PHP How to access argument of function inside anonymous function?

Question

I need to access the first argument to a function inside an anonymous function passed as the second argument to said function. For example:

<?php
function a($arg, $func) {
    echo $func();
}
a("argument 1", function () use($arg) {return $arg;});
?>

The above will return the following error:

Notice: Undefined variable: arg in path/to/file.php on line 5

While the desired result would be:

argument 1

Chetan Ameta · Accepted Answer · 2016-03-07 05:06:38Z

2

Try below solution:

function a($arg, $func) {
    echo $func($arg);
}
a("argument 1", function ($arg) {return $arg;});

output

argument 1

answered Mar 7, 2016 at 5:04

Chetan Ameta

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