Array.push() is not a function - attempting to push numbers to an array, but the array has changed type

Question

I'm 3 months into Javascript, and find it has no respect for its own declarations.

I have declared an empty array, and want to push a number to it, so that it returns an array of numbers (i.e., separable). However, I am told .push() is not a function. Concat() will not work either, and of course += ruins the algorithm.

How do I get it to accept each next value into an array? I have tried using 'new Array()', but it does not assist.

Similar to, but not the same as, Array.push throwing TypeError for array.push is not a function

In brief:

const fn = (n) => {
  let factors = [];
  let index = 1;

  while (n % 2 == 0){
    let out = 2 ** index;
    factors.push(out);
    index++;
    n /= 2;
  }
}

This returns:

Uncaught TypeError: factors.push is not a function

(I have left out a lot of code that does not affect the issue.)

Edit update: WIP. Apparently, the loop this is enclosed in has an effect, as do other variable declarations.

With any luck, I will return with a different question, having solved the initial problem.

The problem could not be found with this code.

Hyunjune Kim
– Hyunjune Kim

2022-01-20 08:17:40 +00:00
Commented Jan 20, 2022 at 8:17 — Hyunjune Kim
– Hyunjune Kim, Commented Jan 20, 2022 at 8:17

Jay Ayerson · Accepted Answer · 2022-01-21 07:17:24Z

0

Nested loops seems to have been the actual problem, which means others could not reproduce.

const fn = (m, n) => {
let factors = [];
let index = 1;
for (let i = m; i <= n; i++) {
while (i % 2 == 0){
  let out = 2 ** index;
  factors.push(out);
  index++;
  n /= 2;
  } 
}}

For unknown reasons, this causes a type mismatch. The declaration of the array appears to be the cause.

This problem has been abandoned as the initiating function was completed by other methods.

answered Jan 21, 2022 at 7:17

Jay Ayerson

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Comments

DUDEZKIE · Accepted Answer · 2022-01-20 08:15:35Z

-1

Here! its working well, what might went wrong on your side?

const fn = (n) => {
  let index = 1;
  let factors = [];
  while (n % 2 == 0){
    let out = 2 ** index;
    factors.push(out);
    index++;
    n /= 2;
  }
  return factors
}


 document.write(fn(400));

answered Jan 20, 2022 at 8:15

DUDEZKIE

4151 gold badge5 silver badges13 bronze badges

3 Comments

Quentin Over a year ago

This reads more like a game of spot-the-difference than an answer. Why should the changes you made solve push not being a function? I couldn’t reproduce the issue in the first place, and your changes seem of take place only after the call to push that the OP says doesn’t work.

Jay Ayerson Over a year ago

I have discovered that apparently, wrapping it all in a for() loop makes a difference. I do not understand why, but I believe I shall edit the question.

Jay Ayerson Over a year ago

However, since I'm getting a different error by only wrapping it in a for loop, we'll call that progress.

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