Hi LarsLai,
That's not actually going to be much use, I'm afraid ...
With your definition, there is no 'return', so cannot be used to assign ...
$number = 20;
$x = plusMinus($number); // Won't work ... no return in plusMinus()
// And since you haven't passed by reference ...
echo $number;// Still prints 20;
To fix for use with references ...
function plusMinus(&$number) { $number *= -1; }
Then ...
plusMinus($number); // This will work
echo $number;// Now prints -20;
Paul 🙂