根據條款21的說明。實現了static版本的程式碼

class Rational{
public:
    Rational(){}
    Rational(int numerator/* = 0*/, int denominator = 5)
    {
        numerator1 = numerator;
        denominator1 = denominator;
    }
public:
    int numerator1;
    int denominator1;
};

bool operator == (const Rational& lhs, const Rational& rhs)
{
    if (lhs.numerator1*rhs.denominator1 == rhs.numerator1*lhs.denominator1)
        return true;
    else
        return false;
}

const Rational& operator*(const Rational& lhs, const Rational& rhs)
{
    static Rational result;
    result.numerator1 = lhs.numerator1*rhs.numerator1;
    result.denominator1 = lhs.denominator1*rhs.denominator1;
    return result;
}

int main(int argc, char *argv[])
{
    Rational oneEight(1, 8);
    Rational oneHalf(1, 2);
    Rational onefour(1, 4);
    Rational onefive(1, 5);

    Rational one = oneEight*oneHalf;
    Rational two = onefour*onefive;

    if( one == two )
        qDebug()<<"true";
    else
        qDebug()<<"false";

    if((oneEight*oneHalf)==(onefour*onefive))
        qDebug()<<"this is true";
    else
        qDebug()<<"this is false";
}
 

結果

第二個if判斷語句中。在operator==被呼叫前，已有兩個operat*呼叫式起作用，每一個都返回reference指向operator*內部定義的static Rational物件。因此operator==被要求將“operator*內的static Rational物件值”拿來和“operator*內的static Rational物件值”比較，結果一定是一樣的。（兩次operator*呼叫的確各自改變了static Rational物件值，但由於它們返回的都是 reference，因此呼叫端看到的永遠是static Rational物件的“現值”。）