Question

I’m using Type Hinting on a class that gets passed an object into its constructor. The passed object (Bar) exists in its own namespace. This is all fine until I swap out the passed object with a fake one (for unit testing purposes). My test Bar object resides in its own namespace and now an error is (quite rightly) thrown.

use Some\Place\Bar
class Foo {
    // Passing a test Bar into here now throws error.
    public function __construct(Bar $bar) {
        $this->bar = $bar;
    }

}

I’m new to PHP so unsure on what is possible. Is there a way round this or do I just need to drop Type Hinting from the class (which is my current solution)?

Answer 1

2 of the several options :

pass an object in the constructor that implement an interface IBar :

 interface IBar{
  function baz();
 }

 class Bar implements IBar{
    function baz(){ return $this->getSomethingExpensiveFromDatabase();}
 }

 class FakeBar implements IBar{
    function baz(){ return "baz";}
 }

 class Foo {
 function __construct(IBar $bar) ...

or dont use type hinting

function __construct($bar) 

the point is , the signature of a method should never depend on an implementation , always an interface.

EDIT : so you just need to import the interface then the fake class wherever it comes from. FakeBar can then mock Bar.