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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?
Share Improve this question edited Nov 5, 2018 at 0:23 Константин Ван asked Nov 5, 2018 at 0:20 Константин ВанКонстантин Ван 14.7k8 gold badges74 silver badges86 bronze badges 5 |2 Answers
Reset to default 26It'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
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0
and1
? – Константин Ван Commented Nov 5, 2018 at 0:25