bubbleSort and swap method not working [Java]

Question

In the main, whenever I call this method to sort the array, it stops as if it's waiting for a response. Any idea why it's not working?

public void bubbleSort(){
    boolean finished = false;
    boolean swapOccurred = false;
    int i = 0;
    int next = i + 1;

    while (finished == false)
    {

        for (; i < theArray.length - 1; i++)
        {
            if (theArray[i] > theArray[next])
            {
                swapValues(theArray[i], theArray[next]);
                swapOccurred = true;
            }
        }
        if (swapOccurred == false)
        {
            finished = true;
        }
    }
}

private void swapValues(int i, int next) {

    int temp;

    temp = i;
    i = next;
    next = temp;

}

I'm sorry! How much of the main will I add in? I only create a new object and call the bubbleSort(); method. — Markopolo, Feb 28 '15 at 01:48
Also consider adding a tag for the language this is in. We can guess, but why should we need to? — Jongware, Feb 28 '15 at 02:24
Sorry again! I'm pretty new to this, didn't realise. Updated, thanks! — Markopolo, Feb 28 '15 at 11:28

score 2 · Answer 1 · edited Feb 28 '15 at 02:50

2

If it is java, then

The next index is never updated so it stays 1 all the time.
The swap method has no side effect, you want to swap the elements in theArray, but in java methods arguments are passed as values so the next and i variables change the value only inside the swap method, they have nothing to do with the cells of theArray array.

edited Feb 28 '15 at 02:50

Siddharth

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answered Feb 28 '15 at 01:57

Zielu

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I'm confused,sorry. Do I need to return a value to recognise the change. How do I swap the elements of the array? Thank you – Markopolo Feb 28 '15 at 11:33
inside the loop by simple: int tmp = theArray[i]; theArray[i]=theArray[next]; theArray[next] = tmp if you want call the method, then it has to take 3 argu, the array, and the two indexes swap(int[] arr, int i, int j) {in tmp = arr[i], arr[i]= arr[j]; arr[j]=tmp;} – Zielu Feb 28 '15 at 13:22

score 1 · Answer 2 · edited Feb 28 '15 at 02:50

1

The problem is with your swap.

In C, arguments are passed by value, so when you do the swap, the values being passed in aren't affected, so nothing is happening. You need to pass in pointers to the numbers instead (and then dereference them inside):

private void swapValues(int *i, int *next) {
    int temp;

    temp = *i;
    *i = *next;
    *next = temp;
}

And then call it with the address of the variables to swap:

swapValues(&theArray[i], &theArray[next]);

Edit: Ha - @Zielu saw Java, I saw C. Same problem in both cases, and as @Zielu points out, you also need to increment next. I believe you actually just want to use [i+1] as your index in place of next.

edited Feb 28 '15 at 02:50

Siddharth

9,349
16
86
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answered Feb 28 '15 at 01:59

Dov Rosenberg

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I actually had i + 1 before the next variable but I was experimenting to see if it worked but substituting with the next variable. No luck – Markopolo Feb 28 '15 at 11:37
Next never changes, so you can get rid of it. What you need is i+1: swapValues(theArray[i],theArray[i+1]) - but you still have the problem of the swap doing nothing. Simplest solution in this case is to put the swap code in the for loop. Or, you could put the numbers to swap in an object and pass it to the swap, since objects are passed by reference. See the top answer to [this question](http://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value), as I think it will help your confusion – Dov Rosenberg Feb 28 '15 at 12:34

bubbleSort and swap method not working [Java]

2 Answers2