CODEWITHSUNDEEP
c++objectdynamic-cast
Johan Hjalmarsson
Johan Hjalmarsson

Reputation: 3493

dynamic cast doesn't recognize member

I'm working on a project where I have 2 classes: Room and EventRoom EventRoom inherits from Room and have a few more members.

In my code I do this(tmpPtr is a Room-pointer):

if(eventRoom)
     tmpPtr = dynamic_cast<EventRoom*>(tmpPtr);

and later when I try this:

if(line == "false;")
     tmpPtr->setComplete(false);

I get compilation errors. setComplete is a member of EventRoom

Short version: I want to create objects of type Room, and in some cases EventRoom. The code currently works for Room only, but 90% of the code would be identical for EventRoom. Any way of using the same code? (with dynamic_cast or something similiar)

Upvotes: 0

Views: 89

Answers (2)

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171117

The code which has to work with both Room and EventRoom (that is, it only works with the Room interface), has to work through a pointer statically typed Room*.

The code which uses specifics of EventRoom has to work through a pointer statically typed EventRoom*. So example code could look like this:

void someFunction(Room *myRoom) {
  // doRoomStuff must be a function declared in Room.
  myRoom->doRoomStuff();

  // Try to determin if the room is an event room or not. This will
  // occur at runtime.
  EventRoom *myEventRoom = dynamic_cast<EventRoom*>(myRoom);

  if (myEventRoom) {
    // doEventRoomStuff is a function declared in EventRoom. You cannot call
    // myRoom->doEventRoomStuff even if 'myRoom' points to an object that is
    // actually an EventRoom. You must do that through a pointer of type
    // EventRoom.
    myEventRoom->doEventRoomStuff();

    // doRoomStuff2 is a function that is declared in Room. Since every 
    // EventRoom is also a room, every EventRoom can do everything that
    // rooms can do.
    myEventRoom->doRoomStuff2();
  }

  myRoom->doRoomStuff3();
}

You can access Room members through a EventRoom* variable, but not vice versa.

Upvotes: 1

juanchopanza
juanchopanza

Reputation: 227390

You need tmpPtr to be an EventRoot pointer.

EventRoom* tmpPtr;

if(eventRoom)
     tmpPtr = dynamic_cast<EventRoom*>(tmpPtr);

You can only call Room public methods on a Room pointer. You cannot call EventRoom-only methods.

Upvotes: 3

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