Ternary Operator Similar To ?:
I am trying to avoid constructs like this:
val result = this.getClass.getSimpleName
if (result.endsWith("$")) result.init else result
Ok, in this example the then
and else
branch are simple, but you can image complex ones.
I built the following:
object TernaryOp {
class Ternary[T](t: T) {
def is[R](bte: BranchThenElse[T,R]) = if (bte.branch(t)) bte.then(t) else bte.elze(t)
}
class Branch[T](branch: T => Boolean) {
def ?[R] (then: T => R) = new BranchThen(branch,then)
}
class BranchThen[T,R](val branch: T => Boolean, val then: T => R)
class Elze[T,R](elze: T => R) {
def :: (bt: BranchThen[T,R]) = new BranchThenElse(bt.branch,bt.then,elze)
}
class BranchThenElse[T,R](val branch: T => Boolean, val then: T => R, val elze: T => R)
implicit def any2Ternary[T](t: T) = new Ternary(t)
implicit def fct2Branch[T](branch: T => Boolean) = new Branch(branch)
implicit def fct2Elze[T,R](elze: T => R) = new Elze(elze)
}
Defined that, I can replace the above simple example with:
this.getClass.getSimpleName is {s: String => s.endsWith("$")} ? {s: String => s.init} :: {s: String => s}
But how can I get rid of the s: String =>
? I want something like that:
this.getClass.getSimpleName is {_.endsWith("$")} ? {_.init} :: {identity}
I guess the compiler needs the extra stuff to infer types.
Solution 1:
From Tony Morris' Lambda Blog:
I hear this question a lot. Yes it does. Instead of
c ? p : q
, it is writtenif(c) p else q
.This may not be preferable. Perhaps you’d like to write it using the same syntax as Java. Sadly, you can’t. This is because
:
is not a valid identifier. Fear not,|
is! Would you settle for this?c ? p | q
Then you’ll need the following code. Notice the call-by-name (
=>
) annotations on the arguments. This evaluation strategy is required to correctly rewrite Java’s ternary operator. This cannot be done in Java itself.case class Bool(b: Boolean) { def ?[X](t: => X) = new { def |(f: => X) = if(b) t else f } } object Bool { implicit def BooleanBool(b: Boolean) = Bool(b) }
Here is an example using the new operator that we just defined:
object T { val condition = true import Bool._ // yay! val x = condition ? "yes" | "no" }
Have fun ;)
Solution 2:
We can combine How to define a ternary operator in Scala which preserves leading tokens? with the answer to Is Option wrapping a value a good pattern? to get
scala> "Hi".getClass.getSimpleName |> {x => x.endsWith("$") ? x.init | x}
res0: String = String
scala> List.getClass.getSimpleName |> {x => x.endsWith("$") ? x.init | x}
res1: String = List
Is this adequate for your needs?