错误：C++ 中的“控件可能到达非 void 函数的末尾”

Question

我了解错误消息的基本信息；它提醒我，这段代码可能根本不会返回任何东西。但是，我不明白为什么。我的代码很完美，不是吗？

rational operator / (const rational &lhs, const rational &rhs)
{
if(rhs.numerator() != 0)
{
    int numerator = lhs.numerator() * rhs.denominator();
    int denominator = lhs.denominator() * rhs.numerator();
    rational quotient(numerator, denominator);
    return quotient;
}
else cout << "error" << endl;
}   //this is where error is occurring 

Answer 1

用刻有 cmets 的修正。

rational operator / (const rational &lhs, const rational &rhs)
{
  if(rhs.denominator() != 0) {
    int numerator = lhs.numerator() * rhs.denominator() + 
                    rhs.numerator() * lhs.denominator();
    int denominator = lhs.denominator() * rhs.denominator();
    rational quotient(numerator, denominator);
    return quotient;
  }
  else {
   // what should be returned?
   // let's return lhs, as caller is expecting something, lhs is better than nothing
   // ideally should raise exception, and program shopuld not continue further
   cout << "error" << endl; 
   return lhs; 
  }
}

Answer 2

所以它很简单，你的函数需要返回一个类型rational。如果失败：

if(rhs.numerator() != 0) 

那么您将不会返回该类型的任何内容。因此发出警告。