Extend class and pass all constructor arguments to super - class

How would you extend a class using CoffeeScript, but have the construction arguments passed to super?
Eg:
class List extends Array
# Some other stuff to make it work...
list = new List(1,2,3)
console.log list
[1, 2, 3]

In general, this would work without additional code; the parent constructor is used unless expressly overridden:
class A
constructor: ->
console.log arg for arg in arguments
class B extends A
new B('foo') # output: 'foo'
And the problem isn't that Array doesn't have a constructor method:
coffee> Array.constructor
[Function: Function]
The problem is just that Array is just plain weird. While arrays are "just objects" in principle, in practice they're stored differently. So when you try to apply that constructor to an object that isn't an array (even if it passes the instanceof Array test), it doesn't work.
So, you can use Acorn's solution, but then you may run into other problems down the road (especially if you pass a List to something that expects a true array). For that reason, I'd recommend implementing List as a wrapper around an array instance, rather than trying to use inheritance from a native object type.
While we're on the subject, one very important clarification: When you use super by itself, that does pass all arguments! This behavior is borrowed from Ruby. So
class B extends A
constructor: ->
super
will pass along all arguments to A's constructor, while
class B extends A
constructor: ->
super()
will invoke A's constructor with no arguments.

class List extends Array
constructor: ->
#push arguments...
toString: ->
#join('-')
list = new List(1, 2)
list.push(3)
list.toString()
=>
'1-2-3'

Using extends in CoffeeScript expects the superclass to be in CoffeeScript too. If you're using a non-CS class, e.g. Array in the original question, then you may encounter problems.
This solved the general case for me. It's a bit of a hack because it uses _super which probably isn't intended to be used in the compiled JS.
class MyClass extends SomeJsLib
constructor: ->
_super.call #, arg1, arg2
Or if you just want to pass thru arguments from the caller:
class MyClass extends SomeJsLib
constructor: ->
_super.apply #, arguments

In my exploration of javascript I required a generic way to create a class with a dynamic number of constructor arguments. As mentioned this won't work for array as far as I know, it will only work for coffee-script style classes.
Calling a specific function with a dynamic number of arguments is easy enough through .apply
args = [1, 2, 3]
f = measurement.clone
f.apply measurement, args
A class can extend a class saved in a variable. As a result we can write a function that returns new sub-classes.
classfactory = (f) ->
class extension extends f
Putting it all together we can create a function that returns new sub-classes in which we apply arguments to the super class's constructor.
classwitharguments = (f, args) ->
class extension extends f
constructor: () ->
extension.__super__.constructor.apply #, args
To use this new factory
args = [1, 2, 3]
instance = new (classwitharguments Measurement, args)
Thoughts? Comments? Suggestions? Limitations I didn't think about? Let me know.

Related

Why this map function does not give traits' simple names

I try to get names of all trait a class extends using getInterfaces which returns an array of trait's names. When I manually access each member of the array, the method getName returns simple names like this
trait A
trait B
class C() extends A, B
val c = C()
val arr = c.getClass.getInterfaces
arr(0).getName // : String = A
arr(1).getName // : String = B
However, when I use map function on arr. The resulting array contains a cryptic version of trait's names
arr.map(t => t.getName) // : Array[String] = Array(repl$.rs$line$1$A, repl$.rs$line$2$B)
The goal of this question is not about how to get the resulting array that contains simple names (for that purpose, I can just use arr.map(t => t.getSimpleName).) What I'm curious about is that why accessing array manually and using a map do not yield a compatible result. Am I wrong to think that both ways are equivalent?
I believe you run things in Scala REPL or Ammonite.
When you define:
trait A
trait B
class C() extends A, B
classes A, B and C aren't defined in top level of root package. REPL creates some isolated environment, compiles the code and loads the results into some inner "anonymous" namespace.
Except this is not true. Where this bytecode was created is reflected in class name. So apparently there was something similar (not necessarily identical) to
// repl$ suggest object
object repl {
// .rs sound like nested object(?)
object rs {
// $line sounds like nested class
class line { /* ... */ }
// $line$1 sounds like the first anonymous instance of line
new line { trait A }
// import from `above
// $line$2 sounds like the second anonymous instance of line
new line { trait B }
// import from above
//...
}
}
which was made because of how scoping works in REPL: new line creates a new scope with previous definitions seen and new added (possibly overshadowing some old definition). This could be achieved by creating a new piece of code as code of new anonymous class, compiling it, reading into classpath, instantiating and importing its content. Byt putting each new line into separate class REPL is able to compile and run things in steps, without waiting for you to tell it that the script is completed and closed.
When you are accessing class names with runtime reflection you are seeing the artifacts of how things are being evaluated. One path might go trough REPLs prettifiers which hide such things, while the other bypass them so you see the raw value as JVM sees it.
The problem is not with map rather with Array, especially its toString method (which is one among the many reasons for not using Array).
Actually, in this case it is even worse since the REPL does some weird things to try to pretty-print Arrays which in this case didn't work well (and, IMHO, just add to the confusion)
You can fix this problem calling mkString directly like:
val arr = c.getClass.getInterfaces
val result = arr.map(t => t.getName)
val text = result.mkString("[", ", ", "]")
println(text)
However, I would rather suggest just not using Array at all, instead convert it to a proper collection (e.g. List) as soon as possible like:
val interfaces = c.getClass.getInterfaces.toList
interfaces .map(t => t.getName)
Note: About the other reasons for not using Arrays
They are mutable.
Thet are invariant.
They are not part of the collections hierarchy thus you can't use them on generic methods (well, you actually can but that requires more tricks).
Their equals is by reference instead of by value.

Can we create auxiliary constructor in Scala with List?

I am trying to create a Request object where I can get a single String as input or a list of String as Input. I tried to look through various answers but I don't think I have found any use of List in constructor.
I am trying to do something like this.
class GetRequest(val url: String) {
def this(val urlList: List[String]){
}
}
Is it something to do with List's immutability?
Scala enforces to have one main constructor and as much as you need
auxiliary constructors. There's a rule: each constructor must invoke one of the previously defined constructors.
This should work for you:
class GetRequest(val urlList: List[String]) {
def this(url: String) {
this(List(url))
}
}
An auxiliary constructor isn't really necessary when making a distinction between a single passed parameter and a collection of the same type. That's what the varargs syntax is for.
class GetRequest(val urls: String*)
new GetRequest("s")
new GetRequest("a","b","c")
new GetRequest(List("x","y","z"):_*)
In the constructor code urls can be treated as a Seq[String] of zero or more elements.

Pre-process parameters of a case class constructor without repeating the argument list

I have this case class with a lot of parameters:
case class Document(id:String, title:String, ...12 more params.. , keywords: Seq[String])
For certain parameters, I need to do some string cleanup (trim, etc) before creating the object.
I know I could add a companion object with an apply function, but the LAST thing I want is to write the list of parameters TWICE in my code (case class constructor and companion object's apply).
Does Scala provide anything to help me on this?
My general recommendations would be:
Your goal (data preprocessing) is the perfect use case of a companion object -- so it is maybe the most idiomatic solution despite the boilerplate.
If the number of case class parameters is high the builder pattern definitely helps, since you do not have to remember the order of the parameters and your IDE can help you with calling the builder member functions. Using named arguments for the case class constructor allows you to use a random argument order as well but, to my knowledge, there is not IDE autocompletion for named arguments => makes a builder class slightly more convenient. However using a builder class raises the question of how to deal with enforcing the specification of certain arguments -- the simple solution may cause runtime errors; the type-safe solution is a bit more verbose. In this regard a case class with default arguments is more elegant.
There is also this solution: Introduce an additional flag preprocessed with a default argument of false. Whenever you want to use an instance val d: Document, you call d.preprocess() implemented via the case class copy method (to avoid ever typing all your arguments again):
case class Document(id: String, title: String, keywords: Seq[String], preprocessed: Boolean = false) {
def preprocess() = if (preprocessed) this else {
this.copy(title = title.trim, preprocessed = true) // or whatever you want to do
}
}
But: You cannot prevent a client to initialize preprocessed set to true.
Another option would be to make some of your parameters a private val and expose the corresponding getter for the preprocessed data:
case class Document(id: String, title: String, private val _keywords: Seq[String]) {
val keywords = _keywords.map(kw => kw.trim)
}
But: Pattern matching and the default toString implementation will not give you quite what you want...
After changing context for half an hour, I looked at this problem with fresh eyes and came up with this:
case class Document(id: String, title: String, var keywords: Seq[String]) {
keywords = keywords.map(kw => kw.trim)
}
I simply make the argument mutable adding var and cleanup data in the class body.
Ok I know, my data is not immutable anymore and Martin Odersky will probably kill a kitten after seeing this, but hey.. I managed to do what I want adding 3 characters. I call this a win :)

Scala class constructor local parameters

Can I pass arguments to Scala class constructor that are not stored into class itself?
I want to achieve functionality which in Java could be written as follows:
class A {
private final SomethingElse y;
public A(Something x) {
y = x.derive(this);
}
}
I.e. class constructor takes parameter that is later transformed to another value using reference to this. The parameter is forgotten after constructor returns.
In Scala I can do:
class A(x: Something) {
val y = x.derive(this)
}
But it means that x is stored in the class, which I want to avoid. Since x.derive method uses reference to this, I can not make the transformation in companion object.
But it means that x is stored in the class, which I want to avoid.
If you don't reference constructor argument anywhere except the constructor itself, field won't be created. If you reference x e.g. in toString(), Scala will automatically create and assign private val for you.
Use javap -c -private A to verify what kind of fields are actually created.
BTW you pass this inside a constructor, which means a.derive() gets a reference to possibly non-initialized instance of A. Be careful!

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.