我是 C++ 的新手(我一生都在 C 中度过,所以我认为是时候花一些时间学习一门新语言以丰富我的知识了 :))。我有一个名为“Rational”的类,我有它的所有特定函数,用于 getter、setter、构造函数等(这里不相关)。有趣的部分是当我尝试重载 +,-, ,/ 运算符时。我能够在两个 Rational 对象之间成功地做到这一点,例如 Rational a(1,5),b(5,5),c; c = a + b; 所以这一切都很好。现在我正在尝试通过在 Rational 和整数之间进行 +、-、 、/ 来升级我的类,例如 2 + a、10 - b 等。这是我在 Rational 之间重载的(一段)代码:
理性.cc
...
Rational Rational::operator+(Rational B) {
int Num;
int Den;
Num = p * B.q + q * B.p;
Den = q * B.q;
Rational C(Num, Den);
C.simplifierFraction();
return C;
}
Rational Rational::operator-(Rational B) {
int Num;
int Den;
Num = p * B.q - q * B.p;
Den = q * B.q;
Rational C(Num, Den);
C.simplifierFraction();
return C;
}
Rational Rational::operator*(Rational B)
{
int Num;
int Den;
Num = p * B.p;
Den = q * B.q;
Rational C(Num, Den);
C.simplifierFraction();
return C;
}
Rational Rational::operator/(Rational B)
{
int Num;
int Den;
Rational invB = inverse(B);
Num = p * invB.p;
Den = q * invB.q;
Rational C(Num, Den);
C.simplifierFraction();
return C;
}
...
理性.h
Rational operator+(Rational B);
Rational operator-(Rational B);
Rational operator*(Rational B);
Rational operator/(Rational B);
private:
int p;
int q;
protected:
测试鼠.cc
int main() {
...
const Rational demi(1,2);
const Rational tiers(1,3);
const Rational quart(1,4);
r0 = demi + tiers - quart;
r1 = 1 + demi;
r2 = 2 - tiers;
r3 = 3 * quart;
r4 = 1 / r0;
...
所以当我尝试运行 TestRat.cc 它说:
testrat.cc:31: error: no match for ‘operator+’ in ‘1 + r9’
testrat.cc:52: error: passing ‘const Rational’ as ‘this’ argument of ‘Rational Rational::operator+(Rational)’ discards qualifiers
testrat.cc:53: error: no match for ‘operator+’ in ‘1 + demi’
testrat.cc:54: error: no match for ‘operator-’ in ‘2 - tiers’
testrat.cc:55: error: no match for ‘operator*’ in ‘3 * quart’
testrat.cc:56: error: no match for ‘operator/’ in ‘1 / r0’
为了能够完成这项工作,我必须做什么?谢谢!