No E_NOTICE for undefined variables in array?

Question

So.. I'm still confused by this, when creating an array with $array = array(); and then manually setting variables like:

<?php
$array[] = 1;
$array['type'] = 2;
$array['number'] = 3;

I know, this is OK for PHP to do, but then when I echo something like $array['none'] it won't show a E_NOTICE for undefined variables.

Can someone explain me, why?

Answer 1

All this is explained on the config doc page, too:

Enabling E_NOTICE during development has some benefits.

For debugging purposes: NOTICE messages will warn you about possible bugs in your code. For example, use of unassigned values is warned. It is extremely useful to find typos and to save time for debugging.

NOTICE messages will warn you about bad style. For example, $arr[item] is better to be written as $arr['item'] since PHP tries to treat "item" as constant. If it is not a constant, PHP assumes it is a string index for the array.

It's quite simple: When you access a key that doesn't exist, and assign it a new value, PHP will create that key, and add it to the list (or array). But when you try to access a non-existing key, and attempt to echo it's value, PHP won't crash, but it'll let you know that your code contains a possible bug:

$arr = array('foo' => 'bar');
echo $arr['fo'];

This issues a notice because my code may contain a typo. I may expect the key fo to exist, while clearly it doesn't, so I need to work on my code some more.
Another reason why this notice is issued is because lookups for non existing properties/keys are "slow". In order for PHP to know for a fact that the key doesn't exist, the entire array has to be scanned. That, too, is not ideal, though inevitable at times. If you have code that issues tons of E_NOTICE's, chances are that some simple if's, like:

if (!isset($arr['fo']))
{
    $arr['fo'] = '';
}
echo $arr['fo'];

Will, though adding more code, effectively speed up your code. Not in the least because issueing notices isn't free (it's not that expensive, but not free either).

Other benefits:

Notices also let you know when you forgot to quote array keys, for example

echo $arr[foo];
echo $arr['foo'];

Initially, both will echo bar, but let's add 1 line of code to this:

define('foo', 'bar');
echo $arr[foo];
echo $arr['foo'];

This won't, because foo is now a constant, so $arr[foo] amounts to $arr['bar'];, which is an undefined index. Turning off notices, will just echo the string representation of NULL, which is an empty string.

Basically, notices help you. Use them, listen to them, and fix them. If your site is broken, fix it. If you get into the habbit of ignoring these notices, you'll probably set your ini files to a more "forgiving" setting, and grow lazy.
As time progresses, your code will become ever more messy/smelly, until such time you actually have a difficult to trace bug. You'll decide to turn your error reporting to E_STRICT | E_ALL, and won't be able to see the actual notice/warning that points out where your bug actually is, because your screen will be cluttered with E_NOTICE undefined index/variable...

Answer 2

It will. If you have turned on error reporting, it should display a warning similar to the one below:

Notice: Undefined index: none in /path/to/script.php line X.

To check, try the following:

<?php

ini_set('display_errors',1); 
error_reporting(E_ALL);

$array = array();
echo $array['none'];

And, if you want to actually make sure they exist before trying to use them in your code, use isset():

if(isset($array['none'])) {

    // do stuff ...

}

See it live!