These days, I am reading the book "programming in scala". There is one sentence in the book on Page 246, Chapter11, first paragraph:
For example, just as Any is a superclass of every other class, Nothing
is a subclass of every other class.
I understand the first part because every class inherits Any in a direct or indirect way. But I can't understand the latter part of the sentence.
This is class Nothing definition:
abstract final class Nothing extends Any
Conceptually, Nothing is something that is harder to grasp than Any, which we're familiar with from Java and most other object oriented programming. Nothing is Scalas bottom type. The definition from Wikipedia states:
In subtyping systems, the bottom type is the subtype of all types.
(However, the converse is not true—a subtype of all types is not
necessarily the bottom type.) It is used to represent the return type
of a function that does not return a value: for instance, one which
loops forever, signals an exception, or exits.
I think the easiest way to reason about it is to think of it as a way of describing something that never returns. You can't construct an instance of a Nothing type yourself. One particular use of Nothing in Scala is to be able to reason about covariant parametric types:
In Scala, the bottom type is denoted as Nothing. Besides its use for
functions that just throw exceptions or otherwise don't return
normally, it's also used for covariant parameterized types. For
example, Scala's List is a covariant type constructor, so
List[Nothing] is a subtype of List[A] for all types A. So Scala's Nil,
the object for marking the end of a list of any type, belongs to the
type List[Nothing].
Being a subtype of is not the same as being a subclass of. From Subtyping:
Subtyping should not be confused with the notion of (class or object)
inheritance from object-oriented languages; subtyping is a relation
between types (interfaces in object-oriented parlance) whereas
inheritance is a relation between implementations stemming from a
language feature that allows new objects to be created from existing
ones. In a number of object-oriented languages, subtyping is called
interface inheritance.
This is also denoted in the Scala Specification Section §3.5.2 (Conformance) as part of the <: (is subtype of relation), both for value types and type constructors:
For every value type T, scala.Nothing <: T <: scala.Any.
For every type constructor T (with any number of type parameters), scala.Nothing <: T <: scala.Any.
One familiar aspect of subtyping is variance of generic types, where - denotes contravariance and + denotes covariance.
Related
How are primitive types in Scala objects if we do not use the word "new" to instantiate the instances of those primitives? Programming in Scala by Martin Odersky described the reasoning as some enforcing by a "trick" that makes these value classes to be defined abstract and final, which did not quite make sense to me because how are we able to make an instance of these classes if its abstract? If that same primitive literal is to be stored somewhere let's say into a variable will that make the variable an object?
I assume that you use scala 2.13 with implementation of literal types. For this explanation you can think of type and class as synonyms, but in reality they are different concepts.
To put it all together it worth to treat each primitive type as a set of subtypes each of which representing type of one single literal value.
So literal 1 is a value and type at the same time (instance 1 of type 1), and it is subtype of value class Int.
Let's prove that 1 is subtype of Int by using 'implicitly':
implicitly[1 <:< Int] // compiles
The same but using val:
val one:1 = 1
implicitly[one.type <:< Int] // compiles
So one is kind of an instance (object) of type 1 (and instance of type Int at the same time because because Int is supertype of 1). You can use this value the same way as any other objects (pass it to function or assign to other vals etc).
val one:1 = 1
val oneMore: 1 = one
val oneMoreGeneric: Int = one
val oneNew:1 = 1
We can assume that all these vals contain the same instance of one single object because from practical perspective it doesn't actually matter if this is the same object or not.
Technically it's not an object at all, because primitives came form java (JVM) world where primitives are not objects. They are different kind of entities.
Scala language is trying to unify these two concepts into one (everything is classes), so developers don't have to think too much about differences.
But here are still some differences in a backstage. Each value class is a subtype of AnyVal, but the rest of the classes are subtype of AnyRef (regular class).
implicitly[1 <:< AnyVal] //compiles
implicitly[Int <:< AnyVal] // compiles
trait AnyTraint
implicitly[AnyTraint <:< AnyVal] // fails to compail
implicitly[AnyTraint <:< AnyRef] // compiles
And in addition, because of its non-class nature in the JVM, you can't extend value classes as regular class or use new to create an instance (because scala compiler emulates new by itself). That's why from perspective of extending value classes you should think about them as final and from perspective of creating instances manually you should think of them as abstract. But form most of the other perspectives it's like any other regular class.
So scala compiler can kind of extend Int by 1,2,3 .. types and create instances of them for vals, but developers can't do it manually.
I have read this article on covariance/contravariance: http://julien.richard-foy.fr/blog/2013/02/21/be-friend-with-covariance-and-contravariance/
The examples are very clear. However, I am struggling to understand the conclusions drawn at the end:
If you look at the definitions of Run[+A] and Vet[-A] you may notice
that the type Aappears only in the return type of methods of Run[+A]
and only in the parameters of methods of Vet[-A]. More generally a
type that produces values of type A can be made covariant on A (as you
did with Run[+A]), and a type that consumes values of type A can be
made contravariant on A (as you did with Vet[-A]).
From the above paragraph you can deduce that types that only have
getters can be covariant (in other words, immutable data types can be
covariant, and that’s the case for most of the data types of Scala’s
standard library), but mutable data types are necessarily invariant
(they have getters and setters, so they both produce and consume
values).
Producers: If something produces type A, I can imagine some reference variable of type A being set to an object of type A or any subtypes of A, but not supertypes, so it's appropriate that it can be covariant.
Consumers: If something consumes type A, I guess that means type A may be used as parameters in methods. I'm not clear what relationship this has to covariance or contravariance.
It seems from the examples that specifying a type as covariant/contravariant affects how it can be consumed by other functions but not sure how it affects the classes themselves.
It seems from the examples that specifying a type as covariant/contravariant affects how it can be consumed by other functions but not sure how it affects the classes themselves.
It is right that the article focused on the consequences of variance for users of a class, not for implementers of a class.
The article shows that covariant and contravariant types give more freedom to users (because a function that accepts a Run[Mammal] effectively accepts a Run[Giraffe] or a Run[Zebra]). For implementors, the perspective is dual: covariant and contravariant types give them more constraints.
These constraints are that covariant types can not occur in contravariant positions and vice versa.
Consider for instance this Producer type definition:
trait Producer[+A] {
def produce(): A
}
The type parameter A is covariant. Therefore we can only use it in covariant positions (such as a method return type), but we can not use it in contravariant position (such as a method parameter):
trait Producer[+A] {
def produce(): A
def consume(a: A): Unit // (does not compile because A is in contravariant position)
}
Why is it illegal to do so? What could go wrong if this code compiled? Well, consider the following scenario. First, get some Producer[Zebra]:
val zebraProducer: Producer[Zebra] = …
Then upcast it to a Producer[Mammal] (which is legal, because we claimed the type parameter to be covariant):
val mammalProducer: Producer[Mammal] = zebraProducer
Finally, feed it with a Giraffe (which is legal too because the consume method a Producer[Mammal] accepts a Mammal, and a Giraffe is a Mammal):
mammalProducer.consume(new Giraffe)
However, if you remember well, the mammalProducer was actually a zebraProducer, so its consume implementation actually only accepts a Zebra, not a Giraffe! So, in practice, if it was allowed to use covariant types in contravariant positions (like I did with the consume method), the type system would be unsound. We can construct a similar scenario (leading to an absurdity) if we pretend that a class with a contravariant type parameter can also have a method where it is in covariant position (see at the end for the code).
(Note that several programming languages, e.g. Java or TypeScript, have such unsound type systems.)
In practice, in Scala if we want to use a covariant type parameter in contravariant position, we have to use the following trick:
trait Producer[+A] {
def produce(): A
def consume[B >: A](b: B): Unit
}
In that case, a Producer[Zebra] would not expect to get an actual Zebra passed in the consume method (but any value of a type B, lower-bounded by Zebra), so it would be legal to pass a Giraffe, which is a Mammal, which is a super-type of Zebra.
Appendix: similar scenario for contravariance
Consider the following class Consumer[-A], which has a contravariant type parameter A:
trait Consumer[-A] {
def consume(a: A): Unit
}
Suppose that the type system allowed us to define a method where A is in covariant position:
trait Consumer[-A] {
def consume(a: A): Unit
def produce(): A // (does not actually compile because A is in covariant position)
}
Now we can get an instance of Consumer[Mammal], upcast it to Consumer[Zebra] (because of contravariance) and call the produce method to get a Zebra:
val mammalConsumer: Consumer[Mammal] = …
val zebraConsumer: Consumer[Zebra] = mammalConsumer // legal, because we claimed `A` to be contravariant
val zebra: Zebra = zebraConsumer.produce()
However, our zebraConsumer is actually mammalConsumer, whose method produce can return any Mammal, not just Zebras. So, at the end, zebra might be initialized to some Mammal that is not a Zebra! In order to avoid such absurdities, the type system forbids us to define the produce method in the Consumer class.
I am trying to understand the following piece of code. but I don't know what the R#X mean. could someone help me?
// define the abstract types and bounds
trait Recurse {
type Next <: Recurse
// this is the recursive function definition
type X[R <: Recurse] <: Int
}
// implementation
trait RecurseA extends Recurse {
type Next = RecurseA
// this is the implementation
type X[R <: Recurse] = R#X[R#Next]
}
object Recurse {
// infinite loop
type C = RecurseA#X[RecurseA]
}
You may gain type from existing instance of a class:
class C {
type someType = Int
}
val c = new C
type t = c.someType
Or may address to the type directly without instantiating an object: C#someType This form is very usefull for type expressions where you have no space to create intermediate variables.
Adding some clarifications as it was suggested in comments.
Disclaimer: I have only partial understanding of how Scala's type system works. I'd tried to read documentation several times, but was able to extract only patchy knowledges from it. But I have rich experience in scala and may predict compilers behavior on individual cases well.
# called type projection and type projection compliments normal hierarchical type access via . In every type expression scala implicitly uses both.
scala reference gives examples of such invisible conversions:
t ə.type#t
Int scala.type#Int
scala.Int scala.type#Int
data.maintable.Node data.maintable.type#Node
As use see, every trivial usage of type projection actually works on type (that is return with .type) not on an object. The main practical difference (I'm bad with definitions) is that object type is something ephemeral as object itself is. Its type may be changed in appropriate circumstances such as inheritance of an abstract class type. In contrast type's type (the definition of the type projection) is as stable as sun. Types (don't mix them with classes) in scala are not first-class citizens and can not be overridden further.
There are different places suitable for putting type expression into. There are also some places where only stable types are allowed. So basically type projection is more constant for terms of type.
Assume, we have:
class B
class A extends B
trait T
Then it holds:
val a: A with T = new A with T
a.isInstanceOf[B] // result is true !
Is it right to say, the isInstanceOf method checks, if there is at least one type (not all types) which matches the right hand side in a subtype relationship?
At first look, I thought a value with type A with T can not be a subtype of B, because A and T are not both subtypes of B. But it is A or T is a subtype of B -- is that right ?
isInstanceOf looks if there is a corresponding entry in the inheritance chain. The chain of A with T includes A, B and T, so a.isInstanceOf[B] must be true.
edit:
Actually the generated byte code calls javas instanceof, so it would be a instanceof B in java. A little more complex call like a.isInstanceOf[A with T] would be (a instanceof A) && (a instanceof T).
At first look, I thought a value with type A with T can not be a
subtype of B
There's are two misconceptions here. First, that the static type of an instance has any bearing on the result of isInstanceOf: there is none. To be clear, when doing a.isInstanceOf[B], the fact that a is of type A with T is not relevant.
The method isInstanceOf is implemented at the bytecode level by the JVM. It looks at the class information every instance carries, and checks whether B one of the classes (the class of the instance itself and its ancestors), or one of the implemented interfaces. That's the "is-a" relationship: "a is a B".
Technically, isInstanceOf is part of Java's reflection, where it is known as instanceof.
The second misconception is the inheritance can somehow remove a parent type. That never happens: inheritance only adds types, never removes them. The type A with T is an A, a B, a T, an AnyVal and an Any. So even if isInstanceOf did look at the type A with T, it would still return true.
When I compile:
object Test extends App {
implicit def pimp[V](xs: Seq[V]) = new {
def dummy(x: V) = x
}
}
I get:
$ fsc -d aoeu go.scala
go.scala:3: error: Parameter type in structural refinement may not refer to an abstract type defined outside that refinement
def dummy(x: V) = x
^
one error found
Why?
(Scala: "Parameter type in structural refinement may not refer to an abstract type defined outside that refinement" doesn't really answer this.)
It's disallowed by the spec. See 3.2.7 Compound Types.
Within a method declaration in a structural refinement, the type of any value parameter may only refer to type parameters or abstract types that are contained inside the refinement. That is, it must refer either to a type parameter of the method
itself, or to a type definition within the refinement. This restriction does not apply
to the function’s result type.
Before Bug 1906 was fixed, the compiler would have compiled this and you'd have gotten a method not found at runtime. This was fixed in revision 19442 and this is why you get this wonderful message.
The question is then, why is this not allowed?
Here is very detailed explanation from Gilles Dubochet from the scala mailing list back in 2007. It roughly boils down to the fact that structural types use reflection and the compiler does not know how to look up the method to call if it uses a type defined outside the refinement (the compiler does not know ahead of time how to fill the second parameter of getMethod in p.getClass.getMethod("pimp", Array(?))
But go look at the post, it will answer your question and some more.
Edit:
Hello list.
I try to define structural types with abstract datatype in function
parameter. ... Any reason?
I have heard about two questions concerning the structural typing
extension of Scala 2.6 lately, and I would like to answer them here.
Why did we change Scala's native values (“int”, etc.) boxing scheme
to Java's (“java.lang.Integer”).
Why is the restriction on parameters for structurally defined
methods (“Parameter type in structural refinement may not refer
to abstract type defined outside that same refinement”) required.
Before I can answer these two questions, I need to speak about the
implementation of structural types.
The JVM's type system is very basic (and corresponds to Java 1.4). That
means that many types that can be represented in Scala cannot be
represented in the VM. Path dependant types (“x.y.A”), singleton types
(“a.type”), compound types (“A with B”) or abstract types are all types
that cannot be represented in the JVM's type system.
To be able to compile to JVM bytecode, the Scala compilers changes the
Scala types of the program to their “erasure” (see section 3.6 of the
reference). Erased types can be represented in the VM's type system and
define a type discipline on the program that is equivalent to that of
the program typed with Scala types (saving some casts), although less
precise. As a side note, the fact that types are erased in the VM
explains why operations on the dynamic representation of types (pattern
matching on types) are very restricted with respect to Scala's type
system.
Until now all type constructs in Scala could be erased in some way.
This isn't true for structural types. The simple structural type “{ def
x: Int }” can't be erased to “Object” as the VM would not allow
accessing the “x” field. Using an interface “interface X { int x{}; }”
as the erased type won't work either because any instance bound by a
value of this type would have to implement that interface which cannot
be done in presence of separate compilation. Indeed (bear with me) any
class that contains a member of the same name than a member defined in
a structural type anywhere would have to implement the corresponding
interface. Unfortunately this class may be defined even before the
structural type is known to exist.
Instead, any reference to a structurally defined member is implemented
as a reflective call, completely bypassing the VM's type system. For
example def f(p: { def x(q: Int): Int }) = p.x(4) will be rewritten
to something like:
def f(p: Object) = p.getClass.getMethod("x", Array(Int)).invoke(p, Array(4))
And now the answers.
“invoke” will use boxed (“java.lang.Integer”) values whenever the
invoked method uses native values (“int”). That means that the above
call must really look like “...invoke(p, Array(new
java.lang.Integer(4))).intValue”.
Integer values in a Scala program are already often boxed (to allow the
“Any” type) and it would be wasteful to unbox them from Scala's own
boxing scheme to rebox them immediately as java.lang.Integer.
Worst still, when a reflective call has the “Any” return type,
what should be done when a java.lang.Integer is returned? The called
method may either be returning an “int” (in which case it should be
unboxed and reboxed as a Scala box) or it may be returning a
java.lang.Integer that should be left untouched.
Instead we decided to change Scala's own boxing scheme to Java's. The
two previous problems then simply disappear. Some performance-related
optimisations we had with Scala's boxing scheme (pre-calculate the
boxed form of the most common numbers) were easy to use with Java
boxing too. In the end, using Java boxing was even a bit faster than
our own scheme.
“getMethod”'s second parameter is an array with the types of the
parameters of the (structurally defined) method to lookup — for
selecting which method to get when the name is overloaded. This is the
one place where exact, static types are needed in the process of
translating a structural member call. Usually, exploitable static types
for a method's parameter are provided with the structural type
definition. In the example above, the parameter type of “x” is known to
be “Int”, which allows looking it up.
Parameter types defined as abstract types where the abstract type is
defined inside the scope of the structural refinement are no problem
either:
def f(p: { def x[T](t: T): Int }) = p.xInt
In this example we know that any instance passed to “f” as “p” will
define “x[T](t: T)” which is necessarily erased to “x(t: Object)”. The
lookup is then correctly done on the erased type:
def f(p: Object) = p.getClass.getMethod("x", Array(Object)).invoke(p,
Array(new java.lang.Integer(4)))
But if an abstract type from outside the structural refinement's scope
is used to define a parameter of a structural method, everything breaks:
def f[T](p: { def x(t: T): Int }, t: T) = p.x(t)
When “f” is called, “T” can be instantiated to any type, for example:
f[Int]({ def x(t: Int) = t }, 4)
f[Any]({ def x(t: Any) = 5 }, 4)
The lookup for the first case would have to be “getMethod("x",
Array(int))” and for the second “getMethod("x", Array(Object))”, and
there is no way to know which one to generate in the body of
“f”: “p.x(t)”.
To allow defining a unique “getMethod” call inside “f”'s body for
any instantiation of “T” would require any object passed to “f” as the
“p” parameter to have the type of “t” erased to “Any”. This would be a
transformation where the type of a class' members depend on how
instances of this class are used in the program. And this is something
we definitely don't want to do (and can't be done with separate
compilation).
Alternatively, if Scala supported run-time types one could use them to
solve this problem. Maybe one day ...
But for now, using abstract types for structural method's parameter
types is simply forbidden.
Sincerely,
Gilles.
Discovered the problem shortly after posting this: I have to define a named class instead of using an anonymous class. (Still would love to hear a better explanation of the reasoning though.)
object Test extends App {
case class G[V](xs: Seq[V]) {
def dummy(x: V) = x
}
implicit def pimp[V](xs: Seq[V]) = G(xs)
}
works.