Reversing an array - Eloquent javascript chapter 4

Question

I am trying to work through this problem and I am having trouble understanding why the function reverseArrayInPlace isn't doing what I want it to. The console.log in the function will return a reversed array, but when I return the same thing, I don't get the reversed array.

I am a beginner so please dumb down the answers alot. Thanks

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

function reverseArrayInPlace(array){
  console.log(reverseArray(array));
  return reverseArray(array);
}
    

//console.log(reverseArray(["A", "B", "C"]));
// → ["C", "B", "A"];
var arrayValue = [1, 2, 3, 4, 5];
reverseArrayInPlace(arrayValue);
console.log(arrayValue);
// → [5, 4, 3, 2, 1]

Edit: Thanks for the replies. Can I get some feedback on this function:

 function reverseArrayInPlace(array){
  for (i=1; i<array.length; i++){
    var x = array[i];
    array.splice(i,1);
    array.unshift(x);
  }
  return array;
}

It seems to be working for me. Can you see anything wrong with it?

reverseArrayInPlace should actually be called createNewArrayWithValuesReversed (ie, the name of the method lies) — Jamiec
– Jamiec, Commented Feb 19, 2015 at 17:54
There's some good discussion on the same exercise here: stackoverflow.com/questions/27647032/… — Jordan Running
– Jordan Running, Commented Feb 19, 2015 at 17:59

konkked · Accepted Answer · 2015-02-19 18:20:52Z

1

You aren't reversing the array in place

In the function you are creating a new array and then adding the elements in reverse order

in the first line of your function you create a new variable

var x = [];

Unless you want to change your implementation to handle both cases in reverseArray you need to implement a different solution in the reverseArrayInPlace function

You will need to

Determine the element to swap from the front
Determine the element to swap from the back
Know when you have reached the middle of the array and stop

You can optionally return the array if you would like to, but since the change is made in place you shouldn't need to

edited Feb 19, 2015 at 18:20

answered Feb 19, 2015 at 17:59

konkked

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4 Comments

Plato Over a year ago

Actually I think it'll work if he just sets array = reverseArray(array); return array at the end of reverseArrayInPlace

konkked Over a year ago

You aren't actually reversing in place, var x = []; creates a new array, you are returning that at the end, so you aren't doing the change in place

Plato Over a year ago

to me 'in place' simply means 'the input object has been transformed into the output object at the end of the operation', not that there was no temporary array created

konkked Over a year ago

thats not in place, in place means modifying an existing object, you aren't transforming the object by using equal you are replacing the variable that previous lived at array with a new value (that the statement var x=[]; ) created

Prisoner · Accepted Answer · 2015-02-19 17:53:46Z

0

The reverseArrayInPlace() function isn't actually working in place. It is calling reverseArray() which is creating a new array to store the elements. As such, it needs to be returning this value.

answered Feb 19, 2015 at 17:53

Prisoner

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Comments

Plato · Accepted Answer · 2015-02-19 18:05:04Z

0

why not just...

var arrayValue = [1, 2, 3, 4, 5];
arrayValue = arrayValue.reverse();
console.log(arrayValue);

edit: nvm i see you are workign through a tutorial. imo this is the most elegant way though.

answered Feb 19, 2015 at 18:05

Plato

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1 Comment

konkked Over a year ago

I made this mistake too, thought he wanted an elegant solution implementation, but wanted help through tutorial

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