User-Defined Function to Determine a Leap Year : Return Value « Functions « PHP






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User-Defined Function to Determine a Leap Year

 
<?php

    function is_leapyear($year = 2004) {

        $is_leap = (!($year % 4) && (($year % 100) || !($year % 400)));
        return $is_leap;

    }

    $answer = is_leapyear(2000);

    if($answer) {

        echo "2000 is a leap year<BR>";

    } else {

        echo "2000 is not a leap year.<BR>";

    }

    $answer = is_leapyear(); 

    if($answer) {

        echo "2003 is a leap year.<BR>";

    } else {

        echo "2003 is not a leap year.<BR>";

    }

?>








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