Is there any way to force operator precedence in Scala like you do in Haskell with $?
For example, in Haskell, you have:
a b c = ((a b) c)
and
a $ b c = a (b c)
Is there a similar way to do this in Scala? I know Scala doesn't have operators per se, but is there a way to achieve a similar effect?
It is possible to use implicits to achieve a similar effect. For example: (untested, but should be something like this)
object Operator {
class WithOperator[T](that: T) {
def &:[U](f: T => U) = f(that)
}
implicit def withOperator[T](that: T) = new WithOperator(that)
}
Using this system, you can't use the name $, because the name needs to end with a : (to fix the associativity) and the dollar is a normal identifier (not operator identifier), so you can't have it in the same name as a :, unless you separate them with underscores.
So, how do you use them? Like this:
val plusOne = (x: Int) => {x + 1}
plusOne &: plusOne &: plusOne &: 1
infix vs . notation is often used in a similar fashion to control precedence:
a b c d == a.b(c).d
a.b c d == a.b.c(d)
a b c.d == a.b(c.d)
Scala also has a fixed precedence ordering for operators used in infix notation:
(all letters)
|
^
&
< >
= !
:
+ -
* / %
(all other special characters)
Names can usually chosen explicitly to take advantage of this. For example, ~ and ^^ in the standard parsers library.
Scalaz has defined a "pipe" operator, which works similar, but with flipped arguments. See switch function and object with scalaz' |>
