Error: "Control may reach end of non-void function" in c++

Question

I understand the basics of the error message; it is alerting me that this code may not return anything at all. However, I do not understand why. My code is perfect, is it not?

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

"Control reaches the end of a non-void function [*without* returning a value on all execution paths]" — user2864740, Sep 18 '14 at 03:28
You're missing a `return` statement. _"My code is perfect, is it not?"_ Way off ... — πάντα ῥεῖ, Sep 18 '14 at 03:33

score 1 · Answer 1 · answered Sep 18 '14 at 03:23

1

Correction made with inscribed comments.

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; 
  }
}

answered Sep 18 '14 at 03:23

Dr. Debasish Jana

6,980
4
30
69

Could you provide more information on raising an exception? Is this something that I have to code in? – Matt C. Sep 18 '14 at 03:56
Here's an example: http://stackoverflow.com/questions/8480640/how-to-throw-a-c-exception – Dr. Debasish Jana Sep 18 '14 at 06:59

score 0 · Answer 2 · answered Sep 18 '14 at 03:18

0

So its pretty simple right, your function needs to return a type rational. In the case that this fails:

if(rhs.numerator() != 0)

then you will not return anything of that type. Hence the warning.

answered Sep 18 '14 at 03:18

Fantastic Mr Fox

32,495
27
95
175

So I can't return any kind of error message? What if there is no type rational that suits the situation? – Matt C. Sep 18 '14 at 03:24
@MattC. Then you should consider throwing an exception instead. – Fantastic Mr Fox Sep 18 '14 at 03:25
2

@MattC.: You defined your function to return a `rational`, so it has to return a `rational` -- or throw an exception. Or you can change the return type so it can always return a sensible value. – Keith Thompson Sep 18 '14 at 03:25

Error: "Control may reach end of non-void function" in c++

2 Answers2