需要帮助运算符重载，以便我可以显示函数

Question

我一直在试图弄清楚如何重载

这些是我试图重载的行：

cout << "Testing the compare() member function, found:" << endl;
if (r1.compare(r2) == 1)
  cout << r1.display() << " is greater than " << r2.display() << endl;
else if (r1.compare(r2) == 0)
  cout << r1.display() << " is equal to " << r2.display() << endl;
else if (r1.compare(r2) == -1)
  cout << r1.display() << " is less than " << r2.display() << endl;

cout << "Testing the four arithmetic member functions:" << endl;
cout << "r1 + r2 = " << r1.display() << " + " << r2.display() << " = " << r1.add(r2) << endl;
cout << "r1 - r2 = " << r1.display() << " - " << r2.display() << " = " << r1.subtract(r2) << endl;
cout << "r1 * r2 = " << r1.display() << " * " << r2.display() << " = " << r1.multiply(r2) << endl;
cout << "r1 / r2 = " << r1.display() << " / " << r2.display() << " = " << r1.divide(r2) << endl;

每次调用函数时都会发生错误。下面是函数的代码：

void rational::display()
{
   int gcd = GCD();
   if (denom < 0)
   {
      num = -num;
      cout << num / gcd << " / " << denom / gcd << endl;
   }
   else if (num == 0)
   cout << num << endl;
}

rational rational::add(const rational &r2) const
{
   rational sum; 
   sum.num = (num * r2.denom) + (r2.num * denom);
   sum.denom = denom * r2.denom;
   return sum;
}

乘法、除法和减法函数与加法相同，只是更改了符号和变量名以匹配运算。我的重载运算符是这样设置的：

ostream& operator<< (ostream& out, rational &robj)
{
   out << example code << example code;
   return out;
}

任何帮助将不胜感激。这是我第一次发布，所以如果我需要发布更多我的代码，我会的。谢谢！

Answer 1

首先，改变

ostream& operator<< (ostream& out, rational &robj)

到

ostream& operator<< (ostream& out, rational const& robj)

然后，而不是

cout << r1.display() << " is greater than " << r2.display() << endl;

使用

cout << r1 << " is greater than " << r2 << endl;

既然你已经有了上面的operator<<函数，我会去掉display()这个成员函数。它不会（不应该）为您提供比使用 operator<< 函数更多的功能。

给定

rational 1

使用

r.display();

应该给你同样的输出

cout << r << endl;

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有关运算符重载的更多详细信息，请参阅What are the basic rules and idioms for operator overloading?。