解构数组

Question

b is undefined and when assigned 'a' as default values it gets it but when assigned 'c' value it gives throw an Error can someone explain better what's happening here or is this a bug?

const [a, b = c, c] = [1, , 3]

Answer 1

Destructuring into variables evaluates left-to-right in the order of the array (or property list).

Variables declared with const cannot be accessed before the line that initializes them runs.

Your code is equivalent to:

const arr = [1, , 3];
const a = arr[0];
const b = arr[1] === undefined ? arr[1] : c;
const c = arr[2];

Which should make the problem pretty obvious; you're trying to retrieve c before the interpreter has gotten to that point. It's not a bug.

For the output you're looking for, extract c first, so you can use it as a default value for b .

 const arr = [1, , 3]; const c = arr[2]; const [a, b = c] = arr; console.log(a, b, c);

But I'd also recommend avoiding sparse arrays when possible, they can be pretty unintuitive.

Answer 2

It is possible to do this in one line. You can destructure the keys of the array. Get the 2nd index before 1st and set that as the default value. This is purely academic and don't use it in actual code base.

 const { 0: a, 2: c, 1: b = c } = [1, , 3] console.log(a, b, c)

Answer 3

const [a, b = c, c] = [1, , 3]

The statement above could be thought of as

 const a = [1, , 3][0]; const b = [1, , 3][1] || c; // Reference Error const c = [1, , 3][2]; console.log(a, b, c);

Whereas, the statement below

const [a, b = c, c] = [1, , 3]

Could be thought of as

 const a = [1, , 3][0]; const b = [1, , 3][1] || a; const c = [1, , 3][2]; console.log(a, b, c); //1 1 3

So, in the first case c is being referred before is it initialized which gives an error but in the second case a is already present when it is being referred.

Now, if you try the same thing using var , then you wouldn't be getting an error in the first case because of hoisting .

var [a, b = c, c] = [1, , 3]
console.log(a, b, c) //1 undefined 3