Scala secondary constructor "Application does not take parameters" error [duplicate] - scala

I think I see the merit in defining auxiliary constructors in such a way that the primary constructor is the solitary point of entry to the class. But why can't I do something like this?
class Wibble(foo: Int, bar: String) {
def this(baz: List[Any]) = {
val bazLength = baz.length
val someText = "baz length is " ++ bazLength.toString
this(bazLength, someText)
}
}
Is it maybe a way of guaranteeing that the auxiliary constructor doesn't have side effects and/or can't return early?

Auxiliary constructors can contain more than a single invocation of another constructor, but their first statement must be said invocation.
As explained in Programming in Scala, ch. 6.7:
In Scala, every auxiliary constructor must invoke another constructor of
the same class as its first action. In other words, the first statement in every
auxiliary constructor in every Scala class will have the form this(. . . ).
The invoked constructor is either the primary constructor (as in the Rational
example), or another auxiliary constructor that comes textually before the
calling constructor. The net effect of this rule is that every constructor invocation
in Scala will end up eventually calling the primary constructor of the
class. The primary constructor is thus the single point of entry of a class.
If you’re familiar with Java, you may wonder why Scala’s rules for
constructors are a bit more restrictive than Java’s. In Java, a constructor
must either invoke another constructor of the same class, or directly invoke
a constructor of the superclass, as its first action. In a Scala class, only the
primary constructor can invoke a superclass constructor. The increased
restriction in Scala is really a design trade-off that needed to be paid in
exchange for the greater conciseness and simplicity of Scala’s constructors
compared to Java’s.
Just as in Java, one can get round this limitation by extracting the code to be executed before the primary constructor call into a separate method. In Scala it is a bit more tricky than in Java, as apparently you need to move this helper method into the companion object in order to be allowed to call it from the constructor.
Moreover, your specific case is awkward as you have two constructor parameters and - although one can return tuples from a function - this returned tuple is then not accepted as the argument list to the primary constructor. For ordinary functions, you can use tupled, but alas, this doesn't seem to work for constructors. A workaround would be to add yet another auxiliary constructor:
object Wibble {
private def init(baz: List[Any]): (Int, String) = {
val bazLength = baz.length
val someText = "baz length is " ++ bazLength.toString
println("init")
(bazLength, someText)
}
}
class Wibble(foo: Int, bar: String) {
println("Wibble wobble")
def this(t: (Int, String)) = {
this(t._1, t._2)
println("You can execute more code here")
}
def this(baz: List[Any]) = {
this(Wibble.init(baz))
println("You can also execute some code here")
}
}
This at least works, even if it is slightly complicated.
scala> val w = new Wibble(List(1, 2, 3))
init
Wibble wobble
You can execute more code here
You can also execute some code here
w: Wibble = Wibble#b6e385
Update
As #sschaef's pointed out in his comment, this can be simplified using a factory method in the companion object:
object Wobble {
def apply(baz: List[Any]): Wobble = {
val bazLength = baz.length
val someText = "baz length is " ++ bazLength.toString
println("init")
new Wobble(bazLength, someText)
}
}
class Wobble(foo: Int, bar: String) {
println("Wobble wibble")
}
Thus we need no new to create an object anymore:
scala> val w = Wobble(List(1, 2, 3))
init
Wobble wibble
w: Wobble = Wobble#47c130

Related

How does Scala transform case classes to be accepted as functions?

I am trying to understand how a case class can be passed as an argument to a function which accepts functions as arguments. Below is an example:
Consider the below function
def !![B](h: Out[B] => A): In[B] = { ... }
If I understood correctly, this is a polymorphic method which has a type parameter B and accepts a function h as a parameter. Out and In are other two classes defined previously.
This function is then being used as shown below:
case class Q(p: boolean)(val cont: Out[R])
case class R(p: Int)
def g(c: Out[Q]) = {
val rin = c !! Q(true)_
...
}
I am aware that currying is being used to avoid writing the type annotation and instead just writing _. However, I cannot grasp why and how the case class Q is transformed to a function (h) of type Out[B] => A.
EDIT 1 Updated !! above and the In and Out definitions:
abstract class In[+A] {
def future: Future[A]
def receive(implicit d: Duration): A = {
Await.result[A](future, d)
}
def ?[B](f: A => B)(implicit d: Duration): B = {
f(receive)
}
}
abstract class Out[-A]{
def promise[B <: A]: Promise[B]
def send(msg: A): Unit = promise.success(msg)
def !(msg: A) = send(msg)
def create[B](): (In[B], Out[B])
}
These code samples are taken from the following paper: http://drops.dagstuhl.de/opus/volltexte/2016/6115/
TLDR;
Using a case class with multiple parameter lists and partially applying it will yield a partially applied apply call + eta expansion will transform the method into a function value:
val res: Out[Q] => Q = Q.apply(true) _
Longer explanation
To understand the way this works in Scala, we have to understand some fundamentals behind case classes and the difference between methods and functions.
Case classes in Scala are a compact way of representing data. When you define a case class, you get a bunch of convenience methods which are created for you by the compiler, such as hashCode and equals.
In addition, the compiler also generates a method called apply, which allows you to create a case class instance without using the new keyword:
case class X(a: Int)
val x = X(1)
The compiler will expand this call to
val x = X.apply(1)
The same thing will happen with your case class, only that your case class has multiple argument lists:
case class Q(p: boolean)(val cont: Out[R])
val q: Q = Q(true)(new Out[Int] { })
Will get translated to
val q: Q = Q.apply(true)(new Out[Int] { })
On top of that, Scala has a way to transform methods, which are a non value type, into a function type which has the type of FunctionX, X being the arity of the function. In order to transform a method into a function value, we use a trick called eta expansion where we call a method with an underscore.
def foo(i: Int): Int = i
val f: Int => Int = foo _
This will transform the method foo into a function value of type Function1[Int, Int].
Now that we posses this knowledge, let's go back to your example:
val rin = c !! Q(true) _
If we just isolate Q here, this call gets translated into:
val rin = Q.apply(true) _
Since the apply method is curried with multiple argument lists, we'll get back a function that given a Out[Q], will create a Q:
val rin: Out[R] => Q = Q.apply(true) _
I cannot grasp why and how the case class Q is transformed to a function (h) of type Out[B] => A.
It isn't. In fact, the case class Q has absolutely nothing to do with this! This is all about the object Q, which is the companion module to the case class Q.
Every case class has an automatically generated companion module, which contains (among others) an apply method whose signature matches the primary constructor of the companion class, and which constructs an instance of the companion class.
I.e. when you write
case class Foo(bar: Baz)(quux: Corge)
You not only get the automatically defined case class convenience methods such as accessors for all the elements, toString, hashCode, copy, and equals, but you also get an automatically defined companion module that serves both as an extractor for pattern matching and as a factory for object construction:
object Foo {
def apply(bar: Baz)(quux: Corge) = new Foo(bar)(quux)
def unapply(that: Foo): Option[Baz] = ???
}
In Scala, apply is a method that allows you to create "function-like" objects: if foo is an object (and not a method), then foo(bar, baz) is translated to foo.apply(bar, baz).
The last piece of the puzzle is η-expansion, which lifts a method (which is not an object) into a function (which is an object and can thus be passed as an argument, stored in a variable, etc.) There are two forms of η-expansion: explicit η-expansion using the _ operator:
val printFunction = println _
And implicit η-expansion: in cases where Scala knows 100% that you mean a function but you give it the name of a method, Scala will perform η-expansion for you:
Seq(1, 2, 3) foreach println
And you already know about currying.
So, if we put it all together:
Q(true)_
First, we know that Q here cannot possibly be the class Q. How do we know that? Because Q here is used as a value, but classes are types, and like most programming languages, Scala has a strict separation between types and values. Therefore, Q must be a value. In particular, since we know class Q is a case class, object Q is the companion module for class Q.
Secondly, we know that for a value Q
Q(true)
is syntactic sugar for
Q.apply(true)
Thirdly, we know that for case classes, the companion module has an automatically generated apply method that matches the primary constructor, so we know that Q.apply has two parameter lists.
So, lastly, we have
Q.apply(true) _
which passes the first argument list to Q.apply and then lifts Q.apply into a function which accepts the second argument list.
Note that case classes with multiple parameter lists are unusual, since only the parameters in the first parameter list are considered elements of the case class, and only elements benefit from the "case class magic", i.e. only elements get accessors implemented automatically, only elements are used in the signature of the copy method, only elements are used in the automatically generated equals, hashCode, and toString() methods, and so on.

I need a specific example of how to define a local parameter in the primary constructor of an immutable _case_ class

I have normal Scala class I am wanting to refactor to become an immutable case class. As I'm needing the class to be well-behaved in Set operations, I want all the Scala compiler automatically generated methods provided on a case class. IOW, I am wanting to avoid having to write these various methods; equals, hashCode, toString, etc., as that is very error prone. And I am needing to do this for a raft of classes, so I need a general solution, not just a specific solution anomalous quick fix or hack.
Here's the class with which I am working:
class Node(val identity: String, childrenArg: List[Node], customNodeArg: CustomNode) {
val children: List[Node] = childrenArg
val customNode: CustomNode = customNodeArg
}
As you can see, the class's constructor has three parameters. The first one, identity, is a read-only property. The remaining two, childrenArg and customNodeArg, are just a normal method parameters; i.e. they are only present during the construction of the instance and then disappears altogether from the class instance (unless otherwise captured) upon execution completion of the class constructor.
My first naive attempt to convert this to an immutable case class was this (just removing val from the first parameter):
class Node(identity: String, childrenArg: List[Node], customNodeArg: CustomNode) {
val children: List[Node] = childrenArg
val customNode: CustomNode = customNodeArg
}
However, this resulted in the undesired effect of both the childrenArg and customNodeArg parameters now being elevated to become (read-only) properties (as opposed to leaving them as normal method parameters). And this had the further undesired effect of having them included in the compiler generated equals and hashCode implementations.
How do I mark the immutable case class's constructor parameters childrenArg and customNodeArg such that identity is the only read-only property of the case class?
Any guidance on this; answers, website discussion links, etc., are greatly appreciated.
A second parameter list seems to do the trick:
scala> trait CustomNode
defined trait CustomNode
scala> case class Node(identity: String)(childrenArg: List[Node], customNodeArg: CustomNode)
defined class Node
scala> val n = Node("id")(Nil, null)
n: Node = Node(id)
scala> n.identity
res0: String = id
scala> n.getClass.getDeclaredFields.map(_.getName)
res1: Array[String] = Array(identity)
Case class parameters are vals by default, but you can set them to vars.
case class Node(identity: String, var childrenArg: List[Node], var customNodeArg: CustomNode)
Making them vars gives you getters and setters automatically

Why can auxiliary constructors in Scala only consist of a single call to another constructor?

I think I see the merit in defining auxiliary constructors in such a way that the primary constructor is the solitary point of entry to the class. But why can't I do something like this?
class Wibble(foo: Int, bar: String) {
def this(baz: List[Any]) = {
val bazLength = baz.length
val someText = "baz length is " ++ bazLength.toString
this(bazLength, someText)
}
}
Is it maybe a way of guaranteeing that the auxiliary constructor doesn't have side effects and/or can't return early?
Auxiliary constructors can contain more than a single invocation of another constructor, but their first statement must be said invocation.
As explained in Programming in Scala, ch. 6.7:
In Scala, every auxiliary constructor must invoke another constructor of
the same class as its first action. In other words, the first statement in every
auxiliary constructor in every Scala class will have the form this(. . . ).
The invoked constructor is either the primary constructor (as in the Rational
example), or another auxiliary constructor that comes textually before the
calling constructor. The net effect of this rule is that every constructor invocation
in Scala will end up eventually calling the primary constructor of the
class. The primary constructor is thus the single point of entry of a class.
If you’re familiar with Java, you may wonder why Scala’s rules for
constructors are a bit more restrictive than Java’s. In Java, a constructor
must either invoke another constructor of the same class, or directly invoke
a constructor of the superclass, as its first action. In a Scala class, only the
primary constructor can invoke a superclass constructor. The increased
restriction in Scala is really a design trade-off that needed to be paid in
exchange for the greater conciseness and simplicity of Scala’s constructors
compared to Java’s.
Just as in Java, one can get round this limitation by extracting the code to be executed before the primary constructor call into a separate method. In Scala it is a bit more tricky than in Java, as apparently you need to move this helper method into the companion object in order to be allowed to call it from the constructor.
Moreover, your specific case is awkward as you have two constructor parameters and - although one can return tuples from a function - this returned tuple is then not accepted as the argument list to the primary constructor. For ordinary functions, you can use tupled, but alas, this doesn't seem to work for constructors. A workaround would be to add yet another auxiliary constructor:
object Wibble {
private def init(baz: List[Any]): (Int, String) = {
val bazLength = baz.length
val someText = "baz length is " ++ bazLength.toString
println("init")
(bazLength, someText)
}
}
class Wibble(foo: Int, bar: String) {
println("Wibble wobble")
def this(t: (Int, String)) = {
this(t._1, t._2)
println("You can execute more code here")
}
def this(baz: List[Any]) = {
this(Wibble.init(baz))
println("You can also execute some code here")
}
}
This at least works, even if it is slightly complicated.
scala> val w = new Wibble(List(1, 2, 3))
init
Wibble wobble
You can execute more code here
You can also execute some code here
w: Wibble = Wibble#b6e385
Update
As #sschaef's pointed out in his comment, this can be simplified using a factory method in the companion object:
object Wobble {
def apply(baz: List[Any]): Wobble = {
val bazLength = baz.length
val someText = "baz length is " ++ bazLength.toString
println("init")
new Wobble(bazLength, someText)
}
}
class Wobble(foo: Int, bar: String) {
println("Wobble wibble")
}
Thus we need no new to create an object anymore:
scala> val w = Wobble(List(1, 2, 3))
init
Wobble wibble
w: Wobble = Wobble#47c130

What is the apply function in Scala?

I never understood it from the contrived unmarshalling and verbing nouns ( an AddTwo class has an apply that adds two!) examples.
I understand that it's syntactic sugar, so (I deduced from context) it must have been designed to make some code more intuitive.
What meaning does a class with an apply function give? What is it used for, and what purposes does it make code better (unmarshalling, verbing nouns etc)?
how does it help when used in a companion object?
Mathematicians have their own little funny ways, so instead of saying "then we call function f passing it x as a parameter" as we programmers would say, they talk about "applying function f to its argument x".
In mathematics and computer science, Apply is a function that applies
functions to arguments.
Wikipedia
apply serves the purpose of closing the gap between Object-Oriented and Functional paradigms in Scala. Every function in Scala can be represented as an object. Every function also has an OO type: for instance, a function that takes an Int parameter and returns an Int will have OO type of Function1[Int,Int].
// define a function in scala
(x:Int) => x + 1
// assign an object representing the function to a variable
val f = (x:Int) => x + 1
Since everything is an object in Scala f can now be treated as a reference to Function1[Int,Int] object. For example, we can call toString method inherited from Any, that would have been impossible for a pure function, because functions don't have methods:
f.toString
Or we could define another Function1[Int,Int] object by calling compose method on f and chaining two different functions together:
val f2 = f.compose((x:Int) => x - 1)
Now if we want to actually execute the function, or as mathematician say "apply a function to its arguments" we would call the apply method on the Function1[Int,Int] object:
f2.apply(2)
Writing f.apply(args) every time you want to execute a function represented as an object is the Object-Oriented way, but would add a lot of clutter to the code without adding much additional information and it would be nice to be able to use more standard notation, such as f(args). That's where Scala compiler steps in and whenever we have a reference f to a function object and write f (args) to apply arguments to the represented function the compiler silently expands f (args) to the object method call f.apply (args).
Every function in Scala can be treated as an object and it works the other way too - every object can be treated as a function, provided it has the apply method. Such objects can be used in the function notation:
// we will be able to use this object as a function, as well as an object
object Foo {
var y = 5
def apply (x: Int) = x + y
}
Foo (1) // using Foo object in function notation
There are many usage cases when we would want to treat an object as a function. The most common scenario is a factory pattern. Instead of adding clutter to the code using a factory method we can apply object to a set of arguments to create a new instance of an associated class:
List(1,2,3) // same as List.apply(1,2,3) but less clutter, functional notation
// the way the factory method invocation would have looked
// in other languages with OO notation - needless clutter
List.instanceOf(1,2,3)
So apply method is just a handy way of closing the gap between functions and objects in Scala.
It comes from the idea that you often want to apply something to an object. The more accurate example is the one of factories. When you have a factory, you want to apply parameter to it to create an object.
Scala guys thought that, as it occurs in many situation, it could be nice to have a shortcut to call apply. Thus when you give parameters directly to an object, it's desugared as if you pass these parameters to the apply function of that object:
class MyAdder(x: Int) {
def apply(y: Int) = x + y
}
val adder = new MyAdder(2)
val result = adder(4) // equivalent to x.apply(4)
It's often use in companion object, to provide a nice factory method for a class or a trait, here is an example:
trait A {
val x: Int
def myComplexStrategy: Int
}
object A {
def apply(x: Int): A = new MyA(x)
private class MyA(val x: Int) extends A {
val myComplexStrategy = 42
}
}
From the scala standard library, you might look at how scala.collection.Seq is implemented: Seq is a trait, thus new Seq(1, 2) won't compile but thanks to companion object and apply, you can call Seq(1, 2) and the implementation is chosen by the companion object.
Here is a small example for those who want to peruse quickly
object ApplyExample01 extends App {
class Greeter1(var message: String) {
println("A greeter-1 is being instantiated with message " + message)
}
class Greeter2 {
def apply(message: String) = {
println("A greeter-2 is being instantiated with message " + message)
}
}
val g1: Greeter1 = new Greeter1("hello")
val g2: Greeter2 = new Greeter2()
g2("world")
}
output
A greeter-1 is being instantiated with message hello
A greeter-2 is being instantiated with message world
TLDR for people comming from c++
It's just overloaded operator of ( ) parentheses
So in scala:
class X {
def apply(param1: Int, param2: Int, param3: Int) : Int = {
// Do something
}
}
Is same as this in c++:
class X {
int operator()(int param1, int param2, int param3) {
// do something
}
};
1 - Treat functions as objects.
2 - The apply method is similar to __call __ in Python, which allows you to use an instance of a given class as a function.
The apply method is what turns an object into a function. The desire is to be able to use function syntax, such as:
f(args)
But Scala has both functional and object oriented syntax. One or the other needs to be the base of the language. Scala (for a variety of reasons) chooses object oriented as the base form of the language. That means that any function syntax has to be translated into object oriented syntax.
That is where apply comes in. Any object that has the apply method can be used with the syntax:
f(args)
The scala infrastructure then translates that into
f.apply(args)
f.apply(args) has correct object oriented syntax. Doing this translation would not be possible if the object had no apply method!
In short, having the apply method in an object is what allows Scala to turn the syntax: object(args) into the syntax: object.apply(args). And object.apply(args) is in the form that can then execute.
FYI, this implies that all functions in scala are objects. And it also implies that having the apply method is what makes an object a function!
See the accepted answer for more insight into just how a function is an object, and the tricks that can be played as a result.
To put it crudely,
You can just see it as custom ()operator. If a class X has an apply() method, whenever you call X() you will be calling the apply() method.

Scala singleton factories and class constants

OK, in the question about 'Class Variables as constants', I get the fact that the constants are not available until after the 'official' constructor has been run (i.e. until you have an instance). BUT, what if I need the companion singleton to make calls on the class:
object thing {
val someConst = 42
def apply(x: Int) = new thing(x)
}
class thing(x: Int) {
import thing.someConst
val field = x * someConst
override def toString = "val: " + field
}
If I create companion object first, the 'new thing(x)' (in the companion) causes an error. However, if I define the class first, the 'x * someConst' (in the class definition) causes an error.
I also tried placing the class definition inside the singleton.
object thing {
var someConst = 42
def apply(x: Int) = new thing(x)
class thing(x: Int) {
val field = x * someConst
override def toString = "val: " + field
}
}
However, doing this gives me a 'thing.thing' type object
val t = thing(2)
results in
t: thing.thing = val: 84
The only useful solution I've come up with is to create an abstract class, a companion and an inner class (which extends the abstract class):
abstract class thing
object thing {
val someConst = 42
def apply(x: Int) = new privThing(x)
class privThing(x: Int) extends thing {
val field = x * someConst
override def toString = "val: " + field
}
}
val t1 = thing(2)
val tArr: Array[thing] = Array(t1)
OK, 't1' still has type of 'thing.privThing', but it can now be treated as a 'thing'.
However, it's still not an elegant solution, can anyone tell me a better way to do this?
PS. I should mention, I'm using Scala 2.8.1 on Windows 7
First, the error you're seeing (you didn't tell me what it is) isn't a runtime error. The thing constructor isn't called when the thing singleton is initialized -- it's called later when you call thing.apply, so there's no circular reference at runtime.
Second, you do have a circular reference at compile time, but that doesn't cause a problem when you're compiling a scala file that you've saved on disk -- the compiler can even resolve circular references between different files. (I tested. I put your original code in a file and compiled it, and it worked fine.)
Your real problem comes from trying to run this code in the Scala REPL. Here's what the REPL does and why this is a problem in the REPL. You're entering object thing and as soon as you finish, the REPL tries to compile it, because it's reached the end of a coherent chunk of code. (Semicolon inference was able to infer a semicolon at the end of the object, and that meant the compiler could get to work on that chunk of code.) But since you haven't defined class thing it can't compile it. You have the same problem when you reverse the definitions of class thing and object thing.
The solution is to nest both class thing and object thing inside some outer object. This will defer compilation until that outer object is complete, at which point the compiler will see the definitions of class thing and object thing at the same time. You can run import thingwrapper._ right after that to make class thing and object thing available in global scope for the REPL. When you're ready to integrate your code into a file somewhere, just ditch the outer class thingwrapper.
object thingwrapper{
//you only need a wrapper object in the REPL
object thing {
val someConst = 42
def apply(x: Int) = new thing(x)
}
class thing(x: Int) {
import thing.someConst
val field = x * someConst
override def toString = "val: " + field
}
}
Scala 2.12 or more could benefit for sip 23 which just (August 2016) pass to the next iteration (considered a “good idea”, but is a work-in-process)
Literal-based singleton types
Singleton types bridge the gap between the value level and the type level and hence allow the exploration in Scala of techniques which would typically only be available in languages with support for full-spectrum dependent types.
Scala’s type system can model constants (e.g. 42, "foo", classOf[String]).
These are inferred in cases like object O { final val x = 42 }. They are used to denote and propagate compile time constants (See 6.24 Constant Expressions and discussion of “constant value definition” in 4.1 Value Declarations and Definitions).
However, there is no surface syntax to express such types. This makes people who need them, create macros that would provide workarounds to do just that (e.g. shapeless).
This can be changed in a relatively simple way, as the whole machinery to enable this is already present in the scala compiler.
type _42 = 42.type
type Unt = ().type
type _1 = 1 // .type is optional for literals
final val x = 1
type one = x.type // … but mandatory for identifiers