Create lambda two dimensional array

Question

So I've searched through several websites and others questions on this, and none seem to have the answer that works for me. I have code that works, and my programming mentor suggested I change the chained if/else if to using a lambda table instead. I asked about using something of a hash table, and he said using a hash for only 9 items (the real program has 9 if/else statements) would be a waste.

I will post the working code using both the if/else if and the hash table, keeping it limited to 3 items to keep the code short and sweet.

Here is the code for if/else if...

public class testLambda {
    String[] numArray = {"One", "Two", "Three"};

    testLambda(String num){
        if (num.equals(numArray[0])){
            printStringOne();
        } else if (num.equals(numArray[1])){
            printStringTwo();
        } else if (num.equals(numArray[2])){
            printStringThree();
        }
    }

    private void printStringOne(){
        System.out.println("1");
    }

    private void printStringTwo(){
        System.out.println("2");
    }

    private void printStringThree(){
        System.out.println("3");
    }

    public static void main(String[] args) {
        new testLambda("One");
        new testLambda("Three");
        new testLambda("Two");
    }
}

results in system printing...

1
3
2

Here is the code for a hash table

import java.util.HashMap;
import java.util.Map;

public class testLambda {
    String[] numArray = {"One", "Two", "Three"};

    testLambda(String num){
        Map<String, Runnable> printNumber = new HashMap<>();

        printNumber.put(numArray[0], () -> printStringOne());
        printNumber.put(numArray[1], () -> printStringTwo());
        printNumber.put(numArray[2], () -> printStringThree());

        printNumber.get(num).run();
    }

    private void printStringOne(){
        System.out.println("1");
    }

    private void printStringTwo(){
        System.out.println("2");
    }

    private void printStringThree(){
        System.out.println("3");
    }

    public static void main(String[] args) {
        new testLambda("Three");
        new testLambda("One");
        new testLambda("Two");
    }
}

results in system printing...

3
1
2

and now... the lambda. From what I have read, an interface is required. Keep in mind, I can't use extends, because my application is already extending a different class(java doesn't support multiple inheritance) Here is what I have conjured so far (not working):

public class testLambda {
    String[] numArray = {"One", "Two", "Three"};

    public interface PrintNumber{
        void printNumber();
    }

    testLambda(String num){
        PrintNumber[] printNumbers = new PrintNumber[]{
                new PrintNumber() {public void printNumber(){printStringOne();}},
                new PrintNumber() {public void printNumber(){printStringTwo();}},
                new PrintNumber() {public void printNumber(){printStringThree();}}
        };

        for (int n = 0; n < numArray.length; n++){
            if (num.equals(numArray[n])){
                printNumbers[n];
            }
        }
    }

    private void printStringOne(){
        System.out.println("1");
    }

    private void printStringTwo(){
        System.out.println("2");
    }

    private void printStringThree(){
        System.out.println("3");
    }

    public static void main(String[] args) {
        new testLambda("Three");
        new testLambda("Two");
        new testLambda("One");
    }
}

This results in a compile error. Can anyone tell me what I'm doing wrong? I'm really new to the lambda algorithm.

Note that, you can just write new PrintNumber[] { () -> printStringOne(), () -> printStringTwo(), () -> printStringThree()} (same as you did with your Map<.., Runnable>). You do not necessarily need a new interface. Now I don't know why it should get simpler with lambdas (or a lambda table?) and I do not know why it should be a waste to use a Map in this case. I wouldn't place a Runnable in there, but that's another thing. I would rather ask your mentor, what's the point of using a lambda table or why he/she suggested it. — Roland
– Roland, Commented May 15, 2017 at 10:47
You're right! This is much cleaner. Thanks a bunch :) Also, when I meet with him tomorrow, i'll ask him your question and get his response. Also, it looks like the interface is still required for creating the object array CreateCharacter[] createCharacters = new CreateCharacter[]{...};. — Fiddle Freak
– Fiddle Freak, Commented May 15, 2017 at 11:02
Not necessarily. You could also just use Runnable[]. However using your own interface makes sense to make the purpose (and code) clearer. — Roland
– Roland, Commented May 15, 2017 at 11:06
Oh yeah, thanks :) but how would I then call the index of the Runnable array? printNumbers[n].printNumber(); as printNumber() is red out. — Fiddle Freak
– Fiddle Freak, Commented May 15, 2017 at 11:10
nevermind, I found it printNumbers[n].run(); Thanks again :) — Fiddle Freak
– Fiddle Freak, Commented May 15, 2017 at 11:13

Klitos Kyriacou · Accepted Answer · 2017-05-15 10:43:07Z

3

The compilation error is due to this statement:

numArray[n];

which is not a valid statement. What you wanted was this:

    for (int n = 0; n < numArray.length; n++){
        if (num.equals(numArray[n])){
            createCharacters[n].printNumber();
        }
    }

However, this relies on keeping two separate arrays in sync, and is therefore error prone. I suggest using a single HashMap for everything, and getting rid of your original numArray in favour of the HashMap:

public class testLambda {
    Map<String, Runnable> printNumber = new HashMap<>();
    static {
        printNumber.put("One", () -> printStringOne());
        printNumber.put("Two", () -> printStringTwo());
        printNumber.put("Three", () -> printStringThree());
    }

    testLambda(String num){
        printNumber.get(num).run();  // Add some checking here for robustness
    }

By the way, what you call a "lambda table" doesn't necessarily mean it can't be a HashMap. In fact, the above can be called a lambda table. The () -> printStringXXX(); are lambda expressions, and the map is a table of strings to lambdas.

answered May 15, 2017 at 10:43

Klitos Kyriacou

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2 Comments

Fiddle Freak Over a year ago

Sorry, I edited the question and re-wrote the constructor. It's still the statement error, but the intent of use is different. This is my fault.

Fiddle Freak Over a year ago

omg I was so close. I was just missing the .printNumber() in printNumbers[n], making it printNumbers[n].printNumber(); as the correct answer. Thanks :)

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