let bigo = [[1,2,3],[4,5,6],[7,8,9]];
for(let i = bigo.length; i >= 0; i--){
console.log(bigo[i]);
}Result:
> undefined
> [ 7, 8, 9 ]
> [ 4, 5, 6 ]
> [ 1, 2, 3 ]
I keep getting undefined before the result show. What am I doing wrong.
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There is another problem, the index of the array in the computer starts from zero, so your i should be equal to bigo.length - 1 – AmelloAster May 23 '22 at 13:17
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Does this answer your question? [Loop through an array in JavaScript](https://stackoverflow.com/questions/3010840/loop-through-an-array-in-javascript) – 0stone0 May 23 '22 at 13:18
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Probably a typo but you log `bigone` while your variable is `bigo` – Marius ROBERT May 23 '22 at 13:18
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@Ben It's not even the same language – Marius ROBERT May 23 '22 at 13:25
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@MariusROBERT Yeah, I know, I've retracted the flag – Ben May 23 '22 at 13:26
4 Answers
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The last index is length - 1, not length, which is why your first result will be out of bounds and show undefined.
Nicky
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1@0stone0 this question has probably already been asked but you need to find it to close this one – Marius ROBERT May 23 '22 at 13:18
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bigo is 3 items long but does not have anything at index position 3. This is because indices count from 0. You need to write let i = bigo.length - 1
Ben
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Save yourself the hassle and stop using index-based loops.
let bigo = [[1,2,3],[4,5,6],[7,8,9]];
for (let arr of [...bigo].reverse()) {
console.log(arr);
}
connexo
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If bigo.length = 9 the programm will ask for bigone[9], after that it will decrease i with 1. I think you messed up the order.
Erik Leusink
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21) ```bigo.length``` is 3, as it is a nested array. 2) Just stating "I think you messed up the order" is nowhere near specific enough for an answer. Be more specific and/or provide an alteration to the code that would solve the OPs issue. – Ben May 23 '22 at 13:22