Equality of functions in Scala, is functions objects in Scala? - scala

I am reading the book Programming in Scala. In the book, it says that "A function literal is compiled into a class that when instantiated at runtime is a function value". And it mentions that "Function values are objects, so you can store them in variables if you like".
So I try to check the equality between functions. But I failed.
If function is object in Scala, then it should behave like other objects in Scala. Maybe check equality of function is meaningless, so it is disabled?
And will function be compiled into object in Scala?

Lambda are compiled as anonymous classes (not case class, as far as I remember). That means if you do:
val f1: (String) => String = _ => "F1"
val f2: (String) => String = _ => "F2"
Both f1 and f2 are subtype of Function1[String,String], but they are of different anonymous classes, so can't equal.
If you write it as:
case class F(res: String) extends ((String) => String) {
def apply(s: String) = res
}
Then:
val f1: (String) => String = F("A")
val f2: (String) => String = F("A")
f1 == f2 // true

It's not clear what "equality" of functions means. Typically, what people care about is "do these two functions compute the same result?"
This, however, is a well-known undecidable problem, the Function Problem. The actual proof is more complex, obviously, but a simple intuition is: if you could tell whether two functions were equal, then you could solve the Halting Problem by asking "is this function equal to while (true) {}?"
So, we cannot decide whether two functions compute the same result. What we could do, is for example, check whether they contain the exact same code. But that is pretty boring. Just some tiny compiler optimization or renaming a single variable will make two functions that intuitively should be equal not equal.
Ergo, we take the easy way out: two functions are equal if they are identical, otherwise they aren't.

Related

Sequencing functions with multiple arguments in Cats

I was going through "Scala with Cats" by Underscore and stumbled upon the following statement:
Cats provides a cats.syntax.apply module that makes use of Semigroupal and Functor to allow users to sequence functions with multiple arguments.
As far as I understand we can only sequence functions with single arguments because functions can return only single value.
import cats.implicits._
val f1: Int => Int = x => x + 1
val f2: Int => Int = x => x * 4
val f3: Int => String = x => s"x = $x"
f1.map(f2).map(f3)(5)
For the following example, it says: Internally mapN uses the Semigroupal to extract the values from the Option and the Functor to apply the values to the function.
final case class Cat(name: String, born: Int, color: String)
(
Option("Garfield"),
Option(1978),
Option("Orange & black")
).mapN(Cat.apply)
// res10: Option[Cat] = Some(Cat("Garfield", 1978, "Orange & black"))
Here Cat.apply is indeed a function with multiple arguments but it is chained to a function that itself returns a tuple of options i.e., a single value. We could probably make it accept multiple arguments like so:
final case class Cat(name: String, born: Int, color: String)
val f: (String, Int, String) => (Option[String], Option[Int], Option[String]) =
(name, year, color) => (Option(name), Option(year), Option(color))
f("Garfield", 1978, "Orange & black").mapN(Cat.apply)
Now we have functions f and Cat.apply that accept multiple arguments and are chained together. Is it what the above statement was pointing to? But then I can't seem to find a way to chain more functions further. Is the statement applicable to chaining multiple argument functions only at one level? Also, notice here that function f is applied eagerly in contrast to the single argument function chaining example depicted above. Is is possible to apply functions lazily here?
There isn't much explanation I could find anywhere on Semigroupal on internet. Could anyone please explain this statement with an example? TIA.
But then I can't seem to find a way to chain more functions further.
I think we can because we can continue mapping over the context for example
f("Garfield", 1978, "Orange & black")
.mapN(Cat.apply)
.map(_.name) // here is another step in the chain of operations
mapN is product + map behind the scenes; we can reveal the constituent map like so
f("Garfield", 1978, "Orange & black")
.tupled
.map { case (a, b, c) => Cat.apply(a, b, c) }
.map { _.name }
where tupled extension method eventually calls Semigroupal#product.
When they say
...allow users to sequence functions with multiple arguments.
my interpretation is not that we keep chaining with just mapN, but instead we can continue chaining over the context in the general sense of the functor, and if at some point in the chain, usually the beginning, we have multiple values within multiple contexts of the same type, then semigroupal + functor allows us to join the values within the single context and continue chaining.
Also, notice here that function f is applied eagerly in contrast to the single argument function chaining example depicted above. Is is possible to apply functions lazily here?
That is kind of the main selling point of Semigroupal over the Monad, that is, to be able to "eagerly" execute independent operations, join the resulting values within the context and then continue chaining. With monadic chaining even if operations are independent one would still have to wait for the other before continuing the chain.

Why does Scala not have a return/unit function defined for each monad (in contrast to Haskell)?

What is the reason behind the design decision in Scala that monads do not have a return/unit function in contrast to Haskell where each monad has a return function that puts a value into a standard monadic context for the given monad?
For example why List, Option, Set etc... do not have a return/unit functions defined in the standard library as shown in the slides below?
I am asking this because in the reactive Coursera course Martin Odersky explicitly mentioned this fact, as can be seen below in the slides, but did not explain why Scala does not have them even though unit/return is an essential property of a monad.
As Ørjan Johansen said, Scala does not support method dispatching on return type. Scala object system is built over JVM one, and JVM invokevirtual instruction, which is the main tool for dynamic polymorphism, dispatches the call based on type of this object.
As a side note, dispatching is a process of selecting concrete method to call. In Scala/Java all methods are virtual, that is, the actual method which is called depends on actual type of the object.
class A { def hello() = println("hello method in A") }
class B extends A { override def hello() = println("hello method in B") }
val x: A = new A
x.hello() // prints "hello method in A"
val y: A = new B
y.hello() // prints "hello method in B"
Here, even if y variable is of type A, hello method from B is called, because JVM "sees" that the actual type of the object in y is B and invokes appropriate method.
However, JVM only takes the type of the variable on which the method is called into account. It is impossible, for example, to call different methods based on runtime type of arguments without explicit checks. For example:
class A {
def hello(x: Number) = println(s"Number: $x")
def hello(y: Int) = println(s"Integer: $y")
}
val a = new A
val n: Number = 10: Int
a.hello(n) // prints "Number: 10"
Here we have two methods with the same name, but with different parameter type. And even if ns actual type is Int, hello(Number) version is called - it is resolved statically based on n static variable type (this feature, static resolution based on argument types, is called overloading). Hence, there is no dynamic dispatch on method arguments. Some languages support dispatching on method arguments too, for example, Common Lisp's CLOS or Clojure's multimethods work like that.
Haskell has advanced type system (it is comparable to Scala's and in fact they both originate in System F, but Scala type system supports subtyping which makes type inference much more difficult) which allows global type inference, at least, without certain extensions enabled. Haskell also has a concept of type classes, which is its tool for dynamic polymorphism. Type classes can be loosely thought of as interfaces without inheritance but with dispatch on parameter and return value types. For example, this is a valid type class:
class Read a where
read :: String -> a
instance Read Integer where
read s = -- parse a string into an integer
instance Read Double where
read s = -- parse a string into a double
Then, depending on the context where method is called, read function for Integer or Double can be called:
x :: Integer
x = read "12345" // read for Integer is called
y :: Double
y = read "12345.0" // read for Double is called
This is a very powerful technique which has no correspondence in bare JVM object system, so Scala object system does not support it too. Also the lack of full-scale type inference would make this feature somewhat cumbersome to use. So, Scala standard library does not have return/unit method anywhere - it is impossible to express it using regular object system, there is simply no place where such a method could be defined. Consequently, monad concept in Scala is implicit and conventional - everything with appropriate flatMap method can be considered a monad, and everything with the right methods can be used in for construction. This is much like duck typing.
However, Scala type system together with its implicits mechanism is powerful enough to express full-featured type classes, and, by extension, generic monads in formal way, though due to difficulties in full type inference it may require adding more type annotations than in Haskell.
This is definition of monad type class in Scala:
trait Monad[M[_]] {
def unit[A](a: A): M[A]
def bind[A, B](ma: M[A])(f: A => M[B]): M[B]
}
And this is its implementation for Option:
implicit object OptionMonad extends Monad[Option] {
def unit[A](a: A) = Some(a)
def bind[A, B](ma: Option[A])(f: A => Option[B]): Option[B] =
ma.flatMap(f)
}
Then this can be used in generic way like this:
// note M[_]: Monad context bound
// this is a port of Haskell's filterM found here:
// http://hackage.haskell.org/package/base-4.7.0.1/docs/src/Control-Monad.html#filterM
def filterM[M[_]: Monad, A](as: Seq[A])(f: A => M[Boolean]): M[Seq[A]] = {
val m = implicitly[Monad[M]]
as match {
case x +: xs =>
m.bind(f(x)) { flg =>
m.bind(filterM(xs)(f)) { ys =>
m.unit(if (flg) x +: ys else ys)
}
}
case _ => m.unit(Seq.empty[A])
}
}
// using it
def run(s: Seq[Int]) = {
import whatever.OptionMonad // bring type class instance into scope
// leave all even numbers in the list, but fail if the list contains 13
filterM[Option, Int](s) { a =>
if (a == 13) None
else if (a % 2 == 0) Some(true)
else Some(false)
}
}
run(1 to 16) // returns None
run(16 to 32) // returns Some(List(16, 18, 20, 22, 24, 26, 28, 30, 32))
Here filterM is written generically, for any instance of Monad type class. Because OptionMonad implicit object is present at filterM call site, it will be passed to filterM implicitly, and it will be able to make use of its methods.
You can see from above that type classes allow to emulate dispatching on return type even in Scala. In fact, this is exactly what Haskell does under the covers - both Scala and Haskell are passing a dictionary of methods implementing some type class, though in Scala it is somewhat more explicit because these "dictionaries" are first-class objects there and can be imported on demand or even passed explicitly, so it is not really a proper dispatching as it is not that embedded.
If you need this amount of genericity, you can use Scalaz library which contains a lot of type classes (including monad) and their instances for some common types, including Option.
I don't think you're really saying that Scala's monads don't have a unit function - it's rather just that the name of the unit function can vary. That's what seems to be shown in the second slide's examples.
As for why that is so, I think it's just because Scala runs on the JVM, and those function have to be implemented as JVM methods - which are uniquely identified by:
the class they belong to;
their name;
their parameters types.
But they are not identified by their return type. Since the parameter type generally won't differentiate the various unit functions (it's usually just a generic type), you need different names for them.
In practice, they will often be implemented as the apply(x) method on the companion object of the monad class. For example, for the class List, the unit function is the apply(x) method on the object List. By convention, List.apply(x) can be called as List(x) too, which is more common/idiomatic.
So I guess that Scala at least has a naming convention for the unit function, though it doesn't have a unique name for it :
// Some monad :
class M[T] {
def flatMap[U](f: T => M[U]): M[U] = ???
}
// Companion object :
object M {
def apply(x: T): M[T] = ??? // Unit function
}
// Usage of the unit function :
val x = ???
val m = M(x)
Caveat: I'm still learning Haskell and I'm sort of making up this answer as I go.
First, what you already know - that Haskell's do notation desugars to bind:
Borrowing this example from Wikipedia:
add mx my = do
x <- mx
y <- my
return (x + y)
add mx my =
mx >>= (\x ->
my >>= (\y ->
return (x + y)))
Scala's analogue to do is the for-yield expression. It similarly desugars each step to flatMap (its equivalent of bind).
There's a difference, though: The last <- in a for-yield desugars to map, not to flatMap.
def add(mx: Option[Int], my: Option[Int]) =
for {
x <- mx
y <- my
} yield x + y
def add(mx: Option[Int], my: Option[Int]) =
mx.flatMap(x =>
my.map(y =>
x + y))
So because you don't have the "flattening" on the last step, the expression value already has the monad type, so there's no need to "re-wrap" it with something comparable to return.
Actually there is a return function in scala. It is just hard to find.
Scala slightly differs from Haskell in many aspects. Most of that differences are direct consequences of JVM limitations. JVM can not dispatch methods basing on its return type. So Scala introduced type class polymorphism based on implicit evidence to fix this inconvenience.
It is even used in scala standard collections. You may notice numerous usage of CanBuildFrom and CanBuild implicits used in the scala collection api. See scala.collection.immutable.List for example.
Every time you want to build custom collection you should write realization for this implicits. There are not so many guides for writing one though. I recommend you this guide. It shows why CanBuildFrom is so important for collections and how it is used. In fact that is just another form of the return function and anyone familiar with Haskell monads would understand it's importance clearly.
So you may use custom collection as example monads and write other monads basing on provided tutorial.

PartialFunction That Isn't Partial

Is there a reason to use a PartialFunction on a function that's not partial?
scala> val foo: PartialFunction[Int, Int] = {
| case x => x * 2
| }
foo: PartialFunction[Int,Int] = <function1>
foo is defined as a PartialFunction, but of course the case x will catch all input.
Is this simply bad code as the PartialFunction type indicates to the programmer that the function is undefined for certain inputs?
There is no advantage in using a PartialFunction instead of a Function, but if you have to pass a PartialFunction, then you have to pass a PartialFunction.
Note that, because of the inheritance between these two, overloading a method to accept both results in something difficult to use, as the type inference won't work.
The thing is, there are many examples of times when what you need to define on a trait/object/function definition is a PartialFunction but in reality the real implementation may not be one. Case in point, take a look at def collect[B](f: PartialFunction[A,B]):
val myList = thatList collect {
case Right(value) => value
case Left(other) => other.toInt
}
It's clearly not a "real" partial as it is defined for all input. That said, if I wanted to, I could just have the Right match.
However, if I were to have written collect as a full on plain function, then I'd miss out on the desired behavior (that is to be both a filter and a map rolled into one base on when a function is defined.) That's nice behavior and allows for a lot of flexibility when writing my own code.
So I guess the better question is, will you ever want behavior to reflect that a function might not be defined everywhere? If the answer is no, then don't do it.
PartialFunction literals allow pattern matching directly on arguments (e.g. { case (a, b) => ... } instead of _ match { case (a, b) => ... }), which makes code more readable (see #wheaties' answer for another example).
EDIT: apparently this is wrong, see Daniel C. Sobral's comment on his answer. Not deleting, so that the comments still make sense.

Is there any fundamental limitations that stops Scala from implementing pattern matching over functions?

In languages like SML, Erlang and in buch of others we may define functions like this:
fun reverse [] = []
| reverse x :: xs = reverse xs # [x];
I know we can write analog in Scala like this (and I know, there are many flaws in the code below):
def reverse[T](lst: List[T]): List[T] = lst match {
case Nil => Nil
case x :: xs => reverse(xs) ++ List(x)
}
But I wonder, if we could write former code in Scala, perhaps with desugaring to the latter.
Is there any fundamental limitations for such syntax being implemented in the future (I mean, really fundamental -- e.g. the way type inference works in scala, or something else, except parser obviously)?
UPD
Here is a snippet of how it could look like:
type T
def reverse(Nil: List[T]) = Nil
def reverse(x :: xs: List[T]): List[T] = reverse(xs) ++ List(x)
It really depends on what you mean by fundamental.
If you are really asking "if there is a technical showstopper that would prevent to implement this feature", then I would say the answer is no. You are talking about desugaring, and you are on the right track here. All there is to do is to basically stitch several separates cases into one single function, and this can be done as a mere preprocessing step (this only requires syntactic knowledge, no need for semantic knowledge). But for this to even make sense, I would define a few rules:
The function signature is mandatory (in Haskell by example, this would be optional, but it is always optional whether you are defining the function at once or in several parts). We could try to arrange to live without the signature and attempt to extract it from the different parts, but lack of type information would quickly come to byte us. A simpler argument is that if we are to try to infer an implicit signature, we might as well do it for all the methods. But the truth is that there are very good reasons to have explicit singatures in scala and I can't imagine to change that.
All the parts must be defined within the same scope. To start with, they must be declared in the same file because each source file is compiled separately, and thus a simple preprocessor would not be enough to implement the feature. Second, we still end up with a single method in the end, so it's only natural to have all the parts in the same scope.
Overloading is not possible for such methods (otherwise we would need to repeat the signature for each part just so the preprocessor knows which part belongs to which overload)
Parts are added (stitched) to the generated match in the order they are declared
So here is how it could look like:
def reverse[T](lst: List[T]): List[T] // Exactly like an abstract def (provides the signature)
// .... some unrelated code here...
def reverse(Nil) = Nil
// .... another bit of unrelated code here...
def reverse(x :: xs ) = reverse(xs) ++ List(x)
Which could be trivially transformed into:
def reverse[T](list: List[T]): List[T] = lst match {
case Nil => Nil
case x :: xs => reverse(xs) ++ List(x)
}
// .... some unrelated code here...
// .... another bit of unrelated code here...
It is easy to see that the above transformation is very mechanical and can be done by just manipulating a source AST (the AST produced by the slightly modified grammar that accepts this new constructs), and transforming it into the target AST (the AST produced by the standard scala grammar).
Then we can compile the result as usual.
So there you go, with a few simple rules we are able to implement a preprocessor that does all the work to implement this new feature.
If by fundamental you are asking "is there anything that would make this feature out of place" then it can be argued that this does not feel very scala. But more to the point, it does not bring that much to the table. Scala author(s) actually tend toward making the language simpler (as in less built-in features, trying to move some built-in features into libraries) and adding a new syntax that is not really more readable goes against the goal of simplification.
In SML, your code snippet is literally just syntactic sugar (a "derived form" in the terminology of the language spec) for
val rec reverse = fn x =>
case x of [] => []
| x::xs = reverse xs # [x]
which is very close to the Scala code you show. So, no there is no "fundamental" reason that Scala couldn't provide the same kind of syntax. The main problem is Scala's need for more type annotations, which makes this shorthand syntax far less attractive in general, and probably not worth the while.
Note also that the specific syntax you suggest would not fly well, because there is no way to distinguish one case-by-case function definition from two overloaded functions syntactically. You probably would need some alternative syntax, similar to SML using "|".
I don't know SML or Erlang, but I know Haskell. It is a language without method overloading. Method overloading combined with such pattern matching could lead to ambiguities. Imagine following code:
def f(x: String) = "String "+x
def f(x: List[_]) = "List "+x
What should it mean? It can mean method overloading, i.e. the method is determined in compile time. It can also mean pattern matching. There would be just a f(x: AnyRef) method that would do the matching.
Scala also has named parameters, which would be probably also broken.
I don't think that Scala is able to offer more simple syntax than you have shown in general. A simpler syntax may IMHO work in some special cases only.
There are at least two problems:
[ and ] are reserved characters because they are used for type arguments. The compiler allows spaces around them, so that would not be an option.
The other problem is that = returns Unit. So the expression after the | would not return any result
The closest I could come up with is this (note that is very specialized towards your example):
// Define a class to hold the values left and right of the | sign
class |[T, S](val left: T, val right: PartialFunction[T, T])
// Create a class that contains the | operator
class OrAssoc[T](left: T) {
def |(right: PartialFunction[T, T]): T | T = new |(left, right)
}
// Add the | to any potential target
implicit def anyToOrAssoc[S](left: S): OrAssoc[S] = new OrAssoc(left)
object fun {
// Use the magic of the update method
def update[T, S](choice: T | S): T => T = { arg =>
if (choice.right.isDefinedAt(arg)) choice.right(arg)
else choice.left
}
}
// Use the above construction to define a new method
val reverse: List[Int] => List[Int] =
fun() = List.empty[Int] | {
case x :: xs => reverse(xs) ++ List(x)
}
// Call the method
reverse(List(3, 2, 1))

scala loan pattern, optional function param

I have a loan pattern that applies a function n times where 'i' is the incrementing variable. "Occasionally", I want the function passed in to have access to 'i'....but I don't want to require all functions passed in to require defining a param to accept 'i'. Example below...
def withLoaner = (n:Int) => (op:(Int) => String) => {
val result = for(i <- 1 to n) yield op(i)
result.mkString("\n")
}
def bob = (x:Int) => "bob" // don't need access to i. is there a way use () => "bob" instead?
def nums = (x:Int) => x.toString // needs access to i, define i as an input param
println(withLoaner(3)(bob))
println(withLoaner(3)(nums))
def withLoaner(n: Int) = new {
def apply(op: Int => String) : String = (1 to n).map(op).mkString("\n")
def apply(op: () => String) : String = apply{i: Int => op()}
}
(not sure how it is related to the loan pattern)
Edit Little explanation as requested in comment.
Not sure what you know and don't know of scala and what you don't undestand in that code. so sorry if what I just belabor the obvious.
First, a scala program consist of traits/classes (also singleton object) and methods. Everything that is done is done by methods (leaving constructor aside). Functions (as opposed to methods) are instances of (subtypes of) the various FunctionN traits (N the number of arguments). Each of them has as apply method that is the actual implemention.
If you write
val inc = {i: Int => i + 1}
it is desugared to
val inc = new Function1[Int, Int] {def apply(i: Int) = i + 1}
(defines an anonymous class extending Function1, with given apply method and creating an instance)
So writing a function has rather more weight than a simple method. Also you cannot have overloading (several methods with the same name, differing by the signature, just what I did above), nor use named arguments, or default value for arguments.
On the other hand, functions are first classes values (they can be passed as arguments, returned as result) while methods are not. They are automatically converted to functions when needed, however there may be some edges cases when doing that. If a method is intended solely to be used as a function value, rather than called as a method, it might be better to write a function.
A function f, with its apply method, is called with f(x) rather than f.apply(x) (which works too), because scala desugars function call notation on a value (value followed by parentheses and 0 or more args) to a call to method apply. f(x) is syntactic sugar for f.apply(x). This works whatever the type of f, it does not need to be one of the FunctionN.
What is done in withLoaner is returning an object (of an anonymous type, but one could have defined a class separately and returned an instance of it). The object has two apply methods, one accepting an Int => String, the other one an () => String. When you do withLoaner(n)(f) it means withLoaner(n).apply(f). The appropriate apply method is selected, if f has the proper type for one of them, otherwise, compile error.
Just in case you wonder withLoaner(n) does not mean withLoaner.apply(n) (or it would never stop, that could just as well mean withLoaner.apply.apply(n)), as withLoaner is a method, not a value.