Checking whether a number is even or odd: the PHP way! :p

gistfile1.aw

function iseven($n) {
	$lastdigit = substr($n, -1);
	if ($lastdigit == '0')
		return true;
	elseif ($lastdigit == '2')
		return true;
	elseif ($lastdigit == '4')
		return true;
	elseif ($lastdigit == '6')
		return true;
	elseif ($lastdigit == '8')
		return true;
	else
		return isodd($n) != true;
}


function isodd($n) {
	$lastdigit = substr($n, -1);
	if ($lastdigit == '1')
		return true;
	elseif ($lastdigit == '3')
		return true;
	elseif ($lastdigit == '5')
		return true;
	elseif ($lastdigit == '7')
		return true;
	elseif ($lastdigit == '9')
		return true;
	else
		return iseven($n) != true;
}

You are retarded.

I have an additional pair of sarcasm goggles for sale. Interested? :p

Hahaha okay point taken. I am the retard. :(

…

Sent from my iPhone

On Apr 27, 2012, at 9:45 AM, Raynor ***@***.*** wrote: I have an additional pair of sarcasm goggles for sale. Interested? :p --- Reply to this email directly or view it on GitHub: https://gist.github.com/2508693

Well, on the other hand I can't blame you. When it comes to PHP, you're never sure whether someone's serious or not. :)

ha, yes this is true. It's unfortunate how many serious gists (and PHP in general) there are with the outrageously low quality of code you pointed out.

…

On Fri, Apr 27, 2012 at 9:56 AM, Raynor Vliegendhart < ***@***.*** > wrote: Well, on the other hand I can't blame you. When it comes to PHP, you're never sure whether someone's serious or not. :) --- Reply to this email directly or view it on GitHub: https://gist.github.com/2508693

THAT IS SO FRIENDLY OF YOU

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On Fri, Apr 27, 2012 at 10:00 AM, Brandon < ***@***.*** > wrote: Curt's right. Here, I'm feeling friendly today: function is_even($int) { if (!is_int($int)) throw new Exception("not an int"); return ($int%2==0); } --- Reply to this email directly or view it on GitHub: https://gist.github.com/2508693