根据条款21的说明。实现了static版本的代码
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";
}
结果
第二个if判断语句中。在operator==被调用前,已有两个operat*调用式起作用,每一个都返回reference指向operator*内部定义的static Rational对象。因此operator==被要求将“operator*内的static Rational对象值”拿来和“operator*内的static Rational对象值”比较,结果一定是一样的。(两次operator*调用的确各自改变了static Rational对象值,但由于它们返回的都是 reference,因此调用端看到的永远是static Rational对象的“现值”。)