PHP: closure to set variable

Question

In PHP closures are quite useful. But can I use closures to set array elements?

$configs = [
    "y" => "1",
    "x" => function() { return "xxx"; },
];
echo $configs["y"];
echo $configs["x"];  // error

gives

Recoverable fatal error: Object of class Closure could not
be converted to string on line 6 (last line)

Is there a chance to cast the closure or anything the like that the closure works for array initialization?

Working with PHP 7.1.4 on MacOSX

maybe this answer help you https://stackoverflow.com/a/9443941/1779650 — Joe RR, Mar 03 '19 at 19:56
Not really. I want to execute the closure immediately to get the array element initialized. I don't need the closure function later on in the code. — WeSee, Mar 03 '19 at 20:00

Jeto · Accepted Answer · 2019-03-03T21:45:51.230

2

You want an IIFE (Immediately Invoked Function Expression):

$configs = [
  'y' => '1',
  'x' => (function () { return 'xxx'; })()
];

echo $configs['x'];

edited Mar 03 '19 at 21:45

answered Mar 03 '19 at 20:27

Jeto

Can we invoke the closure during definition in line 3? How? – WeSee Mar 03 '19 at 21:06
@WeSee Yes. See my edit for an alternative version using an IIFE (PHP 7 only). – Jeto Mar 03 '19 at 21:15

score 0 · Answer 2 · answered Mar 03 '19 at 19:02

0

I would define the function apart:

function test(){   
    echo "Hello";
}

And then assign the result to a variable:

$a = test();
echo $a; //Returns "Hello"

answered Mar 03 '19 at 19:02

alex55132

In my case that doesn't work. I have a config-Array where array variables are being set... – WeSee Mar 03 '19 at 19:03
1

Then edit your post and show us what do you have so we can help you with more precision ^^ @WeSee – alex55132 Mar 03 '19 at 19:06

2 Answers2