in_array() returns wrong values?

Question

I found a strange error in in_array() PHP function.

Next code returns TRUE but it's wrong value. Anyone can told my what I do wrong?

<?php
var_dump(in_array(0, array('a', 'ab', 'abc')));

Of course I know that I can execute in_array(0, array('a', 'ab', 'abc'), true) but why without this true this function isn't working good.

Because of PHP type juggling. Without the parameter strict to be true PHP does a loosely comparison. — Rizier123, Apr 26 '16 at 18:34
That looks very similar to this comment in the php documentation. Similar issue. Anybody know what's going on? http://php.net/manual/en/function.in-array.php#114117 — Michael Arrison, Apr 26 '16 at 18:37
From the [PHP Docs](http://php.net/manual/en/function.in-array.php) `Searches haystack for needle using loose comparison unless strict is set.` — Mark Baker, Apr 26 '16 at 18:45

score 2 · Answer 1 · answered Sep 13 '16 at 14:32

2

You have to set it strict

var_dump(in_array(0, array('a', 'ab', 'abc'), true));

answered Sep 13 '16 at 14:32

VikingBlooded

score 1 · Accepted Answer · edited Oct 01 '16 at 18:37

1

in_array you have to set type

Syntax:

in_array(search,array,type);

var_dump(in_array(0, array('a', 'ab', 'abc'), true));

Note: If the search parameter is a string and the type parameter is set to TRUE, the search is case-sensitive.

edited Oct 01 '16 at 18:37

marc_s

answered Sep 13 '16 at 14:47

2 Answers2