Avatar of LB1234
LB1234 asked on

Why doesn't this code work?

Don't understand why this isn't working.  When run, I get nothing in the browser.


<?php


function add ($x, $y) {
	$num = $x + $y;
	return $num;
}

add(3,4);

echo $num;


?>

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PHP

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Last Comment
gr8gonzo

8/22/2022 - Mon
Chris Stanyon

$num is only available within your function. You need something like this:


$answer = add(3,4);
echo $answer;

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Chris Stanyon

If you don't need to assign it, then:

echo add(3,4);

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ASKER
LB1234

I thought the whole point of "return" was to make sure something could be used OUTSIDE the function?
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ASKER
LB1234

If I return "$num" shouldn't $num be available outside the variable?
ASKER CERTIFIED SOLUTION
Chris Stanyon

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gr8gonzo