I've been curious about the impact of not having an explicit primary constructor in Scala, just the contents of the class body.
In particular, I suspect that the private or protected constructor pattern, that is, controlling construction through the companion object or another class or object's methods might not have an obvious implementation.
Am I wrong? If so, how is it done?
You can declare the default constructor as private/protected by inserting the appropriate keyword between the class name and the parameter list, like this:
class Foo private () {
/* class body goes here... */
}
Aleksander's answer is correct, but Programming in Scala offers an additional alternative:
sealed trait Foo {
// interface
}
object Foo {
def apply(...): Foo = // public constructor
private class FooImpl(...) extends Foo { ... } // real class
}
Related
I am new to kotlin I have been learning about inheritance in kotlin recently, and then I realised that we cannot mark the visibility of a class as "protected". Correct me if i am wrong, or is there any other way to make a class protected ?
You can mark protected only parts of classes, so that they become accessible only from the derived classes. You can mark protected a member property, a member function or a nested class:
open class X {
protected val v: SomeType = someValue
protected fun f() { }
protected class Y { ... }
}
But you cannot mark protected anything that does not belong to a class (e.g. a top-level class or function), because the modifier would make no sense: a top-level entity is not subject to inheritance, thus there can be no derived class that would access it.
I want the protected fields of a class to be made publicly visible in a subclass:
class MyClass(protected val someDao: SomeDao)
class TestMyClass extends MyClass(mock[SomeDao])
Now I want to access someDao like so:
val testClass = new TestMyClass
testClass.someDao
I've read that in Java and Scala access modifiers can be broadened but not restricted in subclasses (although these are just constructor parameters, not private fields per se). How can I achieve that in this example?
EDIT:
The answers provided suggest making someDao public in the base class. This is not what I want. I want it to remain private, but just change its visibility to public in the TestMyClass class.
As far as I know, an overriden val cannot access the super implementation.
If you can modify the base class, one solution is to explicitely define an accessor method, which can be overriden; something like this:
class MyClass(_someDao: SomeDao) {
protected def someDao = _someDao
}
class TestMyClass extends MyClass(mock[SomeDao]) {
override def someDao = super.someDao
}
If you cannot change the base class, one workaround would be to define a public accessor method with a different name:
class MyClass(protected val someDao: SomeDao)
class TestMyClass extends MyClass(mock[SomeDao]) {
def someDao2 = someDao
}
Add var or val to parameters declaration:
class MyClass(val someDao: SomeDao)
In this case only getter function will be generate for someDao field. So when you write testClass.someDao you retrieve not the someDao as field you use their getter function.
Good example about visibility of class parameters in Scala demostrated at this article.
Regarding to your last updates
Set package scope for parameter declaration:
class MyClass(private[lastPackageOfSomeDao] val someDao: SomeDao)
Could someone explain why scala would allow a public variable, to satisfy the implementation of an abstract declared Protected item? My first assumption is that the compiler would complain, but I created a small test to see if this worked, and to my surprise it does. Is there an advantage to this? (perhaps this is normal in OOP?) Any methods to avoid the accidental pitfall?
object NameConflict extends App {
abstract class A {
protected[this] var name:String
def speak = println(name)
}
class B(var name:String) extends A { //notice we've declared a public var
}
val t = new B("Tim")
t.speak
println(t.name) // name is exposed now?
}
It's normal and as in Java. Sometimes it's desirable to increase the visibility of a member.
You can't do it the other way around and turn down visibility in a subclass, because the member can by definition be accessed through the supertype.
If invoking a method has terrible consequences, keep the method private and use a template method that can be overridden; the default implementation would invoke the dangerous method.
abstract class A {
private[this] def dangerous = ???
final protected def process: Int = {
dangerous
template
}
protected def template: Int = ???
}
class B extends A {
override def template = 5
}
A Java library class I'm using declares
protected getPage(): Page { ... }
Now I want to make a helper Scala mixin to add features that I often use. I don't want to extend the class, because the Java class has different subclasses I want to extend at different places. The problem is that if I use getPage() in my mixin trait, I get this error:
Implementation restriction: trait MyMixin accesses protected method getPage inside a concrete trait method.
Is there a solution how to make it work, without affecting my subclasses? And why is there this restriction?
So far, I came up with a work-around: I override the method in the trait as
override def getPage(): Page = super.getPage();
This seems to work, but I'm not completely satisfied. Luckily I don't need to override getPage() in my subclasses, but if I needed, I'd get two overrides of the same method and this work-around won't work.
The problem is that even though the trait extends the Java class, the implementation is not actually in something that extends the Java class. Consider
class A { def f = "foo" }
trait T extends A { def g = f + "bar" }
class B extends T { def h = g + "baz" }
In the actual bytecode for B we see
public java.lang.String g();
Code:
0: aload_0
1: invokestatic #17; //Method T$class.g:(LT;)Ljava/lang/String;
4: areturn
which means it just forwards to something called T$class, which it turns out is
public abstract class T$class extends java.lang.Object{
public static java.lang.String g(T);
Code:
...
So the body of the code isn't called from a subclass of A at all.
Now, with Scala that's no problem because it just omits the protected flag from bytecode. But Java enforces that only subclasses can call protected methods.
And thus you have the problem, and the message.
You cannot easily get around this problem, though the error message suggests what is perhaps the best alternative:
public class JavaProtected {
protected int getInt() { return 5; }
}
scala> trait T extends JavaProtected { def i = getInt }
<console>:8: error: Implementation restriction: trait T accesses
protected method getInt inside a concrete trait method.
Add an accessor in a class extending class JavaProtected as a workaround.
Note the last line.
class WithAccessor extends JavaProtected { protected def myAccessor = getInt }
trait T extends WithAccessor { def i = myAccessor }
works.
In Scala, a val can override a def, but a def cannot override a val.
So, is there an advantage to declaring a trait e.g. like this:
trait Resource {
val id: String
}
rather than this?
trait Resource {
def id: String
}
The follow-up question is: how does the compiler treat calling vals and defs differently in practice and what kind of optimizations does it actually do with vals? The compiler insists on the fact that vals are stable — what does in mean in practice for the compiler? Suppose the subclass is actually implementing id with a val. Is there a penalty for having it specified as a def in the trait?
If my code itself does not require stability of the id member, can it be considered good practice to always use defs in these cases and to switch to vals only when a performance bottleneck has been identified here — however unlikely this may be?
Short answer:
As far as I can tell, the values are always accessed through the accessor method. Using def defines a simple method, which returns the value. Using val defines a private [*] final field, with an accessor method. So in terms of access, there is very little difference between the two. The difference is conceptual, def gets reevaluated each time, and val is only evaluated once. This can obviously have an impact on performance.
[*] Java private
Long answer:
Let's take the following example:
trait ResourceDef {
def id: String = "5"
}
trait ResourceVal {
val id: String = "5"
}
The ResourceDef & ResourceVal produce the same code, ignoring initializers:
public interface ResourceVal extends ScalaObject {
volatile void foo$ResourceVal$_setter_$id_$eq(String s);
String id();
}
public interface ResourceDef extends ScalaObject {
String id();
}
For the subsidiary classes produced (which contain the implementation of the methods), the ResourceDef produces is as you would expect, noting that the method is static:
public abstract class ResourceDef$class {
public static String id(ResourceDef $this) {
return "5";
}
public static void $init$(ResourceDef resourcedef) {}
}
and for the val, we simply call the initialiser in the containing class
public abstract class ResourceVal$class {
public static void $init$(ResourceVal $this) {
$this.foo$ResourceVal$_setter_$id_$eq("5");
}
}
When we start extending:
class ResourceDefClass extends ResourceDef {
override def id: String = "6"
}
class ResourceValClass extends ResourceVal {
override val id: String = "6"
def foobar() = id
}
class ResourceNoneClass extends ResourceDef
Where we override, we get a method in the class which just does what you expect. The def is simple method:
public class ResourceDefClass implements ResourceDef, ScalaObject {
public String id() {
return "6";
}
}
and the val defines a private field and accessor method:
public class ResourceValClass implements ResourceVal, ScalaObject {
public String id() {
return id;
}
private final String id = "6";
public String foobar() {
return id();
}
}
Note that even foobar() doesn't use the field id, but uses the accessor method.
And finally, if we don't override, then we get a method which calls the static method in the trait auxiliary class:
public class ResourceNoneClass implements ResourceDef, ScalaObject {
public volatile String id() {
return ResourceDef$class.id(this);
}
}
I've cut out the constructors in these examples.
So, the accessor method is always used. I assume this is to avoid complications when extending multiple traits which could implement the same methods. It gets complicated really quickly.
Even longer answer:
Josh Suereth did a very interesting talk on Binary Resilience at Scala Days 2012, which covers the background to this question. The abstract for this is:
This talk focuses on binary compatibility on the JVM and what it means
to be binary compatible. An outline of the machinations of binary
incompatibility in Scala are described in depth, followed by a set of rules and guidelines that will help developers ensure their own
library releases are both binary compatible and binary resilient.
In particular, this talk looks at:
Traits and binary compatibility
Java Serialization and anonymous classes
The hidden creations of lazy vals
Developing code that is binary resilient
The difference is mainly that you can implement/override a def with a val but not the other way around. Moreover val are evaluated only once and def are evaluated every time they are used, using def in the abstract definition will give the code who mixes the trait more freedom about how to handle and/or optimize the implementation. So my point is use defs whenever there isn't a clear good reason to force a val.
A val expression is evaluated once on variable declaration, it is strict and immutable.
A def is re-evaluated each time you call it
def is evaluated by name and val by value. This means more or less that val must always return an actual value, while def is more like a promess that you can get a value when evaluating it. For example, if you have a function
def trace(s: => String ) { if (level == "trace") println s } // note the => in parameter definition
that logs an event only if the log level is set to trace and you want to log an objects toString. If you have overriden toString with a value, then you need to pass that value to the trace function. If toString however is a def, it will only be evaluated once it's sure that the log level is trace, which could save you some overhead.
def gives you more flexibility, while val is potentially faster
Compilerwise, traits are compiled to java interfaces so when defining a member on a trait, it makes no difference if its a var or def. The difference in performance would depend on how you choose to implement it.