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Javascript – What is happening when you're destructing an array without using square brackets?

Python developer here. I am using JavaScript to run the following command,

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

To my surprise, b and c are undefined while d is the entire a variable. I later realize that I am meant to go let [b, c, d] = a;, however I am just trying to understand what is happening and why they’re undefined.

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2

Answers


  1. 0

    You are not destructuring.

    let b , c, d = a;

    Declares 3 variables and assigns a value to one of them. The line above is identical to:

    let b;
let c;
let d = a;

    This syntax allows you to declare (and optionally assign) several variables in a single statement, e.g.

    let tmp, x = 4, i, y = 2, z = x * 10 + y;

    Unassigned variables are default-initialized to undefined.

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  2. 0

    In Javascript,

    let b, c, d = a;

    defines the variables b, c, and d. It also initializes d with the value of a, or the array [1, 2, 3] but leaves b and c uninitialized.

    When you try to print the values of uninitialized variables to the console in Javascript, it prints undefined to signify the lack of any value.

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