No "Undefined offset" if array is NULL

Question

I'm not sure this is a bug, so can anyone tell why does not php show a notice if you try to access a key which is not defined of an array that is NULL? Documentation says

Converting NULL to an array results in an empty array.

So accessing a key that is not defined must generate notice.

<?php

$a = NULL;
echo $a['1'];

$a = array();
echo $a['1'];
?>

Prints out ony

Notice: Undefined offset: 1 in /var/www/shared/1.php on line 14

Thank you for your help.

Edit: I take the liberty of adding another example script to the question

<?php
$a = NULL;
var_dump('line '.__LINE__, $a['1']);

$a = array();
var_dump('line '.__LINE__, $a['1']);

which prints

string(6) "line 3"
NULL

Notice: Undefined offset: 1 in [...]/test.php on line 6
string(6) "line 6"
NULL

You're not converting NULL to an array; you're treating it as a string when you're trying to access the character at offset 1 — Mark Baker
– Mark Baker, Commented Mar 17, 2014 at 14:12
@MarkBaker: That's what I thought too, but... var_dump($a['1']) => NULL. — Jon
– Jon, Commented Mar 17, 2014 at 14:13
Quoting from the manual: Characters within strings may be accessed and modified by specifying the zero-based offset of the desired character after the string using square array brackets, as in $str[42]. — Mark Baker
– Mark Baker, Commented Mar 17, 2014 at 14:14
It seems to me the answer is simply "well, the null is not converted to anything -- simply indexing into null always returns null no matter the index you use". Why? Because PHP. But the source is too convoluted for me to back this up with facts. — Jon
– Jon, Commented Mar 17, 2014 at 14:23
@MarkBaker: which would be the same problem: the OP would expect a PHP Notice: Uninitialized string offset: 1 in ... error in that case. Which isn't triggered either. — Wrikken
– Wrikken, Commented Mar 17, 2014 at 14:24

Ja͢ck · Accepted Answer · 2014-03-18 00:32:45Z

When you dereference a variable as an array (for reading), the virtual machine will internally call zend_fetch_dimension_address_read, which in turn loads the right value into the result variable, depending on the type of variable you're trying to dereference.

In the case of null (and int / float / bool), it will choose this branch:

case IS_NULL:
  result->var.ptr = &EG(uninitialized_zval);
  PZVAL_LOCK(&EG(uninitialized_zval));
  return;

Whereby EG(uninitialized_zval) is basically the null itself; no notices are raised when this happens. As Amal has pointed out, there's a bug report for this behaviour, but not much has been done about it for almost a year.

Documentation says "Converting NULL to an array results in an empty array."

An array dereference doesn't convert something into an array (unless you use [] or use it in a write context). The conversion that the manual speaks of is most likely an array cast:

var_dump((array)null);
// array(0) { }

Update

After digging into the system a little more, I found that this (partially) makes the following construct work:

$list = ['foo', 'bar', 'baz'];
while (list($key, $value) = each($list)) {
    echo "$key: $value\n";
}

When the end of the array is reached, each() returns false and so internally false[0] and false[1] are evaluated before the list() expression itself yields false.

Amal · Accepted Answer · 2014-03-17 15:01:59Z

It seems this was actually reported as bug. Here's what I found in the comments section (description adjusted to match your variable names):

Your code:

$a = NULL;
echo $a['1'];

$a = array();
echo $a['1'];

Here, the variable $a is initially declared as a variable of type NULL. $a is initially declared as NULL.

In the first echo statement, the attempt to reference an index of 1 in $a is effectively ignored: the variable is not an array to begin with.

$a is then cast into an array. ($a = array();)

In the second echo statement,$a is an empty array, which PHP can enumerate against and ultimately determine that index 1 is not present.

Hence the first echo results in a NULL output, whilst the second results in PHP issuing an E_NOTICE.

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