Initialising object with variable type

Question

I have one Worker interface with one method:

interface Worker {
    public void work()
}

I have 2 classes that implements Worker,

class RoadWorker implements Worker {
    public void setPropertyA() {}
    public void work() {}
}

another one,

class GardenWorker implements Worker {
    public void setPropertyB() {}
    public void work() {}
}

In my Application class - based on some input flag - I want to instantiate one specific type of worker...

class Application {
    // flag
    String whichWorker = "Road";

    // instantiate
    if (whichWorker == "Road") {
        RoadWorker worker = new RoadWorker();
        worker.setPropertyA();
    } else {
        GardenWorker worker = new GardenWorker();
        worker.setPropertyB();
    }

    // use
    worker.work();  <----- OF COURSE THIS DOES NOT WORK (no reference)

So, I tried this -

class Application {
    // flag
    String whichWorker = "Road";
    Worker worker;

    // instantiate
    if (whichWorker == "Road") {
        worker = new RoadWorker();
        worker.setPropertyA(); <----- DOES NOT WORK 
    } else {
        worker = new GardenWorker();
        worker.setPropertyB(); <----- DOES NOT WORK 
    }

    // use
    worker.work();

My question is - how do I design my program to achieve this requirement? I know one crude option is to define worker as Object but then I will have to do lots of lots of type casting that I don't want. Can anyone please suggest?

Do not use == to compare strings!!!!

ajb
– ajb

2014-12-24 18:37:53 +00:00
Commented Dec 24, 2014 at 18:37 — ajb
– ajb, Commented Dec 24, 2014 at 18:37

laune · Accepted Answer · 2014-12-24 18:24:33Z

2

Worker anyworker = null;
if ("Road".equals( whichWorker )) {
    RoadWorker worker = new RoadWorker();
    worker.setPropertyA();
    anyworker = worker;
} else {
    GardenWorker worker = new GardenWorker();
    worker.setPropertyB();
    anyworker = worker;
}

Or pass the specific property values in the subclass constructor.

answered Dec 24, 2014 at 18:24

laune

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1 Comment

hashbrown Over a year ago

Thanks. That looks simple to me. Would you know what is the dis/advantage of this approach over passing values through constructor approach?

Jack · Accepted Answer · 2014-12-24 18:25:05Z

1

Use constructors to set properties or a factory pattern.

The first solution is simpler but has its limits:

interface Worker {
    public void work()
}

class RoadWorker implements Worker {
  RoadWorker(PropertyA property) {
   this.property = property;
  }
}

void foo() {
  Worker worker = null;

  if (whichWorker.equals("road")) {
    worker = new RoadWorker(property);
  }

  worker.work();
}

Two side notes:

don't compare things with ==, use equals instead
use an Enum instead that a string to switch over the worker type, they exist right for this purpose

If you nee more flexibility then a factory method pattern could be more suitable.

answered Dec 24, 2014 at 18:25

Jack

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Comments

forgivenson · Accepted Answer · 2014-12-24 18:27:19Z

You can have the constructors for the worker implementations do the actions specific to those implementations.

RoadWorker:

class RoadWorker implements Worker {
    public RoadWorker() {
        this.setPropertyA();
    }

    public void setPropertyA() {}
    public void work() {}
}

GardenWorker:

class GardenWorker implements Worker {
    public GardenWorker() {
        this.setPropertyB();
    }

    public void setPropertyB() {}
    public void work() {}
}

Then remove those calls to the setProperty methods in your if statement:

class Application {
    // flag
    String whichWorker = "Road";
    Worker worker;

    // instantiate
    if (whichWorker.equals("Road")) {
        worker = new RoadWorker();
    } else {
        worker = new GardenWorker();
    }

    // use
    worker.work();

Note: I changed whichWorker == "Road" to whichWorker.equals("Road"), because for String comparison, you need to use equals instead of ==.

Ali Boutabaa · Accepted Answer · 2014-12-24 18:27:53Z

0

You can cast your worker instance with its proper instanciation type when you set its property.

For instance

if (whichWorker == "Road") {
    worker = new RoadWorker();
    ((RoadWorker)worker).setPropertyA();  
} else {
    worker = new GardenWorker();
    ((GardenWorker)worker).setPropertyB();
}

This should work

answered Dec 24, 2014 at 18:27

Ali Boutabaa

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