额外的限定条件，无法从“int”转换为“Rational”

Question

我想使用以下方法创建名为Rational的代码类：

a、 Rational（）//返回有理数1的默认构造函数。

b、 有理（int n，int d）//命名者是n，分母是d。

c、 int getNumerator（）常量

d、 int getDenominator（）常量

e.有理负数（）const//返回-n/d

F.有理互惠（）const//返回d/n

g、 布尔运算符==（常数有理）

h、 Rational add（const Rational）

i、 有理减法（const有理）

j、 有理乘法（const-Rational）

k、 有理分界（const-Rational）

using namespace std;

class Rational
{
    private:
        int numerator;
        int denominator;

    public:
    Rational()
    {
        numerator = denominator = 1;
    }

    Rational(int n, int d)
    {
        numerator = n;
        denominator = d;
    }

    int getNumerator() const
    {
        return numerator;
    }   

    int getDenominator() const
    {
        return denominator;
    }

    Rational negate() const
    {
        return -numerator/denominator;
    }

    Rational reciprocal() const
    {
        return denominator/numerator;
    }

    bool operator==(const Rational& other)
    {
        if(numerator == other.numerator && denominator == other.denominator)
        {
            return true;
        }
        else
            return false;
    }

    Rational add(const Rational& a) const
    {
        Rational t; // creates Rational object
        t.numerator = a.numerator * denominator;
        t.numerator += a.denominator * numerator;
        t.denominator = a.denominator * denominator;
    return t;
    }

    Rational subtract(const Rational& b) const
    {
        Rational t; // creates Rational object
        t.numerator = b.denominator * numerator;
        t.numerator -= denominator * b.numerator;
        t.denominator = b.denominator * denominator;
    return t;
    }

    Rational multiply(const Rational& m) const
    {
        Rational t; // creates Rational object
        t.numerator = m.numerator * numerator;
        t.denominator = m.denominator * denominator;            
    return t;
    }

    Rational divide(const Rational& d) const
    {
        Rational t; // creates Rational object
        t.numerator = d.denominator * numerator;
        t.denominator = denominator * d.numerator;
    return t;
    }

    void Rational::printRational ()
    {
        if ( denominator == 0 ) // validates denominator
        cout << "\nDIVIDE BY ZERO ERROR!!!" << '\n';
        else if ( numerator == 0 ) // validates numerator
        cout << 0;
        else
        cout << numerator << '/' << denominator;
    } 

    void Rational::printRationalAsDouble()
    {
        cout << static_cast< double >( numerator ) / denominator;
    } 
};

int main() {
    Rational c( 2, 6 ), d( 7, 8 ), x; 
    c.printRational(); 
    cout << " + ";
    d.printRational();
    x = c.add( d ); 
    cout << " = ";
    x.printRational(); 
    cout << '\n';
    x.printRational(); 
    cout << " = ";
    x.printRationalAsDouble(); 
    cout << "\n\n";
    c.printRational(); 
    cout << " - ";
    d.printRational(); 
    x = c.subtract( d ); 

    cout << " = ";
    x.printRational(); 
    cout << '\n';
    x.printRational(); 
    cout << " = ";
    x.printRationalAsDouble(); 
    cout << "\n\n";
    c.printRational(); 
    cout << " x ";
    d.printRational(); 
    x = c.multiply( d );

    cout << " = ";
    x.printRational(); 
    cout << '\n';
    x.printRational(); 
    cout << " = ";
    x.printRationalAsDouble();  
    cout << "\n\n";
    c.printRational(); 
    cout << " / ";
    d.printRational(); 
    x = c.divide( d ); 

    cout << " = ";
    x.printRational(); 
    cout << '\n';
    x.printRational(); 
    cout << " = ";
    x.printRationalAsDouble();  
    cout << endl;
    return 0;
} 

但我认为这些错误

主要的cpp:87:10:错误：成员“printRational”[-fpPermissive]void Rational:：printRational（）上的额外限定“Rational:：”^~~~~~~~

主要的cpp:97:10:错误：额外限定“Rational:：”在成员“PrintRationalAlasDouble”[-fpPermissive]void Rational:：PrintRationalAlasDouble（）上^~~~~~~~

main.cpp：在成员函数'Rational Rational::否定（）const'：main.cpp:35： 26：错误：无法转换'（（-（int）（（const Rational*）this）-

主要的cpp：在成员函数'Rational:：Interactive（）const'中：main。cpp:40:27:错误：无法转换“（（int）（（const Rational*）this）-

Answer 1

将ulualRational::printRational更改为ulualprintRational与printRationalAsDouble相同。直接在类中定义函数时，您不会指定后跟::的类名

另外，在否定中有返回-分子/分母 但这是无效的，因为它无法将其转换为返回类型Rational。你可能想要返回有理数（-分子，分母），并为您的互惠函数提供类似功能。如果您试图返回操作的结果，而不是一个新对象，那么您只需更改返回类型即可。