在 C++ 中重载不同类型

Question

假设我们有以下 class:

class Rational { // Represents rational number. n=1/2=d for example.
public:
  int n = 0;
  int d = 1;
};

Rational x = Rational();
x.n = 1;
x.d = 2;

是否可以进行重载，使 3 * x 给出 3/2 而不是错误？

我老师说重载只发生在相同类型的对象之间，但是为什么我们可以在ostream类型的cout和Rational类型的对象之间进行重载而不是 int 和 Rational?

类型

Answer 1

你可以这样写

Rational operator *( const Rational &r, int x )
{
    return { r.n * x, r.d };
}

Rational operator *( int x, const Rational &r )
{
    return { r.n * x, r.d };
}

您可以为用户定义的类型重载运算符。对于二元运算符，至少有一个操作数必须是用户定义的类型。

来自 C++ 20 标准（12.4.2.3 表达式中的运算符）

2 If either operand has a type that is a class or an enumeration, a user-defined operator function can be declared that implements this operator or a user-defined conversion can be necessary to convert the operand to a type that is appropriate for a built-in operator. In this case, overload resolution is used to determine which operator function or built-in operator is to be invoked to implement the operator. Therefore, the operator notation is first transformed to the equivalent function-call notation as summarized in Table 15 (where @ denotes one of the operators covered in the specified subclause). However, the operands are sequenced in the order prescribed for the built-in operator (7.6).

在 C++ 中重载不同类型

Overloading different types in C++

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operator-overloading

overloading