Change global array values by changing the reference returned from a function

Question

It's difficult to explain in word what I'm trying to do, so I'm providing a minimal example, see comments:

$g = array( 
    'a' => array(1, 2, 3), 
    'b' => array(4, 5, 6)
); // A global array

function &search($key) {
    global $g;
    return $g[$key];
}

$a = search('b'); // Now $a should be a reference to $g['b'], right?

$a[2] = 666;
print_r($a); // Ok changed
print_r($g); // Why not changed?

Tested on PHP 5.6.4.

Reason for what I'm trying to do is the fact the search function is obviously more complex in my use case (not just a key indexing!), and after the result has been found, it's handy to work on results: the original array is nested at various levels.

Because you're returning by value, not by reference; so modifying the returned value doesn't modify the global value — Mark Baker
– Mark Baker, Commented Jun 20, 2016 at 13:36
@MarkBaker Why you say I'm returning by value? Shouldn't &search return by reference? — lorenzo-s
– lorenzo-s, Commented Jun 20, 2016 at 13:37

iainn · Accepted Answer · 2016-06-20 13:39:52Z

4

From the manual:

Note: Unlike parameter passing, here you have to use & in both places - to indicate that you want to return by reference, not a copy, and to indicate that reference binding, rather than usual assignment, should be done for $myValue.

Your code just needs an extra ampersand (the reference "binding" one referred to above), as follows:

$a =& search('b');

answered Jun 20, 2016 at 13:39

iainn

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2 Comments

lorenzo-s Over a year ago

Greeeeeat! Thank you! I tough it was sufficient to use & just in the function declaration!

samayo Over a year ago

I have given up on this, when it comes to php. I could never get references, this way. For example, why is the purpose then of doing function &search($key){} ? What does that ampersand do in this case?

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