c++ - 错误: “Control may reach end of non-void function” in c++

我了解错误消息的基本知识；提醒我，此代码可能根本不返回任何内容。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

最佳答案

用题词注释进行更正。

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;
  }
}

关于c++ - 错误: “Control may reach end of non-void function” in c++，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/25903619/