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const myOldArr = [2, 3, 6];
const newArr = myOldArr.pop().shift();

Why does doing this give me a TypeError? Is there a way to do this without a loop? I want the new array to be stored inside a variable so the remaining numbers that don't get removed are stored inside an array of some sort.

Cody Gray - on strike
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user12159018
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    `pop` [returns the _element_ that you popped](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/pop). It is not an array that you can then use `shift` on. – Andy Oct 07 '19 at 10:52
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2 Answers2

1

.pop() returns the last element form your array and so .shift() won't work,

If you want to keep the myOldArr and create a new one without the first and last element, you can do this by using .slice(1, -1):

const myOldArr = [2, 3, 6];
const myNewArr = myOldArr.slice(1, -1);
console.log(myNewArr);

Or, if you don't mind moifying the original array, you can do this by not chaining both methods:

const myOldArr = [2, 3, 6];
myOldArr.pop()
myOldArr.shift();
console.log(myOldArr);
Nick Parsons
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0

Use them separately

Array.pop() and Array.shift() will return a value of array your trying to use shift in number 6 so it will throw error 

const myOldArr = [2, 3, 6];
myOldArr.pop();//last 
myOldArr.shift();//first
console.log(myOldArr)
Ravi Makwana
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ellipsis
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