Class name before variable in PHP function argument

Question

In lots of PHP scripts I see things such as this:

public function __construct(Container $ci) {

}

I understand what a constructor does and how to pass a variable. However I'm not sure what Container means in this example? Is this the equivalent of $ci = new Container; ?

type hinting

Mark Baker
– Mark Baker

2016-10-19 09:36:01 +00:00
Commented Oct 19, 2016 at 9:36 — Mark Baker
– Mark Baker, Commented Oct 19, 2016 at 9:36
@Faraz There is no PHP 6. Only PHP 5 and 7.

Charlotte Dunois
– Charlotte Dunois

2016-10-19 10:11:16 +00:00
Commented Oct 19, 2016 at 10:11 — Charlotte Dunois
– Charlotte Dunois, Commented Oct 19, 2016 at 10:11
@CharlotteDunois my bad php 7. current documentation

Faraz
– Faraz

2016-10-19 10:14:32 +00:00
Commented Oct 19, 2016 at 10:14 — Faraz
– Faraz, Commented Oct 19, 2016 at 10:14

Rax Weber · Accepted Answer · 2016-10-19 09:36:53Z

8

No, it is just type hinting the method's $ci parameter, meaning that you should pass an argument declared as an instance of Container like so:

$cont = new Container();

$obj = new YourClass($cont);

answered Oct 19, 2016 at 9:36

Rax Weber

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2 Comments

Maytyn Over a year ago

Not should but have to.

peter Over a year ago

If you do not pass a Container object to the constructor, it will ,throw a fatal error(Uncaught TypeError) and abort. See myprogrammingnotes.com/class-name-function-arguments.html

Uri · Accepted Answer · 2022-03-09 12:33:53Z

0

For those who are looking for "Type Hinting" and using the broken links above, in PHP this is now known as Type Declaration.

https://www.php.net/manual/en/language.types.declarations.php

answered Mar 9, 2022 at 12:33

Uri

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1 Comment

Community Over a year ago

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