Php - understanding create_function() - passing simple variable - php

Question

First time I am trying to use the dynamic create_function, and up to now, not much success :-)
My function is this :
function o99_brsa_custom_widgets() {
global $wp_meta_boxes;
global $o99_brsa_options;
for($i=0; $i> count($o99_brsa_options[content]); $i++) {
$widgt_id = 'o99_dashboard_widget_dyn' . $i;
$widgt_name = 'obmek99 widget name' . $i;
$out = $o99_brsa_options[content][$i];
$f = create_function(' $out ',' global $out; echo $out;');
do_the_widgets($widgt_id, $widgt_name, $f);
}
}
The do_the_widgets() action is accepting only a direct echo and prints the content of the widget.
The $o99_brsa_options[content] is a verified array with $i elements (each is content) .
The strange thing is that the $i is working on the $widgt_id and $widgt_name but on the create_function() I get the same value printed in all widgets . ( echo $out )
It seems that I do not know how to pass a simple variable to the new function ( I am using global inside create_function(), but it helps little as for now .
So, what is my mistake / misunderstanding / misuse now :-) ??

Answer 1

create_function was during the stone age , when kaᵠ used pen and paper to write applications, when PeeHaa埽 got a beer because he wrote hello would, The world is better now please use closures
$f = function ($out) {
echo $out;
};
$f("Welcome");
You would thank me one day, But you can only use create_function if you are Gordon (The machine from the past sent here to torment us) he wrote this
$fn = create_function(
'$x',
'return $x; } $foo = 42; function foo($int) { return $int; '
);
See Live Demo