Re: No notices for undefined index

>> print $a[0]; // prints 5 >> print $a[100]; // Notice: Uninitialized string offset: 100 Yup, this should happen when 5 is treated as an array of characters. In other words as a string. $a = '5'; echo $a[0]; echo $a[100]; gives you the expected result regarding the original question, i think that the interpreter is prefilling the variable with null $a = 5; var_dump(isset($a[0])); var_dump($a[0]); since $a[0] is already assigned (to null) the interpreter is not throwing a notice $b = null; var_dump(isset($b)); var_dump($b);