繼承ostringstream時的歧義

我有一個簡單的結構，它繼承自std::ostringstream，以便更好地處理資料庫的一些值。如果我只是繼承并添加一個簡單的建構式來設定雙精度的寬度，它就可以正常作業。

struct myostream : std::ostringstream {
    myostream() noexcept( false ) {
        this->precision( 6 );
        this->setf( ios_base::fixed, ios_base::floatfield );
    }
};

現在，我正在嘗試添加一個operator<<for type double，以處理無窮大和 NaN。

myostream &operator<<( double val ) {
    if ( std::isnan( val ) ) {
        *static_cast<std::ostringstream *>( this ) << "'NaN'";
    } else if ( std::isinf( val ) ) {
        *static_cast<std::ostringstream *>( this ) << ( val < 0 ? "'-Infinity'" : "'Infinity'" );
    } else {
        *static_cast<std::ostringstream *>( this ) << val;
    }
    return *this;
}

當我在添加此功能，我只是得到模棱兩可的函式呼叫無處不在，每一個型別（bool，char，char *，int，等）。

2>...\include\myostream.hpp(13,14): message : could be 'myostream &myostream::operator <<(double)' (compiling source file ...)
2>C:\Program Files\Microsoft Visual Studio\2022\Preview\VC\Tools\MSVC\14.30.30704\include\ostream(694,32): message : or       'std::basic_ostream<char,std::char_traits<char>> &std::operator <<<char,std::char_traits<char>>(std::basic_ostream<char,std::char_traits<char>> &,char)' (compiling source file ...)
2>C:\Program Files\Microsoft Visual Studio\2022\Preview\VC\Tools\MSVC\14.30.30704\include\ostream(775,31): message : or       'std::basic_ostream<char,std::char_traits<char>> &std::operator <<<std::char_traits<char>>(std::basic_ostream<char,std::char_traits<char>> &,char)' (compiling source file ...)
2>C:\Program Files\Microsoft Visual Studio\2022\Preview\VC\Tools\MSVC\14.30.30704\include\ostream(856,32): message : or       'std::basic_ostream<char,std::char_traits<char>> &std::operator <<<char,std::char_traits<char>>(std::basic_ostream<char,std::char_traits<char>> &,_Elem)'
2>        with
2>        [
2>            _Elem=char
2>        ] (compiling source file ...)
2>C:\Program Files\Microsoft Visual Studio\2022\Preview\VC\Tools\MSVC\14.30.30704\include\ostream(898,31): message : or       'std::basic_ostream<char,std::char_traits<char>> &std::operator <<<std::char_traits<char>>(std::basic_ostream<char,std::char_traits<char>> &,signed char)' (compiling source file ...)
2>C:\Program Files\Microsoft Visual Studio\2022\Preview\VC\Tools\MSVC\14.30.30704\include\ostream(909,31): message : or       'std::basic_ostream<char,std::char_traits<char>> &std::operator <<<std::char_traits<char>>(std::basic_ostream<char,std::char_traits<char>> &,unsigned char)' (compiling source file ...)
2>C:\Program Files\Microsoft Visual Studio\2022\Preview\VC\Tools\MSVC\14.30.30704\include\thread(275,26): message : or       'std::basic_ostream<char,std::char_traits<char>> &std::operator <<<char,std::char_traits<char>>(std::basic_ostream<char,std::char_traits<char>> &,std::thread::id)' (compiling source file ...)
2>...: message : or       'built-in C   operator<<(int, int)'
2>...: message : while trying to match the argument list '(myostream, char)'

所以，我的問題是：如何operator<<在不增加歧義的情況下覆寫某些型別？這甚至是可能的，還是我需要創建一個回傳字串的函式？

編輯：添加using std::ostringstream::operator<<;只會增加歧義的數量。

uj5u.com熱心網友回復：

這是一種選擇：

struct myostream : std::ostringstream {
    using std::ostringstream::operator<<;    // bring in all the member overloads
    
    myostream() noexcept( false ) {
        this->precision( 6 );
        this->setf( ios_base::fixed, ios_base::floatfield );
    }

    // call the base class member function explicitly in here:
    myostream &operator<<( double val ) {
        if ( std::isnan( val ) ) {
            std::ostringstream::operator<<("'NaN'");
        } else if ( std::isinf( val ) ) {
            std::ostringstream::operator<<( val < 0 ? "'-Infinity'" : "'Infinity'" );
        } else {
            std::ostringstream::operator<<(val);
        }
        return *this;
    }    
};

// front-end for all the free operator<< functions:
// (this could be narrowed down to not match on T:s that isn't supported)
template<class T>
myostream& operator<<(myostream& m, const T& x) {
    static_cast<std::ostringstream&>(m) << x;
    return m;
}

uj5u.com熱心網友回復：

當 of 的型別T不是double諸如 時char，會隱式轉換為doubleand call myostream::operator<<，但由于ostream::operator<<有char多載，myostream也會隱式轉換為ostream并呼叫 base 的operator<<，所以這里有歧義。

您可以使用 SFINAE 來防止doublein myostream::operator<<when Tis not a的隱式轉換double：

template<class T>
inline std::enable_if_t<std::is_same_v<T, double>, myostream&>
operator<<(myostream& os, T val ) {
  if ( std::isnan( val ) ) {
      static_cast<std::ostringstream &>( os ) << "'NaN'";
  } else if ( std::isinf( val ) ) {
      static_cast<std::ostringstream &>( os ) << ( val < 0 ? "'-Infinity'" : "'Infinity'" );
  } else {
      static_cast<std::ostringstream &>( os ) << val;
  }
  return os;
}

演示。

標籤：C

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