6 ms·
The difference starts to show when you want to partially apply some values. // process1 :: (Item -> Bool) -> [Item] -> [OtherType] // process2 :: [Item
by CrossEye 11y ago
The difference starts to show when you want to partially apply some values.
// process1 :: (Item -> Bool) -> [Item] -> [OtherType]
// process2 :: [Item] -> [OtherType]
process1(isOk, items); //=> [otherType1, otherType2, ...]
process2(items); //=> [otherType1, otherType2, ...]
var processR1 = seq(filter, map(toOtherType), flatten);
var processR2 = processR1(isOk);
var processT1 = (isOk, items) => items.filter(isOk).map(toOtherType).flatten()
var processT2 = items => processT1(isOk, items)
That's why I'm wondering if there's some easier way to do this.
- icebraining 11y agoI believe you can use partial() from this post[1] to do something like this: var processT1 = (isOk, items) => items.filter(isOk).map(toOtherType).flatten(); var processT2 = partial(processT1, isOk); http://benalman.com/news/2012/09/partial-application-in-javascript/#partial-application-from-the-left http://benalman.com/news/2012/09/partial-application-in-java...