Use Keyword in Functions - PHP

Use keyword in functions - PHP

The use of "use" is correct in this case too.

With closures, to access variables that are outside of the context of the function you need to explicitly grant permission to the function using the use function. What it means in this case is that you're granting the function access to the $tax and $total variables.

You'll noticed that $tax was passed as a parameter of the getTotal function while $total was set just above the line where the closure is defined.

Another thing to point out is that $tax is passed as a copy while $total is passed by reference (by appending the & sign in front). Passing by reference allows the closure to modify the value of the variable. Any changes to the value of $tax in this case will only be effective within the closure while the real value of $total.

How and where should I use the keyword use in php

In PHP, the keyword use is used in 3 cases:

As class name alias - simply declares short name for a class (must be declared outside of the class definition)
(manual: Using namespaces: Aliasing/Importing )
To add a trait to a class (must be declared inside (at the top) of the class definition)
(manual: Traits)
In anonymous function definition to pass variables inside the function
(manual: Anonymous functions)

PHP | What does 'use function function_name;' (where function is a PHP built-in function) do?

Ok is actually pretty easy, the main issue was, that I did not read properly the documentation or I just miss the part, is true though that the use keyword is widely used in PHP, besides with different functionality all together.
Either way, thanks for whom helped me figured out in the comment section[ (@riggsfolly, @peterm, @eis, @david) of the question and after good reading and testing I'd decided to post the answer my self.

Quick Answer

The use function [some_function] is a namespace which defines / import that function for a particular place, this place being the namespaced section givin possibility to have same named function from different namespace and even modifying the name with alias, both for functions and namespace all together.

It can be a user defined namespace, or the global namespace

In this case, calling built-in function this way merely for style / clarity but also may PHP Performance Tip: Use fully-qualified function calls when working with namespaces

Detailed Answer

So the usekeyword is scattered with these functionalities:

Namespace Import / Include: Modules (as packages), Classes, Traits, Interfaces, functions, constants
Use outer scope of variables in closure functions
Inject traits in Classes.

So my question refers to the first point Import / Include (Functions).

The namespace are somehow more evident when using Classes, as soon as a namespace is declared on the script, that script becomes relative to that namespace and not the global anymore. Hence, if you then call any classes that aren't defined in the very same script without specify the namespace, will result in error, example:

<?php    

    namespace A\B\C;
    class Exception extends \Exception
    {
    }
    
    $a = new Exception('hi'); // $a is an object of class A\B\C\Exception
    $b = new \Exception('hi'); // $b is an object of class Exception
    
    $c = new ArrayObject; // fatal error, class A\B\C\ArrayObject not found

?>

But even more clear example is this:

<?php
namespace A\B\C;
$b = new Exception('hi'); // Uncaught Error: Class "A\B\C\Exception"

By removing the namespace, it goes back to the global context

<?php
$b = new Exception('hi'); // no errors..

However, this is not happening with the const and functions, that's why is little less obvious and will silently fall back to the outer scope if not found.

For functions and constants, PHP will fall back to global functions or constants if a namespaced function or constant does not exist.

<?php
namespace A\B\C;

const E_ERROR = 45;
function strlen($str)
{
    return \strlen($str) - 1;
}

echo E_ERROR, "\n"; // prints "45"
echo INI_ALL, "\n"; // prints "7" - falls back to global INI_ALL

echo strlen('hi'), "\n"; // prints "1"
if (is_array('hi')) { // prints "is not array"
    echo "is array\n";
} else {
    echo "is not array\n";
}
?>

Finally, for some example regarding my question:
File functions.php
<?php

namespace MyApp\Package;

//use function var_dump;
function var_dump($var)
{
    echo '[', __NAMESPACE__, '] don\'t wnat to do that...<br/>';
}
function explode($separator, $some_string)
{
    echo '[', __NAMESPACE__, '] opps, bomb failed :(...<br/>';
}

File index.php

<?php

namespace MyApp;
require_once 'functions.php';

use MyApp\Package; // Demostrate that we can 'import' the entire package namespace
use function MyApp\Package\var_dump; // Import the function only

class A
{
    public static function say() {
        $a = "a";
        $b = "b";
        $data = ['a' => $a, 'b' => $b];
        echo var_dump($data); // --> Calling the namespaced function
    }
    public static function sayReal()
    {
        $a = "a";
        $b = "b";
        $data = ['a' => $a, 'b' => $b];
        echo \var_dump($data); // -> with \ we make the lookup fully-qualified and falls back to global
    }
    public static function explodeBadBomb($bomb="something to explode")
    {
        Package\explode(" ",$bomb); // -> demostrate that we can namespaced in withing blocks and statements
    }
    public static function explodeGooodBomb($bomb="something to explode")
    {
        echo print_r(\explode(" ", $bomb)); // again calling the built-in
    }
}
A::say();               // [MyApp\Package] don't wnat to do that...
A::sayReal();           // array(2) { ["a"]=> string(1) "a" ["b"]=> string(1) "b" }
A::explodeBadBomb();    // MyApp\Package] opps, bomb failed :(...
A::explodeGooodBomb();  // Array ( [0] => something [1] => to [2] => explode )

Documentation

Using namespaces: fallback to global function/constant
Namespaces
FAQ: things you need to know about namespaces

More on performance improvement

PHP RFC: declare(function_and_const_lookup='global')
[RFC] "use global functions/consts" statement

Using use within a function?

It's because you cannot declare it from within a function. From PHP: Using Namespaces:

The use keyword must be declared in the outermost scope of a file (the global scope) or inside namespace declarations. This is because the importing is done at compile time and not runtime, so it cannot be block scoped.

You'll have to move it outside of any function or class.

Purpose of PHP use keyword in array_map() function?

The callback passed to array_map() doesn't have access to outside variables so they must be passed using use.

You can read more about anonymous functions in the PHP documentation.

what is the use keyword in php symfony

This:

$def=function($name, $class, $args=[]) use ($container){
   return $container->doStuff($name, $class);
};

is roughly the same than:

$def = my_function($container, $name, $class);

function my_function($container, $name, $class, $args=[]) {
  return $container->doStuff($name, $class);
}

Use Keyword in Functions - PHP