Bug or feature in referencing arrays?

Anon

This caught me out:

unset($tarray)
unset($undefined)
$tarray["one"] = 1;
$undefined =& $tarray["two"];
while (list($key, $value) = each($tarray)) {
echo $key . "|" . $value . "<br>";
}

That code will generate this output:

one|1
two|

So you can see that the line

$undefined =& $tarray["two"];

creates a key "two" with an unset value in the array. This caught me out because I had a routine that checked to see if a key/value was in an array like this:

$tmp =& $tarray[$key];
if (! empty($tmp)) { do stuff }

This test inserted the orphaned key into the array if it didn't already exist and generated an error miles away in the code, which puzzled me since I thought I had cleverly protected against improper values being added to the array.

I think it's a bug, personally.

Anon

Then report it! www.php.net, under "support".

-Ben

Anon

I don't think it's a bug. If you reference an undefined variable, it seems logical that PHP creates this variable first.
If you want to check the existence of a key in an array, proceed like this :

if(isset($a["key"]))
{
//Do something
}

Anon

Olivier wrote:

I don't think it's a bug. If you reference an undefined variable, it seems logical that PHP creates this variable first.

But it doesn't do it in any other circumstances. This code:

unset($undefined);
unset($not_there);
$undefined =& $not_there;

produces precisely nothing. Both variables remain unset. So the array behaviour is inconsistent.

Anon

Olivier wrote:

I don't think it's a bug. If you reference an undefined variable, it seems logical that PHP creates this variable first.

It turns out that there's a difference between a non-existent variable and an undefined variable. As soon as you refer to a variable it is created with a NULL value. So even testing for a variable's existence alters its nature. All a bit quantum physics😉