PHP - How to change reference of a function's pointer parameter from the definition

Question

I'm using PHP 7.4. From the function call, I need to change the reference of a variable passed to it by reference. This is the simplified version of my code to make it clear and more focus on the problem.

function f(&$p){
    $p['x'] = [];
    $p = &$p['x'];
    //$p represents $a['x'] here
}
$a = [];
$p = &$a;
f($p);
//$p revert to $a here
$p['y'] = 3;
echo json_encode($a);

I expected the function to change the reference of variable $p from pointing to $a to $a['x']. Within the definition scope, it did. But the reference reverted back to $a after the call.

So from the above code, instead of this {"x" : {"y" : 3}}, I get this {"x" : {}, "y" : 3} as the result.

I assume that a function's pointer parameter can only be used to change the value not the reference. But is there any way to do the same for a reference considering that reference is also a type of value?.

Why would you not just pass $a by reference? what is echo json_encode($p); ? — futureweb
– futureweb, Commented Feb 2, 2022 at 20:46
@futureweb to show the example that matches exactly the problem i'm having and passing $a doesn't show it. php.net/manual/en/function.json-encode.php — stackunderflow
– stackunderflow, Commented Feb 2, 2022 at 20:54
What has json_encode got to do with it ? you are passing an array by reference &$p which is copied from an array $a then somehow you expect $a to be equal to $p because you say $p = &a you are passing $a by ref to $p then passing $p by ref to the function I don't think this will update $a — futureweb
– futureweb, Commented Feb 2, 2022 at 21:07
No of course the result will be ``` {"x" : {}, "y" : 3}``` you set x to an empty array then json_encode it — futureweb
– futureweb, Commented Feb 2, 2022 at 21:11

Developer · Accepted Answer · 2022-02-02 21:11:12Z

1

I realized what you needed, your solution is this.

function &f(&$p){
    $p['x'] = [];
    $p = &$p['x'];
    return $p;
}
$a = [];
$p = &$a;
$p = &f($p);
$p['y'] = 3;
echo json_encode($a);

answered Feb 2, 2022 at 21:11

Developer

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2 Comments

stackunderflow Over a year ago

thank you for your answer. actually i need more straight forward solution which doesn't involve reassigning because the parameter is already a pointer. i take your answer for now but it would be nice if you'd update it with better answer later

Developer Over a year ago

Actually I agree with your concept, because it is logical and as we know references in 'C' have working as you have described. But PHP blocks the propagation of this assignment $p=&$p['x']; through input parameters between 2 different contexts. I think in PHP, the maximum what we can do in this case with references, is what I wrote.

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