javascript - ES6 array destructuring assignment default assignment

Question

When I saw Ruan Yifeng’s ES6 tutorial on destructuring assignment and default values, I didn’t quite understand this place.
Original link

Note that ES6 uses the strict equality operator (===) internally to determine whether a position has a value. Therefore, if an array member is not strictly equal to undefined, the default value will not take effect.

function f() {
  console.log('aaa');
}

let [x = f()] = [1];

The book says that the above code is equivalent to the following code

let x;
if ([1][0] === undefined) {
  x = f();
} else {
  x = [1][0];
}

May I ask where this [1][0] comes from? Shouldn't it be like this?

let x;
if (1 === undefined) {
  x = f();
} else {
  x = 1;
}

Answer 1

When deconstructing an array, the principle is as follows: put one or more variables into array A, and then make this array A equal to another array B. Then during destructuring, the value of a certain position in array A will be equal to the corresponding position of array B. value.

let [x = f()] = [1];

The meaning of this code is to first create an array A. The first item in array A is x, and then there is an array B, B = [1].
Then let A = B. The final effect is A[ 0] = B[0], that is, x=B[0], that is, x=[1][0].
So when judging whether it is equal to undefined, do this

if([1][0] === undefined)

Answer 2

The 1 in [1] on the right corresponds to x, that is, [1][0]corresponds to x

Answer 3

Deconstruct, deconstruct, deconstruct. . . So the purpose is to untie the things on the right side of the equal sign, so we must untie [1].

So let [x]=[1], then x is [1][0], which is 1. So in fact, the assignment of x is judged based on [1][0].

I don’t know if I understand what I’m saying, but I’d better give you the documentation:

https://developer.mozilla.org...