### re: 15 must-know JavaScript array methods in 2020 VIEW POST

re: It's not wrong. I've just followed the way flatMap() work. It applies map() first and flat() after. Therefore for the example of flat() and map() i...

Try it in JSFiddler, Chrome, it does not work, it is wrong !!

I've not tested with your array. You're right. Thanks for your comment.

Sorry, but NOW it's incorrect - it was correct before. That's beacuse the flat part in flatMap() works differently, than .flat(1) alone, it only works for [[1], [2]], but not for [[1, 2]].

``````[[1, 2], 3, 4].flatMap(x => [x * 2])
// [NaN, 6, 8]

[[1, 2], 3, 4].map(x => [x * 2]).flat(1)
// [NaN, 6, 8]
``````

See? flatMap will put the NaN there as well, try it for yourself.

I'd say it's either a bug in the implementation or on MDN description:

The flatMap() method first maps each element using a mapping function, then flattens the result into a new array. It is identical to a map() followed by a flat() of depth 1, but flatMap() is often quite useful, as merging both into one method is slightly more efficient.

and later:

The flatMap method is identical to a map followed by a call to flat of depth 1.

Example on MDN:

``````let arr1 = [1, 2, 3, 4];

arr1.map(x => [x * 2]);
// [[2], [4], [6], [8]]

arr1.flatMap(x => [x * 2]);
// [2, 4, 6, 8]

// only one level is flattened
arr1.flatMap(x => [[x * 2]]);
// [[2], [4], [6], [8]]
``````

Which translates to this one-liner:

``````[1, 2, 3, 4].map(x => [x * 2]).flat(1);
// [2, 4, 6, 8]
``````

and not the other way around:

``````[1, 2, 3, 4].flat(1).map(x => [x * 2]);
// [[2], [4], [6], [8]]
``````
code of conduct - report abuse