php - check in constructor if class function is called by class instance

Question

Is it possible to do something like this one:

class conditional
    {

        function __construct()
        {
            if (launch() is called) {
                //return True 
                //function is called by class instance
            }else{
                //return False
                //launch() is not called
            }
        }

        public function launch(){
            echo "function is launcher";
        }
    }

I am trying to use a condition in class constructor to know if the class function is called by class instance or object as $class_instance->launch(). If the function is not called by class instance then it should run return else condition.

$class_instance->launch() doesn't trigger $class_instance::__construct method as you expect in the code above — RomanPerekhrest
– RomanPerekhrest, Commented Jan 21, 2017 at 16:17
@RomanPerekhrest So is there any other possible way to implement this one? — Syed Aqeel
– Syed Aqeel, Commented Jan 21, 2017 at 16:27

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

2

I don't think you can do what you want to do (nor I understand why would want to).

The constructor will run when you create the instance, before you are able to call any of the instance methods. Unless you have a way to delve into the future and know at instantiation time if any of the instance methods will be called later on, no dice :)

But you can do something that more or less looks like that using the __call magic method, even if probably doesn't make much sense in most scenarios.

E.g., a very simplified implementation:

class Magical {

   public function __call($method, $args) {

       switch($method) {
          case "launch":
             echo "magical launch() was called\n";
             $this->magicLaunch();
             break;
          default:
             throw new \Exception("Method not implemented");
       }
  }

  private function magicLaunch() {
      echo "magic!";
  }
}


// this is the only point where the constructor gets called
$magic = new Magical();

// since the launch method doesn't exist, the magic method __call is invoked
$magic->launch();

Outputs:

magical launch() was called

magic!

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jan 21, 2017 at 16:39

yivi

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1 Comment

Syed Aqeel Over a year ago

Yes this is the thing that I am finding... Thank you

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