Discussion on: JavaScript Interview Question #18: What's the sum of two booleans in JavaScript?

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You can use this property of booleans to implement an exclusive or:

const xor = (...bools) =>
    bools.reduce(
        (total, bool) => total + bool,
        0,
    ) === 1

xor(true, false) // true
xor(true, true, false) // false
xor(false, false) // false
xor(true) // true

Coderslang: Become a Software Engineer • Feb 22 '21

That's clever. Thank you!

willsmart • Feb 22 '21

Or count things cleanly...

isEven = v => !(v & 1)

 // count the even numbers in an array
[1,2,3,4,5,6,7].reduce((acc, v)=> acc + isEven(v), 0) // <- 3

lionel-rowe • Feb 22 '21 • Edited

Now I think about it, filter with length would be simpler in both cases

const xor = (...bools) =>
    bools.filter(x => x).length === 1

const countEven = arr =>
    arr.filter(isEven).length

willsmart • Feb 22 '21 • Edited

A hell of a lot cleaner, in theory a bit slower too.
filter allocates, fills and returns an array, which we're then instantly throwing away after getting its length. reduce functions more like a for loop, iterating over each element and accumulating a sum.

That's in theory though. JS performance is fickle thing, and on perf.link it seems your filter method is both cleaner looking, and faster for arrays with fewer than 10000 items.
Perf.link link
Go figure! Good to know

What we really want is something like ruby's count, which only returns the filtered size...

isEven = -> (v) { (v&1)==0 }
[1,2,3,4,5,6,7].count(&isEven) # <- 3

😎