[christophilus]

Push returns the length of the array, though, so that won't work.

[eyelidlessness]

Comma operator to the rescue:

    (acc.push(2*n), acc)

In an expression position, the push statement will be executed, then its return will be discarded, and the final expression will be the result of the expression. Is it “better”? Almost certainly not. But it lets you stay in expression syntax while executing statements. (Much more useful for logging than meaningful runtime side effects IMO, but I think it should be more widely known in general.)

Edit: and I’m glad to see another reference to it down thread!

[ReleaseCandidat]

Ah, sorry, so you have to concat the arrays using `concat`.

    [1, 2, 3, 4, 5].reduce((acc, n) => n % 2 === 1 ? acc.concat([2*n]) : acc, [])

[jasonkillian]

Wouldn't recommend doing this - if the original array is of significant length this'll get quite slow because `acc.concat` has to create a brand new array of slightly longer length on each iteration it's called. Better to just use `push` like you suggested before and then return the array if you want to use `reduce`.

[ReleaseCandidat]

Yes, of course, that's why I used `push` at first.

[jacobolus]

Use the comma operator: (acc.push(2*n), acc) will return acc. Or e.g.

  [1, 2, 3, 4, 5].reduce((acc, n) => (n % 2 ? acc.push(2*n) : null, acc), [])

[wk_end]

If you're just iterating through the array and mutating an object on each iteration, just use a for loop.

[jacobolus]

Obviously you can alternately write:

  let input = [1, 2, 3, 4, 5], output = [];
  for (let i = 0; i < input.length; ++i) {
    let n = input[i];
    if (n % 2) output.push(2*n);
  }
  return output;

But in some circumstances the other style can be more convenient / legible. The immediate question was about pushing to an array and then returning the array, for which the comma operator can be handy.

[spiralx]

Surely the spread operator is nicer here?

    [1, 2, 3, 4, 5].reduce((acc, n) => n % 2 === 1 ? [ ...acc, 2 * n ] : acc, [])

[v413]

This is not efficient. Each iteration creates a new array instance due to the spread operator.

[spiralx]

`acc.concat()` also creates a new array instance, so I don't get your point.