# Using ostream with overloading arithmetic operators for class

I have class `Rational`

, its fragment:

```
class Rational {
private:
int numerator;
int denominator;
void saveAsIrreducible();
gcd(int x, inty);
public:
Rational(int numerator, int denominator=1);
Rational operator*(Rational &r);
friend ostream& operator<<(ostream &output, Rational &r);
}
```

And implementation:

```
Rational::Rational(int numerator, int denominator) {
this->numerator = numerator;
this->denominator = denominator;
this->saveAsIrreducible();
}
Rational Rational::operator*(Rational &r) {
Rational r2 = Rational((this->numerator*r.numerator), (this->denominator*r.denominator));
return r2;
}
ostream& operator<<(ostream &output, Rational &r) {
output << r.numerator;
if(r.denominator!=1)
output << "|" << r.denominator;
return output;
}
void Rational::saveAsIrreducible() {
int greatestCommonDivisor = gcd(numerator, denominator);
numerator = numerator/greatestCommonDivisor;
denominator = denominator/greatestCommonDivisor;
if(numerator > 0 && denominator < 0) { /* for example for 3|-2, after arithmetic operations */
numerator = -numerator;
denominator = -denominator;
}
}
int Rational::gcd(int x, int y) {
while(y!=0) {
int r = x%y;
x = y;
y = r;
}
return x;
}
```

I have problem with using operator `*`

with `cout`

. For example, it works:

```
Rational r1(3,5), r2(3,6);
Rational r3 = r1*r2;
cout << r3 << endl;
```

Output is correct. But when I try do this:

```
Rational r1(3,5), r2(3,6);
cout << r1*r2 << endl;
```

I get a lot of errors from the compiler:

```
Rational.cpp|186|error: no match for 'operator<<' (operand types are 'std::ostream {aka std::basic_ostream<char>}' and 'Rational')|
Rational.cpp|186|error: invalid initialization of non-const reference of type 'Rational&' from an rvalue of type 'Rational'|
```

How to fix this problem?

### 1 Answer

The overloaded `operator<<`

takes `Rational&`

as its argument.

So in the case of the below statements:

```
Rational r3 = r1*r2;
cout << r3 << endl;
```

`r3`

is an lvalue (which contains the result of `r1*r2`

) whose reference is passed to `operator<<`

and it succeeds.

But when you do the below:

```
cout << r1*r2 << endl;
```

You are not passing a reference to `operator<<`

because `operator*`

is returning a temporary. So it fails.

P.W
posted this

## Have an answer?

JD