PHP interface in extended class not recognized as instance

Question

I have an abstract class A which is extended by class B which implements the interface I.

abstract class A {
  public function test(){
    return $this->getX();
  }
  abstract protected function getX();
}

class B extends A implements I {
  public function Test() {
    $x = this->getX();
    if (!$x instanceof I) {
      throw new RuntimeException("not an instance of I");
    }
  }

  public function getX() {
    $aCoordinates = array('x' => 1, 'y' => 4, 'z' => 5); 
    return $aCoordinates;
  }
}

interface I {}

The RuntimeException is always thrown, despite $x is an instance of I. Anyone an idea why this happens?

getX() returns an array. getX() is just an example for a function returning an array.

If I look at this correctly, $x here will be set to 10, which is not an instance of I, but just a normal integer, right? — p0358
– p0358, Commented Nov 18, 2018 at 22:52
I've updated my example code. Actually $x ist just the result of getX() — Jake
– Jake, Commented Nov 18, 2018 at 22:55
What does getX return? Whatever it is, it's not an instance of I either — Joni
– Joni, Commented Nov 18, 2018 at 22:58
@Jake if $x = getX(), therefore $x = array. An array is not an instance of your interface. This is why your exception is being thrown. — Gherkin
– Gherkin, Commented Nov 18, 2018 at 23:10

Rasclatt · Accepted Answer · 2018-11-19 07:05:36Z

1

The getX() method doesn't return any data So the value of $x is null; to have an object instance of I interface you need to put return $this in getX() method.

edited Nov 19, 2018 at 7:05

Rasclatt

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answered Nov 18, 2018 at 23:39

Dena

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2 Comments

Jake Over a year ago

actually it does. I just forgot it in the example code.

Dena Over a year ago

Now getX() returns an array and of course its not instance of I interface. You need put return $this; instead of return $aCoordinates;

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