PHP Adding references to array

Question

I'm trying to add integer references to an array in PHP but for some reason it doesn't work. I am completely confused as to why.

Simplifying things, the code is:

<?php
$myArray = array( 1 => true, 2 => true, 3 => true );
$param_ref = array();
foreach($myArray as $key => $value) {
 $param_ref[] = &$key;
}
var_dump($param_ref);
?>

I expect the output to be:

array(3) { 
 [0] => &int(1) 
 [1] => &int(2) 
 [2] => &int(3) 
} 

But the actual output is:

array(3) { 
 [0] => &int(3) 
 [1] => &int(3) 
 [2] => &int(3) 
} 

With some closer inspection, it seems the array's ($param_ref) values are being overwritten on each iteration of the loop.

Any idea what's going on?

• Alma Do: Why do that? Just use array_keys(). And - yes, in your case you'll end with multiple references to same (end-of-loop) value - see answers below

Answer 1

$key is being changed each iteration of the loop, so your reference to $key is always going to be a reference to the current value of $key, which (at the end of the loop) is three.... so all three references to $key are pointing to the one and only instance of $key, which has a value of 3

Consider as:

Iteration #1

$key is assigned the value 1; $param_ref[0] is a reference to $key, so it's pointing to a variable with a value of 1.

Iteration #2

$key now has a value of 2, ; $param_ref[1] is a reference to $key, so it's pointing to a variable with a value of 2.... but $param_ref[0] is also a reference to $key, so it's pointing to a variable which now has a value of 2

Iteration #3

$key now has a value of 3, ; $param_ref[2] is a reference to $key, so it's pointing to a variable with a value of 3.... but both $param_ref[0] $param_ref[1] are also referencing $key, so they're pointing to a variable which now has a value of 3

Answer 2

You fought PHP, and PHP won. :-)

What happened is that by using the expression &$key you caused PHP to regard the value pointed to by $key as a reference. This in turn causes reassignments to $key (which happen on every iteration of the loop) to be visible through all variables that were assigned the value &$key at any point in the past.

That is, after the first iteration the resulting array is

array(1) { 
 [0] => &int(1) 
} 

After the second iteration, the array is

array(2) { 
 [0] => &int(2) 
 [1] => &int(2) 
} 

and so on.

How to get the expected result: Simply unset($key) at the end of the loop:

foreach($myArray as $key => $value) {
 $param_ref[] = &$key;
 unset($key);
}

• Alma Do: I'd rather use array_keys() for part "How to get the expected result"
• Jon: As an aside, I recently happened to answer another question where the unexpected result was caused by exactly this same internal mechanism. It might be useful as additional material, especially since the answer there has more detail and considers several alternatives.
• SweetSour: Unfortunately this is now giving me the result of: array(3) { [0]=> int(1) [1]=> int(2) [2]=> int(3) } whereas I require an array of references: array(3) { [0]=> &int(1) [1]=> &int(2) [2]=> &int(3) }
• sectus: @user3576985 , keys cannot be used as reference. So, "expected result" is useless.
• sectus: because keys are not zval

Answer 3

Your $param_ref values are not being overwritten, it's how a foreach works in general. Also the cause of much talked-about foreach($foo as &$bar) issue, which you can read if you search. The thing is, when you do = &$key;, you don't create a reference to something $key is pointing at, you create reference to $key value. When this value changes, so does all references to $key.

Anyways, as been pointed out numerous times, you would do yourself a favor if you accepted the fact that PHP's references are terribly broken, and just forget of their existence whatsoever.