++ 前缀运算符重载不适用于 <<

Question

这里有一点新的 C++ 开发人员。我有一个类Rational 用于表示有理数并允许用户对它们执行算术和关系运算。我重载了所有运算符并且它们都正常工作，除非我尝试将++/-- 前缀运算符（例如--Rational）或+/- 一元运算符与<< 输出运算符组合在一起，如：

Rational num(7, 2); // initializes rational to 7/2 (fraction)
cout << --num; // should change val to 5/2 (decrements by one)
// error: no match for 'operator<<'(operand types are 'std::ostream' {aka 'std::basic_ostream<char>'} and 'Rational')

或：

Rational num(1, 10);
cout << -num; // should change to -1/10
// error: no match for 'operator<<'(operand types are 'std::ostream' {aka 'std::basic_ostream<char>'} and 'Rational')

我不明白为什么，因为我重载了所有这些运算符以返回 Rational 对象，并且我重载了 << 运算符以接受 Rational 对象。有趣的是，当我尝试这个时：

Rational num(5, 3);
cout << num;

它按预期工作。那么有人可以告诉我这里发生了什么吗？

类的相关代码：

class Rational
{
private:
    // Instance variables declarations
    int numerator, denominator;
public:
    // Constructors declarations
    Rational(int numer_val, int denom_val = 1);
    Rational();

    // Unary operators declarations
    friend Rational operator +(const Rational &num);
    friend Rational operator -(const Rational &num);
    friend Rational operator ++(Rational &num);
    friend Rational operator --(Rational &num);
    // I/O operators
    friend ostream& operator <<(ostream &outs, Rational &num);
};

// Constructors definitions

Rational::Rational(int numer_val, int denom_val) : numerator(numer_val)
{
   set_denominator(denom_val); // irrelevant
   simplify(); // irrelevant
}
Rational::Rational() : Rational(0) { }

// Unary operators definitions
Rational operator +(const Rational &num)
{
    return Rational(+num.numerator, num.denominator);
}
Rational operator -(const Rational &num)
{
    return Rational(-num.numerator, num.denominator);
}
Rational operator ++(Rational &num)
{
    num.numerator += num.denominator;
    return num;
}
Rational operator --(Rational &num)
{
    num.numerator -= num.denominator;
    return num;
}
// I/O operators
ostream& operator <<(ostream &outs, Rational &num)
{
    outs << to_string(num.numerator) + "/" + to_string(num.denominator);
    return outs;
}

Answer 1

operator<< 的第二个参数绑定到左值引用：

ostream& operator <<(ostream &outs, Rational & num)
                                          // ^

但是您的所有运算符都返回 r 值。例如，当你这样做时：

cout << -num;

您正在尝试将左值引用绑定到右值，这是不允许的。

您可以通过更改 operator<< 以接受 const 引用来解决此问题：

ostream& operator <<(ostream &outs, Rational const & num)
                                          // ^^^^^^^