The reason I used is_int, is because 1.4 is divisible by 2 but according
to wiki, it isn't an even number.
Justin Giboney
What about error handling the case where a sting is passed to the function?
This function handles the '3', and the 'your name'
function evenOrOdd ($number) {
if (is_numeric($number)) {
$evenOrOdd = 'odd';
if (($number % 2) == 0)
{
$evenOrOdd = 'even';
}
print $evenOrOdd."\n";
}
else {
throw new Exception('Argument is invalid, it must be a number');
}
}
- Kirk
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