类数学操作重载，适用于也具有强制转换运算符重载的类型

问：

作为我的 uni 任务的一部分，我需要为有理数编写类、覆盖数学运算符、比较运算符等。但我还需要将强制转换为 和 类型。这是我的类的简化代码：shortintlong

class RationalNumber {
  long long numerator, divider;
public:
  RationalNumber() : numerator(0), divider(1) {}
  RationalNumber(long long numerator, long long divider = 1) : numerator(numerator), divider(divider) {}
  
  // Let's take only one math operator
  RationalNumber operator*(const RationalNumber& other) {
    return { numerator * other.numerator, divider * other.divider };
  }

  // And one cast operator
  operator int() {
    return numerator / divider;
  }
  //...
};

问题是，如果我现在想将 RationalNumber 对象乘以 ，我会收到一个错误，说这个操作不明确：int

int main() {
  int integer = 2;
  RationalNumber rational(1, 2);
  
  rational * integer;
  // Well, ambiguous:
  // 1. Cast 'rational' to int and multiply. Or
  // 2. Cast 'integer' to RationalNumber and RationalNumber::operator*
  return 0;
}

添加专门用于单个的构造函数 -- 不起作用。确实有效的东西 -- 添加 .这种方法的问题是：我需要覆盖所有数学运算、比较运算等，因为我已将运算符转换为 int -- .intRationalNumber::operator*(int)intoperator int

好的，但即使我会这样做，我尝试使用编译器的第二个也会用大量关于转换和潜在数据丢失的警告轰炸我（认为它会编译）：unsignedunsignedint

class RationalNumber {
  //...
public:
  //...
  RationalNumber operator*(int othNum) {
    return { numerator * othNum, divider };
  }
  //...
};

int main() {
  unsigned int uinteger = 2;
  RationalNumber rational(1, 2);

  rational * uinteger; // Well... AAAAAAA
  return 0;
}

test.cpp: In function ‘int main()’:
test.cpp:29:14: warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second:
   29 |   rational * integer;
      |              ^~~~~~~
test.cpp:14:18: note: candidate 1: ‘RationalNumber RationalNumber::operator*(int)’
   14 |   RationalNumber operator*(int othNum) {
      |                  ^~~~~~~~
test.cpp:29:14: note: candidate 2: ‘operator*(int, unsigned int)’ (built-in)
   29 |   rational * integer;
      | 

为了完全满足编译器的要求，我需要添加.现在让我们记住，我需要重载 3 个转换运算符和十几个数学/比较运算符。这意味着我需要为 EACH 运算符重载创建 6 个额外的函数。RationalNumber::operator*(unsigned)

那么，我可以以某种方式强制编译器始终为我的类转换值吗？