PHP: Passing array by reference fail

Question

I've the following code function:

function foo(&$vett) {

    $vettore = $vett;

    $vettore[] = "ciao";

    var_dump($vettore);
}

$v = array();

foo($v);

var_dump($v);

When I dump the final array is empty. Have you any idea of what could be?

So if you pass something by reference - what's the point of assigning passed value to another one? — u_mulder
– u_mulder, Commented Jun 21, 2018 at 7:05
When you fixed it, you can post your own answer and accept it. Beside the reputation, your solution may be of some help for other users. — PJProudhon
– PJProudhon, Commented Jun 21, 2018 at 7:23

Charis · Accepted Answer · 2018-06-21 07:08:44Z

1

Because $v never modified. Inside the function you assign the variable into another variable. So nothing ever happen to the old $vett

try something like:

function foo(&$vett) {
    $vett[] = "ciao";
    echo __LINE__;
    var_dump($vett);
}

$v = array();
foo($v);
var_dump($v);

answered Jun 21, 2018 at 7:08

Charis

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Comments

u_mulder · Accepted Answer · 2018-06-21 07:07:03Z

0

Correct version is:

function foo(&$vett) {
    $vett[] = "ciao";
}

$v = array();
foo($v);
var_dump($v);

answered Jun 21, 2018 at 7:07

u_mulder

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