reversing an array in javascript

Question

I have a Javascript function called reverseArray that takes an array as the argument and returns a new array which has the same values as the input array in reverse order. I want to create a function called reverseArryInPlace which would change the value of the input array to the reverse order.

function reverseArray(inputArray) {
  var outputArray = [];
  for (var i = inputArray.length - 1; i >= 0; i--)
    outputArray.push(inputArray[i]);
  return outputArray;
}

function reverseArrayInPlace(inPlaceInputArray) {
  inPlaceInputArray = reverseArray(inPlaceInputArray);
  console.log('Inside reverseArrayInPlace: ' + inPlaceInputArray);
  return inPlaceInputArray;
}

var arrayValue = [1, 2, 3, 4, 5];
reverseArrayInPlace(arrayValue);
console.log('Outside reverseArrayInPlace: ' + arrayValue);
// Expected Value: [5, 4, 3, 2, 1]

Here is the result I get when I execute this chunk of code:

Inside reverseArrayInPlace: 5,4,3,2,1
Outside reverseArrayInPlace: 1,2,3,4,5

Within the reverseArrayInPlace function the arrayValue variable has been reversed as expected. Why is it that when I reference the same variable outside the reverseArrayInPlace function, it is back to the original order?

Answer 1

If you want to reverse it in place, you have to reverse it in place.

function reverseArrayInPlace(array) {
  for (let i = 0, j = array.length - 1; i < j; i++, j--) 
    [array[i], array[j]] = [array[j], array[i]];
}

const a = [1,2,3,4,5];
reverseArrayInPlace(a);
console.log(a);

Answer 2

The underlying problem here is that primitives are passed by value in JavaScript. See the following question for details:

Javascript by reference vs. by value

As a simple example, here is a function that attempts to mutate a string that was passed to it:

var outsideValue = 'foo'

function mutate(value) {
  value = 'fish'
}

mutate(outsideValue);
console.log(outsideValue);

However, the console output is foo.

This happens because the value variable within the mutate function is a variable that has a reference to outsideValue when the function is initially invoked. When it is assigned the new value within the function body, it merely changes the value variable to reference a new string. As a result, the outsideValue is untouched.

Check this answer for an example that reverses in place:

https://stackoverflow.com/a/43486686/249933

Notice, that it does not re-assign the function argument.

Answer 3

As others have said, arrays are objects so they're passed by reference. You have to modify the original array, not create a new one.

So here's another version of reverseInPlace, it uses shift to remove the last element from the array and splice to insert it in the new location:

function reverseInPlace(arr) {
  var i = arr.length;
  while (i--) {
    arr.splice(i, 0, arr.shift());
  }
}

var arr = [1,2,3,4,5];
console.log('Before: ' + arr.join());

reverseInPlace(arr);
console.log('After:  ' + arr.join());

For some fun, you can also leverage sort:

NB: this only works in some browsers, it's dependent on the built-in sort algorithm which is implementation dependent.

function reverseInPlace(arr) {
  arr.sort(() => 1);
}

var arr = [47, 95, 80, 62, 8, 34, 31, 17, 62, 17, 85, 72, 51, 20, 68, 60, 30, 84, 7, 34];
console.log('Before: ' + arr.join());

reverseInPlace(arr);
console.log('After : ' + arr.join());

Answer 4

function reverseArray(inputArray) {
  var outputArray = [];
  for (var i = inputArray.length - 1; i >= 0; i--)
    outputArray.push(inputArray[i]);
  return outputArray;
}

function reverseArrayInPlace(inPlaceInputArray) {
  inPlaceInputArray = reverseArray(inPlaceInputArray);
  console.log('Inside reverseArrayInPlace: ' + inPlaceInputArray);
  return inPlaceInputArray;
}
var arrayValue = [1, 2, 3, 4, 5];
**arrayValue = reverseArrayInPlace(arrayValue);**
alert('Outside reverseArrayInPlace: ' + arrayValue);
// Expected Value: [5, 4, 3, 2, 1]

//

your code is correct your just need to replace one line and write arrayValue = reverseArrayInPlace(arrayValue); instead of reverseArrayInPlace(arrayValue); Then arratValue will print expected values