乎遇到了其他错误。因此，虽然我手边没有 MSYS2，但我确实尝试使用 Visual Studio 在 WinOS 上编译代码。就我而言，至少，你的 main()由于 VS STL 实现引起的歧义，立即失败:

int main()
{
    MyStream stream;
    stream << "Hello World" << std::endl;
}

无需怀疑您当前的方法，因为评论中已经出现了这种情况。但只是为了回答您的问题；那么一种方法是实现另一个 operator<<对于 MyStream处理 char*类型:

#include <iostream>
#include <ostream>
#include <sstream>

class MyStream:
    private std::stringbuf,
    public std::ostream
{
public:
    MyStream() noexcept:
        std::stringbuf(),
        std::ostream(this)
    {}

    ~MyStream()
    {
        if (pptr() - pbase() < 2)
        { sync(); }
    }

protected:
    virtual int sync() override
    {
        // make sure there actually is something to sync
        if (pptr() - pbase() < 2)
        { return 0; }

        const std::string message{str()};
        const std::size_t length{message.length()};

        // update the pointers for the next sync
        setp(pbase() + length, epptr());

        std::cout << message;

        return std::stringbuf::sync();
    }
};

MyStream& operator<<(MyStream& out, const char* str)
{
    static_cast<std::ostream&>(out) << str;
    return out;
}

MyStream& operator<<(MyStream& out, bool boolean) noexcept
{ return out << (boolean ? "yes" : "no"); }


int main()
{
    MyStream stream;
    bool b=1;      
    stream << b << " oook " << b << std::endl;    
}

或者，由于 MyStream实际上是源自ostream然后我们可以使用任何 operator<<对于 ostream带有显式强制转换的类型。例如:

#include <iostream>
#include <ostream>
#include <sstream>

class MyStream:
    private std::stringbuf,
    public std::ostream
{
public:
    MyStream() noexcept:
        std::stringbuf(),
        std::ostream(this)
    {}

    ~MyStream()
    {
        if (pptr() - pbase() < 2)
        { sync(); }
    }

protected:
    virtual int sync() override
    {
        // make sure there actually is something to sync
        if (pptr() - pbase() < 2)
        { return 0; }

        const std::string message{str()};
        const std::size_t length{message.length()};

        // update the pointers for the next sync
        setp(pbase() + length, epptr());

        std::cout << message;

        return std::stringbuf::sync();
    }
};

std::ostream& operator<<(MyStream& out, bool boolean) noexcept
{
    return static_cast<std::ostream&>(out) << (boolean ? "yes" : "no");
}

int main()
{
    MyStream stream;
    bool b=1;

    stream << b << " ooOok ";
    stream << b << std::endl;
}

希望输出的位置:

yes ooOok yes