how can empty slots appear in arrays?

Question

I'm quite new to javascript and I can't find out at which point an undefined object is inserted into my array:

function foo(arrA){
  if(!arrA || arrA[0] === 0)
       return undefined;

  let arrB = [];
  let prevEntry = 1;

  for(let i=0; i < arrA.length; i++) {
      if (2*prevEntry == arrA[i]) {
          prevEntry *= 2;
          arrB[i] = (Math.pow(arrA[i], 2)).toString();
      }
  }

  return arrB;
}

foo([2,4,6,8,9,15,16,27]);

As output I get an array with powers of 2, as expected, but it's interlaced with undefined objects and I just don't see why.

(7) ["4", "16", undefined × 1, "64", undefined × 2, "256"]

Thanks for your help in advance! :-)

if means you dont set a value into ArrB at each ArrA. So there are empty positions — Jonas Wilms
– Jonas Wilms, Commented Jul 1, 2017 at 11:53
Whenever you don't know where it crashes, stick to a try-catch block with a 'break' in it or print something in the log like the variable 'i'. And as already mentioned - pushing will solve it. — Garamaru
– Garamaru, Commented Jul 1, 2017 at 12:06

Jonas Wilms · Accepted Answer · 2017-07-01 11:55:15Z

1

You may want to push:

for(let i=0; i <arrA.length; i++){
  if (2*prevEntry == arrA[i]){
    prevEntry *= 2;
    arrB.push((Math.pow(arrA[i], 2)).toString());
  }
}

answered Jul 1, 2017 at 11:55

Jonas Wilms

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Sebastian Krysiak · Accepted Answer · 2017-07-01 12:01:32Z

0

When you add element to array at specific index, every not defined previously index becomes initialized with undefined value.

For example when you have

var array = [];
array[3] = 1

console log will provide an output

[undefined, undefined, undefined, 1]

so the solution is , as Jonas wrote, to simply push the elements to array instead of inserting them at specific index.

answered Jul 1, 2017 at 12:01

Sebastian Krysiak

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2 Comments

Milli Over a year ago

oh my god... i am sorry, that's such a stupid mistake! I should have seen that myself.

RandomB Over a year ago

In node.js it looks differently: [ <3 empty items>, 1 ] and in Opera - [empty × 3, 1]

Hassan Imam · Accepted Answer · 2017-07-01 12:09:45Z

0

Due to this line 2*prevEntry == arrA[i], your code was skipping some of the indexes in arrA and Since the indexes of both arrA and arrB were same, arrB had no value i.e. undefined there. Instead of inserting square in arrB by index, use push(), this will only insert squared number.

function foo(arrA){
  if(!arrA || arrA[0]===0)
    return undefined;

  let arrB =[];
  let prevEntry = 1;

  for(let i=0; i <arrA.length; i++){
    if (2*prevEntry == arrA[i]){
      prevEntry *= 2;
      arrB.push(Math.pow(arrA[i], 2)).toString();
    }
  }
  return arrB;
}
console.log(foo([2,4,6,8,9,15,16,27]));

answered Jul 1, 2017 at 12:09

Hassan Imam

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