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When I do

const arrayLike = { 1:'foo', 2:'bar', length:2 };
Array.from(arrayLike);

I get

[ undefined, 'foo' ]

When I do

const arrayLike = { 1:'foo', 2:'bar', length:3 };
Array.from(arrayLike)

I get

[ undefined, 'foo', 'bar' ]

• Why do I not see bar when I set the length?
• Why do I see only my elements only when I hard-set the length to 3?
• Why is the first element showing up as undefined?

When I do

const arrayLike = { 1:'foo', 2:'bar', length:2 };
Array.from(arrayLike);

I get

[ undefined, 'foo' ]

When I do

const arrayLike = { 1:'foo', 2:'bar', length:3 };
Array.from(arrayLike)

I get

[ undefined, 'foo', 'bar' ]

• Why do I not see bar when I set the length?
• Why do I see only my elements only when I hard-set the length to 3?
• Why is the first element showing up as undefined?

• javascript
• arrays
• ecmascript-6

Share Improve this question Follow edited Jul 11, 2017 at 17:18 Evan Carroll asked Jul 11, 2017 at 17:02 Evan CarrollEvan Carroll 1 3

• Possible duplicate of: How do I create an array in javascript whose index starts from 1 – Washington Guedes Commented Jul 11, 2017 at 18:03
• @WashingtonGuedes nothing to do with array.from. that question is just challenging wanting to implement a 1-index array, not convert a 1-index object to an array. – Evan Carroll Commented Jul 11, 2017 at 18:07
• Still related. Explains the issue of this post. – Washington Guedes Commented Jul 11, 2017 at 18:11

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 9

Array's are 0 indexed in JavaScript

All of these questions are answered pretty simply. Array.from assumes that the array-like object you're giving it is zero-indexed. If it isn't you'll get an array that is zero-indexed with all elements that aren't set, including 0 set to undefined. So in your example,

const arrayLike = { 1:'foo', 2:'bar', length:2 };
Array.from(arrayLike);

Is essentially the same as,

const arrayLike = { 0:undefined, 1:'foo', 2:'bar', length:2 };
Array.from(arrayLike);

Because the length is less than the max elements, it essentially stops iterating at that point and the false-length functions as a truncation. What you want is likely,

const arrayLike = { 0:'foo', 1:'bar', length:2 };
Array.from(arrayLike);

Or if you're getting the Array-like from a source that is 1-indexed, set length appropriately so as not to lose the last element, and then ditch the first element from the result (the equivalent of shift).

const arrayLike = { 1:'foo', 2:'bar', length:3 };

let arr = Array.from(arrayLike).splice(1);

let arr = Array.from(arrayLike);
arr.shift();


let [, ...arr] = Array.from(arrayLike);

Array.from(obj) does not return undefined, it returns a new array using properties from a given array-like object obj.

And the default values of each index position created is undefined when not given.

The undefined value you are seeing is at index 0 of the created array.

Anyways, it is not a problem, you can still access the indexes 1 and 2 and they will return the values you expect them to return.

You can also verify the default behavior of created arrays with:

console.log(new Array(10))

which logs (index positions 0..9) (length 10):

[undefined x 10]

In JavaScript, you will not see any array with the first index other than 0. Independently how the array was created.

---

Another alternative to what you are doing is to set the prototype:

const arrayLike = { 1:'foo', 7:'bar' };

//Dynamic set the length
arrayLike.length = Math.max(...Object.keys(arrayLike)) + 1;

//Get built-in features of Array to your arrayLike
Object.setPrototypeOf(arrayLike, Array.prototype);

//Do something
arrayLike.forEach((x, i) => console.log("%i: %s", i, x))

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