Having Fun With |+|
To start with, let's have some fun with |+| operator.
Let's start
scala> 6 |+| 7
res0: Int = 13
Well, not that interesting. It's just an addition. What about
scala> 6 + "9"
res2: String = 69
scala> 6 |+| "9"
<console>:14: error: type mismatch;
found : java.lang.String("9")
required: Int
6 |+| "9"
Not bad. |+| somehow protects you from adding integer to String. OK, not bad, but, not that fun. What about this:
scala> some(6) |+| some(9)
res9: Option[Int] = Some(15)
All right, that starts to be interesting. Give me more:
scala> some(6) |+| some(9) |+| some(10)
res10: Option[Int] = Some(25)
scala> some(6) |+| some(9) |+| some(10) |+| none[Int] |+| some(6)
res11: Option[Int] = Some(31)
scala> some(6) |+| some(9) |+| some(10)
res10: Option[Int] = Some(25)
scala> some(6) |+| some(9) |+| some(10) |+| none[Int] |+| some(6)
res11: Option[Int] = Some(31)
Not bad at all. Want something more than that. Some String ?
scala> "Hello" |+| "World"
res25: java.lang.String = HelloWorld
scala> some("Hello") |+| some("World")
scala> "Hello" |+| "World"
res25: java.lang.String = HelloWorld
scala> some("Hello") |+| some("World")
res26: java.lang.String = Some(HelloWorld)
What else do you have ? List ?
scala> List(2,4) |+| List(4, 5)
res28: List[Int] = List(2, 4, 4,5)
Cool. Boolean?
scala> val b = true
b: Boolean = true
scala> val c = true
c: Boolean = true
scala> b |+| c
<console>:24: error: value |+| is not a member of Boolean
b |+| c
^
Ouch. Why ? Well, because |+| can be interpreted in conjunction or disjunction (note that, actually this works in Scalaz 6, not sure if scalaz seven will make it work as in scalaz 6). To fix this, let's do the following:
scala> val a = Conjunction(true)
a: scalaz.package.@@[Boolean,scalaz.Tags.Conjunction] = true
scala> val b = Conjunction(true)
b: scalaz.package.@@[Boolean,scalaz.Tags.Conjunction] = true
scala> val c = Conjunction(false)
c: scalaz.package.@@[Boolean,scalaz.Tags.Conjunction] = false
scala> a |+| b
res34: scalaz.package.@@[Boolean,scalaz.Tags.Conjunction] = true
scala> a |+| c
res35: scalaz.package.@@[Boolean,scalaz.Tags.Conjunction] = false
scala> a |+| b |+| c
res36: scalaz.package.@@[Boolean,scalaz.Tags.Conjunction] = false
All right. That makes sense. What about Disjunction ? Well, just do the same.
scala> val e = Disjunction(true)
e: scalaz.package.@@[Boolean,scalaz.Tags.Disjunction] = true
scala> val f = Disjunction(true)
f: scalaz.package.@@[Boolean,scalaz.Tags.Disjunction] = true
scala> val g = Disjunction(false)
g: scalaz.package.@@[Boolean,scalaz.Tags.Disjunction] = false
scala> val h = Disjunction(false)
h: scalaz.package.@@[Boolean,scalaz.Tags.Disjunction] = false
scala> e |+| f
res37: scalaz.package.@@[Boolean,scalaz.Tags.Disjunction] = true
scala> e |+| h
res38: scalaz.package.@@[Boolean,scalaz.Tags.Disjunction] = true
scala> g |+| h
res39: scalaz.package.@@[Boolean,scalaz.Tags.Disjunction] = false
OK. That's cool, but it starts to be boring. Doesn't it? Give me something more spectacular. What about Tuple? All right, what about Tuple?
scala> (6, "Hello", List(4, 2)) |+| (7, "Hello", List(5))
res40: (Int, java.lang.String, List[Int]) = (13,HelloHello,List(4, 2, 5))
So, |+| actually "sums" each corresponding element of two tuples. That's quite cool. What if they are nested? Will it work ?
scala> (some(6), some("Hello"), (some(5), some(3))) |+|
(some(7), none[String], (some(6), none[Int]))
res43: (Option[Int], Option[java.lang.String], (Option[Int], Option[Int])) =
(Some(13),Some(Hello),(Some(11), Some(3)))
Fantastic! Pretty cool.
[ I heard somebody there "you see, this Scalaz guy loves using funny operator. This time, they use |+|. OMG, That's not readable". All right, all right. You can actually change all |+| above with its alias called mappend, hope you're happier now].
Monoid and Semigroup Behind the Scene
Behind the scene, what happen is that Scalaz offers a quite simple but powerful abstraction, called Monoid. Believe me, Monoid is something quite simple: it's a type class with two functions: append and zero:
1 2 trait Monoid[A] { 3 val zero: A 4 def append(s1 : A, s2 : => A):A 5 }
With the following contract (or a law, if you wish)
zero append x = x
x append zero = x
(x append y) append z = x append (y append z)
Scalaz provides several instances of Monoid (we have seen some of them in action above):
- Int, Short, BigInt, BigDecimal, Byte, Short ...
- Boolean conjunction and disjunction
- List and Stream
- String
- Either.LeftProjection and Either.RightProjection.
- ...
What about Option and Tuple ? Actually, Scalaz also provides some derived Monoid, like:
- Option[A] is a monoid if A is monoid
- Tuple[A, B, C, D] is a monoid if A, B, C, and D is monoid
- Map [A,B] is a monoid if B is a monoid
- ...
Map is an interesting example. Let's try this:
scala> val m = Map("UO" -> BigDecimal(40.2),
| "US" -> BigDecimal(50.1),
| "YR" -> BigDecimal(10.1))
scala> val n = Map("UO" -> BigDecimal(10.2),
| "US" -> BigDecimal(40.0),
| "YZ" -> BigDecimal(10.5))
scala> m |+| n
res47: scala.collection.immutable.Map[java.lang.String,scala.math.BigDecimal] = Map(UO -> 50.4, US -
> 90.1, YZ -> 10.5, YR -> 10.1)
As you can see, |+| adds every corresponding element in m and n. If there's no corresponding element, for example "YR" in m and "YZ" in n, then it just puts the element in the result map.
All examples so far use append function of Monoid, but not zero. A type class with only append is called Semigroup. The function zero is useful when we want to fold a list of monoid, for example using suml function also provided by scalaz:
scala> val xs = List(some(2,4), some(1, 3), some(2, 10))
xs: List[Option[(Int, Int)]] = List(Some((2,4)), Some((1,3)), Some((2,10)))
scala> xs.suml
res48: Option[(Int, Int)] = Some((5,17))
Conclusion
Scalaz provides a cool thing called Semigroup and Monoid. With that, we can benefit the operator |+| and sum for their instances like Integer, String, List, and so on. But the main benefit is that Scalaz provides implementation of Monoid for Option, Tuple, and Map. You don't need to know what Monoid or Semigroup are to benefit all this. But, why should you avoid them ? It's a very simple stuff.
I hope you enjoy this very express introduction to Scalaz Seven Monoid. Who said Scalaz is complex ?
I'm preparing more on this subject. So, stay tuned.