Why this code is returning undefined?

Question

var rarr = [];
var reverseArray = function(arr) {
  if(arr[0]) {
     rarr.push(arr[arr.length-1]);
     arr.pop();
     reverseArray(arr);
  } else {
    return rarr;
  }
}

console.log(reverseArray(["A", "B", "C"]));

While debugging value of rarr is ['C','B','A'] at the end, instead it is returning undefined. Thanks in advance.

notwithstanding the `return` bug in the answers given so far, testing `arr[0]` to get loop termination is _horrible_ - it'll break if any element of your array holds a falsey value (e.g. zero) — Alnitak, Apr 25 '16 at 13:05

gurvinder372 · Accepted Answer · 2016-04-25T13:07:01.643

1

While debugging value of rarr is ['C','B','A'] at the end, instead it is returning undefined

You are not returning the value that will be returned by recursive calls

replace

reverseArray(arr);

with

return reverseArray(arr);

Or simply

console.log(["A", "B", "C"].reverse());

edited Apr 25 '16 at 13:07

answered Apr 25 '16 at 13:03

gurvinder372

66,980
10
72
94

Oops.. missed it.. Thanks. Any other optimized way that you can suggest? – kamal kokne Apr 25 '16 at 13:06
@kamalkokne just `["A", "B", "C"].reverse()` – gurvinder372 Apr 25 '16 at 13:07
reverse() is perfect but I was looking for optimized solution w/o reverse method. Thanks – kamal kokne Apr 25 '16 at 13:11
@kamalkokne something like `var reverseArray = function(arr) { var rarr = []; while(arr.length > 0 ) { rarr.push(arr.pop()); } return rarr; }` without recursion. – gurvinder372 Apr 25 '16 at 13:14

score 1 · Answer 2 · answered Apr 25 '16 at 13:04

you forgot a return in your if... try this:

var rarr = [];
var reverseArray = function(arr) {
  if(arr[0]) {
     rarr.push(arr[arr.length-1]);
     arr.pop();
     return reverseArray(arr);
  } else {
    return rarr;
  }
}

console.log(reverseArray(["A", "B", "C"]));

Why this code is returning undefined?

2 Answers2