How to pass an interface as a parameter?

Question

PHP

There is a class and an interface

<?php
class Test
{
  public function __construct(SomeInterface $someInterface){
      $this->someInterface = $someInterface;	
  }
}

interface SomeInterface 
{
  public function foo($tmp){}
}

How to pass an interface to an instance of the Test class?

Answer 1

class Test
{
    public function __construct(SomeInterface $someInterface)
    {
        $this->someInterface = $someInterface;
    }
}

interface SomeInterface
{
    public function foo($bar);
}

class SomeClass implements SomeInterface
{
    public function foo($bar)
    {
    }
}

$a = new SomeClass("bar");

$b = new Test($a);

Answer 2

Interface - nothing. An object that implements an interface is easy. Understand the difference and you will be happy.