The variable b is coming up as undefined and it's being assigned 'a' as the default value when there is no value present. However, when assigned the value of 'c', an error is thrown. Can anyone provide a clear explanation of what is going on here? Is this potentially a bug?
const [a, b = c, c] = [1, , 3]
When destructuring into variables, the evaluation occurs from left to right following the order of the array or property list.
Variables declared with const are not accessible until after the line that initializes them has been executed.
The code provided is essentially:
const arr = [1, , 3];
const a = arr[0];
const b = arr[1] === undefined ? arr[1] : c;
const c = arr[2];
This reveals the issue you're encountering; attempting to access c before it's defined. This behavior is expected and not a bug.
To achieve the desired output, extract the value of c first so that 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);It is advisable to avoid using sparse arrays whenever possible, as they can lead to confusion and unexpected behavior.
There is a clever trick to achieve this in just one line by destructuring the keys of the array. You can assign the 2nd index before the 1st and use that as the default value. However, it's important to note that this method is more for academic purposes and not recommended for actual code implementation.
const { 0: a, 2: c, 1: b = c } = [1, , 3]
console.log(a, b, c)const [a, b = c, c] = [1, , 3]
The scenario depicted above can be viewed as
const a = [1, , 3][0];
const b = [1, , 3][1] || c; // Reference Error
const c = [1, , 3][2];
console.log(a, b, c);In contrast, the following statement
const [a, b = c, c] = [1, , 3]
Can be envisioned as
const a = [1, , 3][0];
const b = [1, , 3][1] || a;
const c = [1, , 3][2];
console.log(a, b, c); //1 1 3Thus, in the initial instance c is referenced before being initialized leading to an error, while in the latter example, a is already defined when accessed.
If you were to perform the same operation using var, you would not encounter an error initially due to hoisting.
var [a, b = c, c] = [1, , 3]
console.log(a, b, c) //1 undefined 3
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