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Comment by christophilus | Hacker News Reader

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

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!

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, [])

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`.

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

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), [])

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

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.

No argument that the comma operator is a neat trick when you need it.

FWIW, it's 2022:

  const output = [];
  for (const n of [1, 2, 3, 4, 5]) {
    if (n % 2) output.push(2 * n);
  }

Minority opinion: please `let` your mutable references. I know `const` doesn’t signal immutability, but we as humans with eyeballs and limited attention span certainly benefit from knowing when a value might change at runtime.

Disagree: virtually everything in JS is mutable, so this almost means "never use the `const` keyword". Pretending that the `const` keyword means something that it doesn't makes things harder for my limited human mind to understand, not easier. Plus using `let` inappropriately makes my linter yells at me, and I usually like to just do whatever my linter tells me.

Anyway, I use TypeScript, so if I really want to assert that my array is immutable (as immutable as stuff in JS-land gets anyway) I just write:

  const input: readonly number[] = [1, 2, 3, 4, 5];

or even

  const input = [1, 2, 3, 4, 5] as const;

I `readonly` and `as const` everything I possibly can. I do know that const doesn’t mean immutable, as I said, but I think it should and I think there’s value in establishing the idiom even if it’s not currently adopted. Because otherwise const basically means nothing unless you’re mutating everything already.

I realize I could be clearer in what I’m asking for: please use const when you use reference types as values, and use let when you intend to mutate the reference. Using const and then changing a value is certainly allowed but it’s confusing and it’s missing an opportunity to signal in the code where changes might happen.

Surely the spread operator is nicer here?

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

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

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

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