从另一个对象获取调用者类实例

Question

Please take a look at this code:

class Foo {

   public $barInstance;

   public function test() {
      $this->barInstance = new Bar();
      $this->barInstance->fooInstance = $this;
      $this->barInstance->doSomethingWithFoo();
   }

}

class Bar {
   public $fooInstance;

   public function doSomethingWithFoo() {
       $this->fooInstance->something();
   }
}

$foo = new Foo();
$foo->test();

Question: is it possible to let the " $barInstance" know from which class it was created (or called) without having the following string: "$this->barInstance->fooInstance = $this;"

Answer 1

In theory, you might be able to do it with debug_backtrace() , which as objects in the stack trace, but you better not do it, it's not good coding. I think the best way for you would be to pass the parent object in Bar's ctor:

class Foo {

    public $barInstance;

    public function test() {
       $this->barInstance = new Bar($this);
       $this->barInstance->doSomethingWithFoo();
    }
}

class Bar {
    protected $fooInstance;

    public function __construct(Foo $parent) {
         $this->fooInstance = $parent;
    }

    public function doSomethingWithFoo() {
        $this->fooInstance->something();
    }
}

This limits the argument to being proper type ( Foo ), remove the type if it's not what you want. Passing it in the ctor would ensure Bar is never in the state when doSomethingWithFoo() would fail.