note 71398 added to language.operators.type

*** w dot bogaerts at kratz dot nl said: instanceof only works with objects. So if you want to check if a variable is an object that implements a certain interface, you need two checks: is_object and instanceof. A fatal error occurs if instanceof is used with other data types. *** I don't exactly know what he meant by that because I was able to use instanceof just fine with other data types. <?php class A{ function __construct(){} } function test($variable) { echo ($variable instanceof A) ? "this is an instance of class A<br/>" : "this is not an object<br/>"; } $a = new A(); //object $b = array("a","b","c"); //array $c = "c"; //string test($a); test($b); test($c); test("d"); //string again ?> results: this is an instance of class A this is not an object this is not an object this is not an object ---- Server IP: 66.207.199.35 Probable Submitter: 216.99.43.20 ---- Manual Page -- http://www.php.net/manual/en/language.operators.type.php Edit -- https://master.php.net/note/edit/71398 Del: integrated -- https://master.php.net/note/delete/71398/integrated Del: useless -- https://master.php.net/note/delete/71398/useless Del: bad code -- https://master.php.net/note/delete/71398/bad+code Del: spam -- https://master.php.net/note/delete/71398/spam Del: non-english -- https://master.php.net/note/delete/71398/non-english Del: in docs -- https://master.php.net/note/delete/71398/in+docs Del: other reasons-- https://master.php.net/note/delete/71398 Reject -- https://master.php.net/note/reject/71398 Search -- https://master.php.net/manage/user-notes.php