PHP OOP Interface typing inheritance

Question

A code like that :

interface entite
{

}

class Foo implements entite
{

}


$foo = new foo;
if( $foo instanceof entite ) echo "he is";

Displays "he is". Foo inherit type "entite" from the interface But when try that :

class FooDeleter implements deleter
{
public function __construct(Foo $Foo)
{

}
}

interface deleter
{
public function __construct(entite $entite);
}

Gives me :

Fatal error: Declaration of FooDeleter::__construct() must be compatible with deleter::__construct(entite $entite)

Why ? How to ? =(

Edit : The unique way is actually define the typed deleter like that :

class FooDeleter implements deleter
{
public function __construct(entite $Foo)
{
    if( $Foo instanceof Foo ) { ... }       
}
}

I'm very surprised PHP lets you define a constructor in an interface. Seems like a terrible idea to me. Related question here - stackoverflow.com/questions/783046/… — Phil
– Phil, Commented Mar 5, 2013 at 3:12
A ctor is an implementation detail and should not go into the interface. — Gordon
– Gordon, Commented Mar 6, 2013 at 9:19

Seth Battin · Accepted Answer · 2013-03-05 03:14:43Z

2

By declaring your FooDeleter constructor with a type hint that is more strict than the interface, you are violating the interface.

If you changed your constructor to

public function __construct(entite $Foo)

...then you could still pass in a Foo object, and the interface would be implemented correctly.

answered Mar 5, 2013 at 3:14

Seth Battin

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Community · Accepted Answer · 2020-06-20 09:12:55Z

1

According to PHP Document:

Note:

The class implementing the interface must use the exact same method signatures as are defined in the interface. Not doing so will result in a fatal error.

Function name and argument numbers and argument type (if specified) are part (all?) of the method signature, so you'll have to declare an exact same method.

You can still use new FooDeleter($foo).

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Mar 5, 2013 at 3:15

Passerby

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