I can't quite understand the differences between [...array].map(...) and [...array.map(...)]. Are they any different? If yes, why and how?
2 Answers
[...array].map(...)will generate a new shallow copy of an array ([...array]) and then.mapover it and return a brand new array as the return value.[...array.map(...)]will.mapover the original array, which will return a new array, and then immediately make a shallow copy of the new array ([...array.map(...)]). I must admit, I don't really understand the use case for this approach-- because.mapitself is returning a brand new array that no other variable should be holding a reference to, I don't see any value in immediately casting a shallow copy of it.
(Also, as pointed out by chazsolo in the comments, if array were not an array, but in fact some other type like a Set, the [...array.map(...)] approach would actually throw an error, as Set instances have no .map method).
2 Comments
chazsolo
I'd mention that
array could be an Iterable, like a Set. The first case would work as expected, the second would throw a TypeError.
Alexander Nied
@chazsolo -- good call out-- updated, thanks!
Well [...array.map(...)] is the same as array.map(...) except that it does an extra unnecessary spread/copy after the mapping is done.
The [...array].map(...) has an small difference, due to the 3rd parameter passed in map (the array it self).
And as mentioned in another comment, by @chazsolo, this case will handle cases where the array variable is not an array, so it does not have the map method, but can be iterated and the spread will create an array out of it.
answered Nov 29, 2021 at 18:54
Gabriele Petrioli
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array.map(...)which does the same thing.