Thread: Not sure why this function doesn't generate a returning temporary warning?

Hi!

Have a look at this. It's an operator overload declared inside a class. It calls a function named GCD(...) which returns the greatest common divisor of two numbers. Then calls the class constructor to create a new class. It reduces a fraction down from say 8/12 to 2/3.

Code:

Fraction Fraction::operator!()
{
	int gcd = GCD(numerator, denominator);
	return Fraction(numerator / gcd, denominator / gcd);
}

Ok, but I was expecting a warning from the compiler about returning a temporary, but it doesn't. I was wondering why the compiler wouldn't generate a warning about returning a temporary here?

Thanks

I understand that better now John thanks very much. It's ok if you're returning a copy of the temporary object, a reference however would be a problem as you'd have a reference to an object that got deleted when the function went out of scope. Thanks