Two initial values for reduce array method?

Question

const result = [1,2,3,4,5].reduce((acc,cur) => (acc[+(cur % 2 !== 0)] += cur, acc),  [0, 0])

console.log(result)

It is supposed to sum up the even values and the odd values in the array but how can we have 2 different initializers for the initial values? There are 2 commas at the end of the reduce method, how do they work in this case?

You are confusing the comma operator that returns the array... — epascarello
– epascarello, Commented Oct 9, 2020 at 19:44
Why are you converting a boolean to a number? [+(boolean expression)] doesn't make sense — Taplar
– Taplar, Commented Oct 9, 2020 at 19:44
@Taplar, otherwise you need an object with true and false keys. — Nina Scholz
– Nina Scholz, Commented Oct 9, 2020 at 19:45
Why have that at all? The mod is already going to return 0 or 1. @NinaScholz — Taplar
– Taplar, Commented Oct 9, 2020 at 19:46

epascarello · Accepted Answer · 2020-10-09 19:47:16Z

3

You are confusing the comma operator in the function. Below it is written without the fat arrow syntax with the implicit return.

var res = [1, 2, 3, 4, 5].reduce(function(acc, cur) {
  acc[+(cur % 2 !== 0)] += cur;
  return acc
}, [0, 0]);

console.log(res)

answered Oct 9, 2020 at 19:47

epascarello

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Comments

Nina Scholz · Accepted Answer · 2020-10-09 19:45:05Z

0

In reality, you have only one start value for the accumulator, an array.

[1, 2, 3, 4, 5].reduce(
    (acc, cur) => (acc[+(cur % 2 !== 0)] += cur, acc),
    [0, 0]
)

answered Oct 9, 2020 at 19:45

Nina Scholz

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