How to call method in lambda expression

Question

I have 3 ways to swap 2 variables (basically 3 different algorithms). Since you can't pass a method as a parameter in Java, I thought this would be a good time to use lambda expressions.

chooseSwapMethod(int selection) {
    if(selection == 1) {
        functionThatUsesSwap(
            (int[] arr, int x, int y) -> {
                int holder = arr[x];
                arr[x] = arr[y];
                arr[y] = holder;
        });
    }
    else if(selection == 2)
    {
        functionThatUsesSwap(
            (int[] arr, int x, int y) -> {
                arr[x] -= arr[y];
                arr[y] = arr[x]-arr[y];
                arr[x] -= arr[y];
        });
    }
    else if(selection == 3) {
                functionThatUsesSwap(
            (int[] arr, int x, int y) -> {
                arr[x] = arr[x]^arr[y];
                arr[y] = arr[y]^arr[x];
                arr[x] = arr[x]^arr[y];
        });
    }
    else {
        throw new IllegalArgumentEarr[x]ception();
    }
}

but in the method functionThatUsesSwap how do you actually use the swap? Am I not understanding lambda expressions clearly? For example

public void functionThatUsesSwap(Swaper s)
{
    int[] someArr = {1, 2};
    s.doSwap(someArr, 0, 1);//this is where I’m stuck
    System.out.println(“a: “+someArr[0]+” b: “+someArr[1]);//this should print out a: 2 b: 1
}

Answer 1

Java is pass by value, that means that in the following:

int a = 5;
int b = 6;
swap(a,b);
System.out.println(a+" "+b);

There is no way for the function swap to change value of a or b and the result will always be 5 6.

What you can do is:

1. pass and arrays of 2 numbers into the swap method, and swap numbers inside that array.
2. make a class to hold 2 numbers and pass that.

Going with possibility 2:

class Pair {
    int a, b;
}

@FunctionalInterface
interface Swapper {
    void swap(Pair p);
}

void main() {
    Pair p = new Pair();
    p.a = 5;
    p.b = 6;
    Swapper swapper = (v -> {
        v.a ^= v.b;
        v.b ^= v.a;
        v.a ^= v.b;
    });
    swapper.swap(p);
    System.out.println(p.a + " " + p.b);
}

Result: 6 5. Note that your claim that you can't pass a method as a parameter in Java isn't entirely true, since you can pass interfaces instead.

EDIT:

There is yet another approach (I haven't thought of this earlier because Interger class is immutable). You can create a mutable (= changable) object vrapper for integer value, something like:

class IntVrapper {
    public int value;
}

Then your swap method could swap data between those two objects.