ES6 Array destructuring weirdness

Question

Can anyone explain, why the following happens with ES6 array destructuring?

let a, b, c
[a, b] = ['A', 'B']
[b, c] = ['BB', 'C']
console.log(`a=${a} b=${b} c=${c}`)

// expected: a=A b=BB c=C
// actual:   a=BB b=C c=undefined

http://codepen.io/ronkot/pen/WxRqXg?editors=0011

Answer 1

As others have said, you're missing semicolons. But…

Can anyone explain?

There are no semicolons automatically inserted between your lines to separate the "two" statements, because it is valid as a single statement. It is parsed (and evaluated) as

let a = undefined, b = undefined, c = undefined;
[a, b] = (['A', 'B']
[(b, c)] = ['BB', 'C']);
console.log(`a=${a} b=${b} c=${c}`);

wherein

• [a, b] = …; is a destructuring assignment as expected
• (… = ['BB', 'C']) is an assignment expression assigning the array to the left hand side, and evaluating to the array
• ['A', 'B'][…] is a property reference on an array literal
• (b, c) is using the comma operator, evaluating to c (which is undefined)

If you want to omit semicolons and let them be automatically inserted where ever possible needed, you will need to put one at the start of every line that begins with (, [, /, +, - or `.

Answer 2

You've fallen into a trap of line wrapping and automatic semicolon insertion rules in JavaScript.

Take this example:

let x = [1, 2]
[2, 1]

It's the interpreted as:

let x = [1, 2][2, 1] // === [1, 2][(2, 1)] === [1, 2][1] === 2

That weird [(2, 1)] thing above is related to how Comma Operator works.

Thus, your example:

let a, b, c
[a, b] = ['A', 'B']
[b, c] = ['BB', 'C']
console.log(`a=${a} b=${b} c=${c}`)

Is interpreted as:

let a, b, c
[a, b] = ['A', 'B'][b, c] = ['BB', 'C']
console.log(`a=${a} b=${b} c=${c}`)

Now, if you insert a semicolon, it will work as you intended:

let a, b, c
[a, b] = ['A', 'B']; // note a semicolon here
[b, c] = ['BB', 'C']
console.log(`a=${a} b=${b} c=${c}`)

Also, it's a good idea to check your code by pasting it into Babel repl to see the generated output:

'use strict';

var a = void 0,
    b = void 0,
    c = void 0;

var _ref = ['A', 'B'][(b, c)] = ['BB', 'C'];

a = _ref[0];
b = _ref[1];

console.log('a=' + a + ' b=' + b + ' c=' + c);

Answer 3

I believe you have forgotten the line breaks ';'. Below is the corrected code. Please try:

let a,b,c
[a, b] = ['A', 'B'];
[b, c] = ['BB', 'C'];
console.log(`a=${a} b=${b} c=${c}`)