scala: anonymous partial function strange syntax - scala

I came across a code similar to this, and it was surprised that it even compiles:
scala> val halfSize: PartialFunction[String, Int] = _.length match {
case even if even % 2 == 0 => even / 2
}
halfSize: PartialFunction[String,Int] = <function1>
scala> List("x", "xx").collect(halfSize)
res1: List[Int] = List(1)
As far as I known, the valid syntax to define a PartialFunction is a case function:
val halfSize: PartialFunction[String, Int] = {
case s if s.length % 2 == 0 => s.length / 2
}
The first code seems more optimized since it calls length only once. But even in the SLS I was not able to find the explanation of the syntax. Is this an undocumented feature of scalac ?

The rules are given in https://www.scala-lang.org/files/archive/spec/2.12/06-expressions.html#anonymous-functions:
The eventual run-time value of an anonymous function is determined by the expected type:
...
PartialFunction[T, U], if the function literal is of the shape x => x match { … }
In this case the literal does have such a shape, as rogue-one's answer explains, so PartialFunction is allowed as the expected type.
(EDIT: actually, it doesn't, since it matches x.length instead of x. This looks like a minor bug, but one which should be fixed by changing the specification.)
A PartialFunction's value receives an additional isDefinedAt member, which is derived from the pattern match in the function literal, with each case's body being replaced by true, and an added default (if none was given) that evaluates to false.
So in this case it ends up with
def isDefinedAt(x: String) = x.length match {
case even if even % 2 == 0 => true
case _ => false
}

val halfSize: PartialFunction[String, Int] = _.length match {
case even if even % 2 == 0 => even / 2
}
The underscore (_) in the above function is just a short hand notation that refers to the single argument of the function. Thus the above snippet is just a short form of
val halfSize: PartialFunction[String, Int] = { (x: String) => x.length match {
case even if even % 2 == 0 => even / 2
}
}

Related

How to create PartialFunction literal after 2.12.5?

Since PartialFunction.apply is deprecated since Scala 2.12.5, how can I create a PartialFunction without declaring a new variable?E.g. I was able to do:
List(0).map(PartialFunction[Int,Int]{
case 0 => 100
case x => x*100
}.orElse{
case 100 => 0
})
How to write the same thing in Scala 2.12.5 in one statement?
The deprecation warning actually tells you what to do: "use an explicit type annotation instead":
val f = ({ case 0 => 100; case x => x * 100 }: PartialFunction[Int, Int])
.orElse[Int, Int]{ case 100 => 0 }
Update
Why the apply was deprecated (probably):
val pf = PartialFunction[Int, Int]{ case 100 => 42 }
pf.isDefinedAt(50)
// res2: Boolean = true
The problem is: it's a partial function that is implicitly cast into normal function, and then again wrapped into partial function. It completely loses all the information about where it is actually defined, so isDefinedAt returns true for all inputs.

Scala Partial Function Type Definition

val even: PartialFunction[Int, String] = PartialFunction[Int, String] {
case i if i % 2 == 0 => i + " is even"
}
val isEven: PartialFunction[Int, String] = {
case i if i % 2 == 0 => i + " is even"
}
val odd: PartialFunction[Int, String] = PartialFunction[Int, String] {
case x if x % 2 == 1 => x + " is odd"
}
val isOdd: PartialFunction[Int, String] = {
case x if x % 2 == 1 => x + " is odd"
}
val tot = even orElse odd
val tot2 = isEven orElse isOdd
println(tot(3))
println(tot2(3))
In this code tot function throws match error while tot2 function works as expected. Only difference between them is the way they're defined. Can anyone explain why such difference in result?
Thanks in advance!!!
At the core the difference is that isDefinedAt on the partial function is not being defined as you would expect on the version using the PartialFunction.apply method. That is actually why this method is now deprecated, PartialFunction.apply is meant to convert a total function into a partial function with isDefinedAt always returning true, which means it will think it is defined for 3 in your example, and try to apply the function instead of falling back to the even function you have supplied as an alternative.
This brings up a common sore point in the community regarding total functions vs partial functions. PartialFunction is a subtype of Function, I guess in the OO design sense that it is a function with an additional method (isDefinedAt) which tells you whether the function is defined for a particular value. Many feel this is a mistake, as in the Liskov sense, Function should be a subtype of PartialFunction because you can use a Function anywhere a PartialFunction is expected, but if you use a PartialFunction where a Function is expected it will compile, then may fail at runtime. My feeling is that it is because Function can be considered to have an implicit isDefinedAt that always returns true, which would allow you to correct the relationship and make Function a subtype of PartialFunction. This comes to a head in PartialFunction.apply which expects a total function and due to this expectation defines isDefinedAt to always return true, but it can’t enforce that expectation, so if you call PartialFunction.apply(somePartialFunction) bad things happen that most programmers won’t expect.
PartialFunction.apply Scaladoc
PartialFunction[Int, String]{...} is syntactic sugar for
PartialFunction[Int, String].apply({...})
To minimize:
val even: PartialFunction[Int, String] = PartialFunction[Int, String]{
case i if i % 2 == 0 => i + " is even"
}
val isEven: PartialFunction[Int, String] = {
case i if i % 2 == 0 => i + " is even"
}
println(even.isDefinedAt(3)) //true
println(isEven.isDefinedAt(3)) //false
In the first two cases you are calling the apply function in the companion object of PartialFunction. I know, that sounds as if it should
work. But it doesn't because PartialFunction.apply should read PartialFunction.fromTotalFunction.
This is a scala language issue and, if I remember correctly (can't find the ticket at the moment, will look later), this apply function will be gone and replaced by fromTotalFunction in Scala 2.13.
UPDATE
The ticket I meant was #6005
It seems that PartialFunction.apply is already deprecated since Scala 2.12.5.

Default empty case for varargs

Say you wish to use pattern matching when calling a method with varargs like so:
def foo(bar: Int*) = ???
val x = false
foo(x match {
case true => 1
case _ =>
})
Running the above code results in type mismatch error, since foo requires an argument of type Int but found Unit instead in the default case. Removing the default case, on the other hand, results in a warning that the match may not be exhaustive, and rightfully so.
My question is, how do I supply an empty default case for the match (which would result in calling foo() without any arguments)?
You could capture the match result in a sequence and represent the lack of arguments as an empty one. Then just splat the result into the parameter:
val x = true
foo((x match {
case true => Seq(1)
case _ => Seq.empty
}):_*)
One option would be to use an Option[Int] instead of an Int:
def foo(bar: Option[Int]*) = ???
val x = false
foo(x match {
case true => Some(1)
case _ => None
})
I think an if-else expression would be less verbose here:
foo(if (x) Some(1) else None)
I'd argue though that if you're matching over a single Boolean there's no point in passing varargs at all.

Magic PartialFunction in Scala

I don't think this code should work, but it does (in Scala 2.10):
scala> ((i: Int) => i.toString match {
| case s if s.length == 2 => "A two digit number"
| case s if s.length == 3 => "A three digit number"
| }): PartialFunction[Int,String]
res0: PartialFunction[Int,String] = <function1>
// other interactions omitted
scala> res0.orElse(PartialFunction((i: Int) => i.toString))
res5: PartialFunction[Int,String] = <function1>
scala> res5(1)
res6: String = 1
How does it work? I would expect a MatchError to be thrown inside res0.
The Scala language specification does not seem to explicitly document how res0 should be interpreted.
The trick is that the compiler is not interpreting your definition as a total function converted to a partial function -- it's actually creating a partial function in the first place. You can verify by noting that res0.isDefinedAt(1) == false.
If you actually convert a total function to a partial function, you will get the behavior you expected:
scala> PartialFunction((i: Int) => i.toString match {
| case s if s.length == 2 => "A two digit number"
| case s if s.length == 3 => "A three digit number"
| })
res0: PartialFunction[Int,String] = <function1>
scala> res0 orElse ({ case i => i.toString }: PartialFunction[Int, String])
res1: PartialFunction[Int,String] = <function1>
scala> res1(1)
scala.MatchError: 1 (of class java.lang.String)
// ...
In this example, PartialFunction.apply treats its argument as a total function, so any information about where it's defined is lost.
orElse is defined on PartialFunction so that the argument is treated as a fallback for the cases when the original is not defined. See the API.
You say that if res0 does not match, you want to try your other pf instead. How this essentially works is:
if (res0.isDefinedAt(1)) {
res0(1)
} else {
other(1)
}
The orElse call creates an instance of OrElse, which inherits from PartialFunction: https://github.com/scala/scala/blob/master/src/library/scala/PartialFunction.scala#L159
When you now call apply on this OrElse, it will call f1.applyOrElse(x, f2): https://github.com/scala/scala/blob/master/src/library/scala/PartialFunction.scala#L162
This will call if (isDefinedAt(x)) apply(x) else f2(x): https://github.com/scala/scala/blob/master/src/library/scala/PartialFunction.scala#L117-L118
Therefore you will only get a MatchError, when neither of the pf's matches.

How is a match word omitted in Scala?

In Scala, you can do
list.filter { item =>
item match {
case Some(foo) => foo.bar > 0
}
}
But you can also do the quicker way by omitting match:
list.filter {
case Some(foo) => foo.bar > 0
}
How is this supported in Scala? Is this new in 2.9? I have been looking for it, and I can figure out what makes this possible. Is it just part of the Scala compiler?
Edit: parts of this answer are wrong; please refer to huynhjl's answer.
If you omit the match, you signal the compiler that you are defining a partial function. A partial function is a function that is not defined for every input value. For instance, your filter function is only defined for values of type Some[A] (for your custom type A).
PartialFunctions throw a MatchError when you try to apply them where they are not defined. Therefore, you should make sure, when you pass a PartialFunction where a regular Function is defined, that your partial function will never be called with an unhanded argument. Such a mechanism is very useful e.g. for unpacking tuples in a collection:
val tupleSeq: Seq[(Int, Int)] = // ...
val sums = tupleSeq.map { case (i1, i2) => i1 + i2 }
APIs which ask for a partial function, like the collect filter-like operation on collections, usually call isDefinedAt before applying the partial function. There, it is safe (and often wanted) to have a partial function that is not defined for every input value.
So you see that although the syntax is close to that of a match, it is actually quite a different thing we're dealing with.
The language specification addresses that in section 8.5. The relevant portions:
An anonymous function can be defined by a sequence of cases
{ case p1 => b1 ... case pn => bn }
If the expected type is scala.Functionk[S1, ..., Sk, R] , the expression is taken to
be equivalent to the anonymous function:
(x1 : S1, ..., xk : Sk) => (x1, ..., xk) match {
case p1 => b1 ... case pn => bn
}
If the expected type is scala.PartialFunction[S, R], the expression is taken to
be equivalent to the following instance creation expression:
new scala.PartialFunction[S, T ] {
def apply(x: S): T = x match {
case p1 => b1 ... case pn => bn
}
def isDefinedAt(x: S): Boolean = {
case p1 => true ... case pn => true
case _ => false
}
}
So typing the expression as PartialFunction or a Function influences how the expression is compiled.
Also trait PartialFunction [-A, +B] extends (A) ⇒ B so a partial function PartialFunction[A,B] is also a Function[A,B].
-- Revised post --
Hmm, I'm not sure I see a difference, Scala 2.9.1.RC3,
val f: PartialFunction[Int, Int] = { case 2 => 3 }
f.isDefinedAt(1) // evaluates to false
f.isDefinedAt(2) // evaluates to true
f(1) // match error
val g: PartialFunction[Int, Int] = x => x match { case 2 => 3 }
g.isDefinedAt(1) // evaluates to false
g.isDefinedAt(2) // evaluates to true
g(1) // match error
It seems f and g behave exactly the same as PartialFunctions.
Here's another example demonstrating the equivalence:
Seq(1, "a").collect(x => x match { case s: String => s }) // evaluates to Seq(a)
Even more interesting:
// this compiles
val g: PartialFunction[Int, Int] = (x: Int) => {x match { case 2 => 3 }}
// this fails; found Function[Int, Int], required PartialFunction[Int, Int]
val g: PartialFunction[Int, Int] = (x: Int) => {(); x match { case 2 => 3 }}
So there's some special casing at the compiler level to convert between x => x match {...} and just {...}.
Update. After reading the language spec, this seems like a bug to me. I filed SI-4940 in the bug tracker.