Php array offset passes 'isset', even though it's not set

Question

The easiest way for me to explain this is to show an example ... Here's a replication of the problem code:

<?php
    $test=array();
    $test['one']='hello';
    if(isset($test['one']['two'][0])) {
        echo 'Apparently it is set ...';
        echo $test['one']['two'][0];
   }
?>

This returns as:

Apparently it is set ...

Warning: Illegal string offset 'two' in C:\test.php on line 6

h

Is this because there are mixed key types? It's just a little anomaly I came across and was wondering if someone could shed some light on it ...

Answer 1

The reason is that, when you dereference a string, it will return a string comprising a single character (assuming the index doesn't exceed the length); the resulting string can be dereferenced again (starting from 5.4 onwards).

For example - link:

$s = 'hello';
$s[0];    // "h"
$s[0][0]; // "h"
// etc. etc.

Illegal indices such as 'two' will cause a notice but it's treated as index 0, except when used inside isset().

Another example:

$s[0][1];    // ""
$s[0][1][0]; // notice: uninitialised string offset: 0

If you don't know beforehand whether a string or array is passed and this is important to you, additional type checks need to take place in between each path.