If I have an instance of Bifunctor[A,A]
bf, a function f : A => A
and a Boolean
value p
:
def calc[A, F[_,_]: Bifunctor](p: Boolean, bf: F[A, A], f: A => A): F[A, A] = {
val BF = implicitly[Bifunctor[F]]
BF.bimap(bf, (a : A) => if (p) f(a) else a, (a : A) => if (!p) f(a) else a)
}
How can I put this more concisely (and expressively)? Basically I am trying to invoke a function on a side of a bifunctor (e.g. a Tuple2
) dependent on some predicate. If the predicate is true, I want to map the LHS and the RHS if it's false
val t2 = (1, 2)
def add4 = (_ : Int) + 4
calc(true, t2, add4) //should be (5,2)
calc(false, t2, add4) //should be (1,6)
Bifunctor
),I seem to be able to use arrows as follows:
def calc[A](p: Boolean, bf: (A, A), f: A => A): (A, A)
= (if (p) f.first[A] else f.second[A]) apply bf
Not all that much nicer:
def calc[A, F[_,_]:Bifunctor](p: Boolean, bf: F[A, A], f: A => A): F[A, A] =
(if (p) (bf :-> (_: A => A)) else ((_:A => A) <-: bf))(f)
A little nicer:
def cond[A:Zero](b: Boolean, a: A) = if (b) a else mzero
def calc[A, F[_,_]:Bifunctor](p: Boolean, bf: F[A, A], f: Endo[A]): F[A, A] =
cond(p, f) <-: bf :-> cond(!p, f)
Some Haskell, just for the language envy:
calc p = if p then first else second
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