Sort numbers using special function in Javascript

Question

I have this way of sorting the array:

const arr = [0,6,8,7,1,2,3];

const sortBubble = () => {
  for (let index = 0; index < arr.length; index++) {
    for (let j = 0; j < arr.length; j++) {
      if(arr[j] > arr[j+1]) {
        const temp = arr[j];
         arr[j] = arr[j + 1]
         arr[j + 1] = temp;
      }
    }
  }

  return arr;
}

console.log(sortBubble())

Also this function:

const arr = [0,6,8,7,1,2,3];

const sortBubble = () => {
  for (let index = 0; index < arr.length; index++) {
    for (let j = 0; j < arr.length; j++) {
      if(arr[j] > arr[j+1]) {
         arr[j] = arr[j + 1]
         arr[j + 1] = arr[j];
      }
    }
  }

  return arr;
}

console.log(sortBubble())

I can't understand, why in the last function i get a different result than in the first function. How you can see in the last function i didn't use: temp variable, but anyway in my vision temp and arr[j] should be equal and i expect in both function the result as in first function, but last one has a different result.
Why the last function acts different if i don't use temp variable?

arr[j] = arr[j + 1] overwrites arr[j]. arr[j + 1] = arr[j]; in the next line is equivalent to arr[j + 1] = arr[j + 1]; — user15388024
– user15388024, Commented Jun 13, 2021 at 20:27
The values in the array are numbers and not objects, that means that when you assign them a value it is the actual value and not the reference to the object (it is not an object). This means that you must save the current value in temp var and if not you overwrite it the original value will be deleted. — Yair I
– Yair I, Commented Jun 13, 2021 at 20:30

Ant · Accepted Answer · 2021-06-13 20:28:37Z

1

If you take a closer look at your second code you are doing the following:

      if(arr[j] > arr[j+1]) {
         arr[j] = arr[j + 1]
         arr[j + 1] = arr[j];
      }

So you set arr[j] equal to arr[j+1], the second line makes no change to the array, as you have over-written arr[j]'s value

answered Jun 13, 2021 at 20:28

Ant

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Guerric P · Accepted Answer · 2021-06-13 20:29:49Z

1

If you want to swap items in an array, you have to use a temporary variable as a buffer to store one of the two values you're swapping:

In your second example, let's say you have this array:

const arr = [5, 6];

Now let's say j equals 0: arr[j] = arr[j + 1] will make your array become: [6, 6]

Then you execute arr[j + 1] = arr[j] which gives [6, 6] and you cannot retrieve 5 anymore as there is no more reference on it.

answered Jun 13, 2021 at 20:29

Guerric P

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user15388024 · Accepted Answer · 2021-06-13 20:30:31Z

0

arr[j] = arr[j + 1] overwrites arr[j]. arr[j + 1] = arr[j]; in the next line is equivalent to arr[j + 1] = arr[j + 1];. You can avoid an explicit temp variable with

if(arr[j] > arr[j+1]) {
  [arr[j], arr[j + 1]] = [arr[j + 1], arr[j]];
}

but of course internally the temporary value has to be stored.

Example:

const arr = [0,6,8,7,1,2,3];

const sortBubble = () => {
  for (let index = 0; index < arr.length; index++) {
    for (let j = 0; j < arr.length; j++) {
      if(arr[j] > arr[j+1]) {
         [arr[j], arr[j + 1]] = [arr[j + 1], arr[j]];
      }
    }
  }

  return arr;
}

console.log(sortBubble())

answered Jun 13, 2021 at 20:30

user15388024

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