Why does PHP throw a "Notice: Undefined offset" error here:
<?php
$MyVar = array();
echo $MyVar[0]; //Notice: Undefined offset: 0
?>
But not here:
<?php
$MyVar = false;
echo $MyVar[0]; //No error
?>
2 Answers
It's ultimately because in your 2nd example $MyVar[0] is null which isn't an error. You could probably reference $MyVar[0][1][2][3] and get the same result.
The first example isn't null it's a missing index in an array so it warns you.
2 Comments
user3163495
The first example does in fact echo NULL, but also throws an error. Try the first example and you'll see "Notice: Undefined offset" followed by "NULL"
Matt
That does make sense as PHP specifically looks out for undefined indexes. Note it's not an error but a notice.
Undefined offset is provided when there are no values in an array and you're trying to reference an uninitialised index.
In case of the 2nd code, you've assigned(initialised) a value to the variable which is not different from variable[0].
If you assign values for 1st two indexes in 1st example, and refer to the 3rd index, you should get the undefined offset error.
I dont think it is about null or not, but a case of assigning values to indexes. If you take a C parlance, array is an equivalent of malloc(upto a certain extent). Thankfully PHP does not crash, just throws an undefined index[offset] error
NULL. The same offset notice applies to strings$MyVar = ''; echo $MyVar[0];Which would result inUninitialized string offset: 0Accessing variables of other types (not including arrays or objects implementing the appropriate interfaces) using [] or {} silently returns NULL.This implies that PHP expects you to verify the variable type prior to accessing it with[]or{}e.g.if (is_array($MyVar))