访客设计模式，如何处理If-else语句

Question

Problem statement:

There are three types of Machines, Machine1 Machine2 and Machine3. There are three types of validators, ValidatorX , ValidatorY , ValidatorZ.

Each Validator validates each machine differently.

Write Java classes to design the above problem.

The program runs from a main function , which already has the list of validators and machines. Each Validator should print a message during validation

"Validator X validating Machine Y";

To design this, First of all I have created one Machine class and 3 different Validator classes.

Machine.java

/**
 * Machine
 * 
 * @author sunny
 *
 */
public class Machine {

    private final String name;

    public Machine(final String name) {
        this.name = name;
    }

    public boolean validate(final Validator validator) {
        return validator.validate(this);
    }

    public String getName() {
        return name;
    }
}

Validator.java

/**
 * This is an abstract class for a Validator
 * 
 * @author sunny
 *
 */
public abstract class Validator {

    private String name;

    public Validator(final String name) {
        this.name = name;
    }

    public boolean validate(final Machine machine) {

        System.out.println("Validator " + getName() + " validating "
                + machine.getName());

        // Here I need to write login corresponding to the machine name
        // I need to write if machine Name is X then do this
        // if machine name is Y then do this.
        // I need to avoid this if else or switch case
        // Is there any other better way to achieve this.

        return true;
    }

    public String getName() {
        return name;
    }
}

ValidatorX.java

/**
 * This is one ValidatorX class that I have created
 * 
 * Similarly I have written ValidatorY and ValidatorZ classes
 * 
 * @author sunny
 *
 */
public class ValidatorX extends Validator {

    public ValidatorX() {
        super("X");
    }

    @Override
    public boolean validate(Machine machine) {
        super.validate(machine);
        return true;
    }
}

TestMain.java

import java.util.ArrayList;
import java.util.List;

/**
 * Test class I have written to test this.
 * 
 * @author sunny
 *
 */
public class TestMain {

    public static void main(String[] args) {

        Validator[] validators = { new ValidatorX(), new ValidatorY(),
                new ValidatorZ() };

        List<Machine> machines = new ArrayList<Machine>();
        machines.add(new Machine("Machine1"));
        machines.add(new Machine("Machine2"));
        machines.add(new Machine("Machine3"));

        for (int i = 0; i < validators.length; i++) {
            for (Machine machine : machines) {
                validators[i].validate(machine);
            }

        }
    }

}

In Validator.java (as you can see in the comments) I need to write login corresponding to the machine name, I need to write if machine Name is Machine1 then do this if machine name is Machine2 then do this. I need to avoid this if else or switch case. Is there any other better way to achieve this.

Answer 1

You might use an action map.

Map<String, Action> actionMap = new HashMap<>();
actionMap.put("Machine1", new SomeActionOne());
actionMap.put("Machine2", new SomeActionTwo());
actionMap.put("Machine3", new SomeActionThree());


for (int i = 0; i < validators.length; i++) {
   for (Machine machine : machines) {
       validators[i].validate(machine);
       Action action = actionMap.get(machine.getName());
       assert action != null;
       action.doSomething();
   }
}

Answer 2

Your use of the visitor pattern is fine, problem is the Machine Class.

From what I see, your machine class represent "depending" on name, different version/type of machine. Therefore you need the if/else to identify the machine and act accordingly.

Your problem coudl be fixe by Abstraction

AbstractMachine - would be the class that represent the shared functions to each of your machine type/categories>

So in it you would have at the very least the following functions :

public boolean Validate (...)
public string GetName()
abstract public  void MachineAction(...) = 0;

Therefore you would have to write up the class Machine1, Machine2, Machine3 and you would be obligated to "overload" MachineAction for each child class.

Your validator's validate function would then look like this :

 public boolean validate(final Machine machine) 
 {
    System.out.println("Validator " + getName() + " validating "
            + machine.getName());

    machine.MachineAction(...);    
    return true;
 }

By design, it is better than having a Dictionnary(Map) of functions and you are sure to always call the right one.

++ I would really give better name than Machine1, Machine2, Machine3 - there is something bigger than a "name" difference here

EDIT::

    MachineA : Child Class (inherits from) Machine

    //Since it inherits from Machine no need to reimplement 
    //Validate/GetName, but we need and it's obligated to 
    //reimplement MachineAction for this subclass.

    override public void MachineAction(...)
    {
       DoThisMachineA();
    }

    private void DoThisMachineA(){...}

while MachineB would look like this :

    MachineB : Child Class (inherits from) Machine

    //Since it inherits from Machine no need to reimplement 
    //Validate/GetName, but we need and it's obligated to 
    //reimplement MachineAction for this subclass.

    override public void MachineAction(...)
    {
       DoThisMachineB();
    }

    private void DoThisMachineB(){...}

** Note that DoThisMachineB() does not exist in the MachineA class, and would not be presented by the intellisense, and would not compile at all, and vice versa.

Yet in the visitor, all you would have to do is Call MachineAction(...) has the visitor knows it's a Machine but not which child-type of machine and that what-ever type of machine you are, you will act properly on a MachineAction(..) call. You can read on this, we call it polymorphism.