Noticed an interesting behavior of Javascript while playing with Array Destructuring eg:
let a = [1, 2, 3];
let b = [one, two, three, four = 4] = a
console.log(one, two, three, four)
console.log(b)When I say a.length output is 3
If I say b.length output is 3.
However, if I say four output is 4.
If four is an element of b then why b.length is shown as 3 why not as 4?
this let b = [one, two, three, four=4] = a does not create a new array with four elements to assign to b.
if you do let b = [one, two] = a, you would still get b.length 3 because it assigns the value a to b without any modification.
remember, you're simply using de-structuring and default assignment feature in here -> [one, two, three, four=4] which does not create a new array.
if you do a === b, you will get true.
four is not an element of b (which is a). You're using the default value =4 feature there; it's not getting assigned into b.
Running your code through the Babel REPL, to de-sugar that destructuring, gives us something like:
var _a, _a$;
var a = [1, 2, 3];
_a = a;
one = _a[0];
two = _a[1];
three = _a[2];
_a$ = _a[3];
four = _a$ === void 0 ? 4 : _a$;
var b = _a;
As you can see, b eventually just gets assigned _a, which is a; the destructuring assignments occur "in-between".
Assignment works right-to-left, and when something is assigned to with = (whether by destructuring or not), the whole expression on the left and right of the = is equivalent to the value of what was assigned (on the right hand side).
There is no intermediate array with
[one, two, three, four=4] =
Rather, that's just assigning to existing variables, without creating an additional expression (other than 4, which is assigned to four).
a = [1,2,3];
let b = [one, two, three, four=4] = a
is like
a = [1,2,3];
let b = (() => {
[one, two, three, four=4] = a;
return a;
})()
In your code, b is the exact same array as a:
a = [1,2,3];
let b = [one, two, three, four=4] = a;
console.log(b === a);
a = b = cis equivalent tob=c, a=cnotb=c, a=b. This means in your case[one, two, three, four=4] = a; let b = a;