Js Functions Composed In The Opposite Order

Starting with a simple function composition const fa = (x => (x + 'a')); const fb = (x => (x + 'b')); fb(fa('x')) I played around and I obtained the following code snippet w

Solution 1:

Let's break this down:

const _fa = fa(y => y)
// _fa = x => (y => y)(x + "a")

To avoid confusing the two x let's name it as x1

// _fa = x1 => (y => y)(x1 + "a")

Now fb would be:

// fb = next => x2 => next(x2 + "b") 

If we call fb with fa(y => y) (ie. _fa), we substitute next with _fa:

_fb = fb(fa(y => y))
// _fb = x2 => (x1 => (y => y)(x1 + "a"))(x2 + "b")

Now lets evaluate _fb with the argument x2 = 'x':

// _fb = (x1 => (y => y)(x1 + "a"))("x" + "b")
// _fb = (x1 => (y => y)(x1 + "a"))("xb")

Notice how x1 => (y => y)(x1 + "a") can be simplified to x1 => x1 + "a". Now we have:

// _fb = (x1 => x1 + "a")("xb")

Now let's evaluate this function (x1 => x1 + "a") with argument x1 = "xb"

// _fb = "xb" + "a"
// _fb = "xba"

Solution 2:

You might not know it, but you're looking at the continuation monad! Also the thrush combinator.

constCont = someValue => next =>
  next (someValue)

constfa = x =>
  Cont (x + "a")

constfb = x =>
  Cont (x + "b")

fa ("x") (fb) (console.log)
// xab

fb ("x") (fa) (console.log)
// xba

fa ("x") (fa) (fa) (fa) (fb) (console.log)
// xaaaab

fb ("x") (fa) (fb) (fa) (fb) (console.log)
// xbabab

In a slightly more involved example, we illustrate liftA2 which takes a binary function and "lifts" it into our Cont context. Now we can take two Cont values and use them with any ordinary function.

constCont = someValue => next =>
  next (someValue)
  
constliftA2 = f => mx =>my =>
  mx (x =>
    my (y =>Cont (f (x, y))))

constmult = (x, y) =>
  x * y
  
const main =
  liftA2 (mult) (Cont (6)) (Cont (7))
  
main (console.log)
// 42

main (x =>Cont (x + 1)) (console.log)
// 43