[RESOLVED] How to declare type Exception as optional function argument?

[JLHeidecker]

try{

}

catch(Exception $e){
myfunction();

}

myfunction(Exception $e){

}

My declaration for myfunction() is obviously wrong. I need to declare $e (which would be an exception object) as an optional parameter/argument of the function myfunction().

anyone know how to do that?

thanks folks! 🆒

[NogDog]

try{

}

catch(Exception $e){
  myfunction($e);

}

myfunction($e = null){
  if($e !== null) {
    if($e instanceof Exception) {
      // $e is an Exception
    }
    else {
      // $e is not an Exception
    }
  }
  else {
    // optional arg $e was not supplied to this function
  }
}

[JLHeidecker]

oh.

duh! 😃

thanks!