Writing dumb code, I've just found something weird.

for(const [[[[[fancy, loop]]]]] in [0, 0]) {
  console.log(fancy, loop);
}

Writing dumb code, I've just found something weird.

for(const [[[[[fancy, loop]]]]] in [0, 0]) {
  console.log(fancy, loop);
}

// Chrome 70.0.3538.77 says:
// 0 undefined
// 1 undefined

It's like assigning 0 and 1 to [[[[[fancy, loop]]]]], which, is array destructurings taking place and supposed to throw an error, isn't it? Or not. It's just my thought getting me confused right now.

Could you please tell me how it is valid and works with no error? What am I missing?

• javascript

• @RobG Yeah, but, how could it destructure 0 and 1? – Константин Ван Commented Nov 5, 2018 at 0:25
• 2 you can always throw the code into a transpiler to see what is happening – Bravo Commented Nov 5, 2018 at 0:25
• 2 babeljs.io/repl/… – Bravo Commented Nov 5, 2018 at 0:26
• 17 I guess you discovered a new question for javascript job interviews. ;-) – RobG Commented Nov 5, 2018 at 0:49
• 4 ^ @RobG for anyone wondering, this should be taken as sarcasm – Pierre Arlaud Commented Nov 5, 2018 at 9:28

2 Answers 2

It's not assigning 0 and 1 to [[[[[fancy, loop]]]]]. You're looping over the keys of [0, 0], because you used in instead of of, and those keys are strings.

The string "0" is an iterable whose sole element is "0". Assigning "0" to [[[[[fancy, loop]]]]] repeatedly unpacks "0" and gets "0", until eventually it gets down to

[fancy, loop] = "0"

at which point the final unpacking assigns "0" to fancy and undefined to loop.

You're using in instead of of so you're getting the properties of the array not the values. For this case you're getting the array indexes as strings (0, 1). You're basically destructuring a string with length of 1 every time. So you always get the first character of every iterated property