Ternary operator on javascript and PHP, different output

Question

I wanted to do Fizzbuzz in php by using unary if, but the output is not what I expect, and I didn't understood why, so I have copy-paste the code to javascript, and now the result is as expected. Why?

<script src="http://code.jquery.com/jquery-latest.js" type="text/javascript"></script>
<script>
$(function(){
papa ='Javascript Output: ';
for($i=1;$i <= 10; $i++){
papa += ($i %5 === 0 && $i %3 === 0) ? 'FizzBuzz' : ($i % 3 === 0) ? 'Fizz' : ($i % 5 === 0) ? 'Buzz' : $i;
$('#result').text(papa);
}
})
</script>
<?php
echo 'PHP Output: ';
for($i=1;$i <= 10; $i++){
$papa ($i %5 === 0 && $i %3 === 0) ? 'FizzBuzz' : ($i % 3 === 0) ? 'Fizz' : ($i % 5 === 0) ? 'Buzz' : $i;
echo $papa;
}
?>
<div id='result'></div>

Output

PHP Output: 12Buzz4BuzzBuzz78BuzzBuzz
Javascript Output: 12Fizz4BuzzFizz78FizzBuzz

kirilloid · Accepted Answer · 2012-03-14 06:42:18Z

7

Ternary operator (which you called "unary if") in javascript uses right associativity and the same operator in php uses left associativity.

Yes, this could be fixed with more parenthesis (as well as any problem relating operator precedence).

answered Mar 14, 2012 at 6:42

kirilloid

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1 Comment

mpen Over a year ago

also note the super cool 2 versions of logical "and" and "or" and their difference in precedence! (in PHP)

tuoxie007 · Accepted Answer · 2012-03-14 07:02:27Z

2

in PHP, change 
    ($i %5 === 0 && $i %3 === 0) ? 'FizzBuzz' : ($i % 3 === 0) ? 'Fizz' : ($i % 5 === 0) ? 'Buzz' : $i;
    to:
    ($i %5 === 0 && $i %3 === 0) ? 'FizzBuzz' : ($i % 3 === 0) ? 'Fizz' : (($i % 5 === 0) ? 'Buzz' : $i);

edited Mar 14, 2012 at 7:02

answered Mar 14, 2012 at 6:40

tuoxie007

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