haskell - What exactly makes Option a monad in Scala? -

i know monads , how use them. don't understand what makes, let's say, option monad?

in haskell monad maybe monad because it's instantiated monad class (which has @ least 2 necessary functions return , bind makes class monad, indeed, monad).

but in scala we've got this:

sealed abstract class option[+a] extends product serializable { ... } trait product extends equals { ... } 

nothing related monad.

if create own class in scala, monad default? why not?

monad concept, abstract interface if will, defines way of composing data.

option supports composition via flatmap, , that's pretty needed wear "monad badge".

from theoretical point of view, should also:

• support unit operation (return, in haskell terms) create monad out of bare value, in case of option some constructor
• respect monadic laws

but not strictly enforced scala.

monads in scala looser concept in haskell, , approach more practical. thing monads relevant for, language perspective, ability of being used in for-comprehension.

flatmap basic requirement, , can optionally provide map, withfilter , foreach.

however, there's no such thing strict conformance monad typeclass, in haskell.

here's example: let's define our own monad.

class mymonad[a](value: a) {   def map[b](f: => b) = new mymonad(f(value))   def flatmap[b](f: => mymonad[b]) = f(value)   override def tostring = value.tostring } 

as see, we're implementing map , flatmap (well, , tostring commodity). congratulations, have monad! let's try out:

scala> {   <- new mymonad(2)   b <- new mymonad(3) } yield + b // res1: mymonad[int] = 5 

nice! not doing filtering, don't need implement withfilter. since we're yielding value, don't need foreach either. implement whatever wish support, without strict requirements. if try filter in for-comprehension , haven't implemented withfilter, you'll compile-time error.