What happens in this code block during php oop? - php

Question

Could someone please explain the third line where Request and $request is used. It would be great if you could provide me a link having an explanation of the same? I just want to know what is happening there.
<?php
class xyz {
public function foo(Request $request){
//some code
}
}

Answer 1

Type hinting:
http://php.net/manual/en/language.oop5.typehinting.php
<?php
// An example class
class MyClass
{
/**
* A test function
*
* First parameter must be an object of type OtherClass
*/
public function test(OtherClass $otherclass) {
echo $otherclass->var;
}
/**
* Another test function
*
* First parameter must be an array
*/
public function test_array(array $input_array) {
print_r($input_array);
}
}
// Another example class
class OtherClass {
public $var = 'Hello World';
}
It throws an error if the argument is not of the type specified:
<?php
// An instance of each class
$myclass = new MyClass;
$otherclass = new OtherClass;
// Fatal Error: Argument 1 must be an object of class OtherClass
$myclass->test('hello');
// Fatal Error: Argument 1 must be an instance of OtherClass
$foo = new stdClass;
$myclass->test($foo);
// Fatal Error: Argument 1 must not be null
$myclass->test(null);
// Works: Prints Hello World
$myclass->test($otherclass);
// Fatal Error: Argument 1 must be an array
$myclass->test_array('a string');
// Works: Prints the array
$myclass->test_array(array('a', 'b', 'c'));
?>

Answer 2

Arguments:
http://php.net/manual/en/functions.arguments.php
Type hints:
http://php.net/manual/en/language.oop5.typehinting.php

Answer 3

An Object of the type Request is being passed to function foo.
It is made available to the function foo in a private variable named $request.

Answer 4

This is a type hint, to tell php to expect an object wich has
$request instanceof Request == true
Please note that this will actually not ensure anything. If $request is null or another object, there will most likely only a catchable fatal error be thrown, and so you have to test for a valid value anyway.

Answer 5

The third line defines a class method called foo that can get a $request argument of type "Request".
This is a security measure for the class developer. Determine that
<?php
class User
{
private $username;
public function get_username()
{
return $this->username;
}
}
class xyz()
{
public function foo(User $currentUser)
{
$currentUser->get_username();
}
}
$x = new xyz();
$u = new User();
$x->foo($u); // That will not produce any error because we pass an Object argument of type User
$name = "my_name";
$x->foo($name); // This will produce an error because we pass a wrong type of argument
?>