Performing an explicit cast based on `Class` instance - scala

Assume there is a class X. At some point in my code, I get an o of type Object and I want to explicitly cast it to X. Normally, I would do val x: X = o.asInstanceOf[X]. However, X is not in scope, all I have is X's instance of Class, like so:
I pass classOf[X] to my function as an argument called classOfX.
Within my function, I would like to do val x: X = o.asInstanceOf[classOfX].
Is there a way to do this in Scala?

Only kind of. java.lang.Class has a cast method that lets you cast a Class[A] to an A. You could write a method like this:
def cast[A](o: Any, clazz: Class[A]): A = clazz.cast(o)
scala> cast("abc", classOf[String])
res10: String = abc
This will "work" for most classes, but not for Scala boxed primitives:
scala> cast(1, classOf[Int])
java.lang.ClassCastException: Cannot cast java.lang.Integer to int
And of course, casting only works modulo type erasure, so this will not immediately throw an exception:
scala> val a = List("a").getClass
a: Class[_ <: List[String]] = class scala.collection.immutable.$colon$colon
scala> cast(List(1), a)
res16: List[String] = List(1)
On the upside, it will also work with classes obtained at run time using getClass, like above.
The downside is that you should be avoiding casting at all costs, because of how error prone it will make the code. It throws type safety out the window. Rarely should anyone find themselves in a situation where they have an object where they don't know what it is, but can somehow tell the compiler what it is using classOf. It is likely that using some form of generics could eliminate a casting problem, but it is impossible to tell without more code.

Related

scala overloading resolution differences between function calls and implicit search

There is a difference in the way the scala 2.13.3 compiler determines which overloaded function to call compared to which overloaded implicit to pick.
object Thing {
trait A;
trait B extends A;
trait C extends A;
def f(a: A): String = "A"
def f(b: B): String = "B"
def f(c: C): String = "C"
implicit val a: A = new A {};
implicit val b: B = new B {};
implicit val c: C = new C {};
}
import Thing._
scala> f(new B{})
val res1: String = B
scala> implicitly[B]
val res2: Thing.B = Thing$$anon$2#2f64f99f
scala> f(new A{})
val res3: String = A
scala> implicitly[A]
^
error: ambiguous implicit values:
both value b in object Thing of type Thing.B
and value c in object Thing of type Thing.C
match expected type Thing.A
As we can see, the overload resolution worked for the function call but not for the implicit pick. Why isn't the implicit offered by val a be chosen as occurs with function calls? If the callers ask for an instance of A why the compilers considers instances of B and C when an instance of A is in scope. There would be no ambiguity if the resolution logic were the same as for function calls.
Edit 2:
The Edit 1 was removed because the assertion I wrote there was wrong.
In response to the comments I added another test to see what happens when the implicit val c: C is removed. In that case the compiler don't complains and picks implicit val b: B despite the caller asked for an instance of A.
object Thing {
trait A { def name = 'A' };
trait B extends A { def name = 'B' };
trait C extends A { def name = 'C' };
def f(a: A): String = "A"
def f(b: B): String = "B"
implicit val a: A = new A {};
implicit val b: B = new B {};
}
import Thing._
scala> f(new A{})
val res0: String = A
scala> implicitly[A].name
val res3: Char = B
So, the overloading resolution of implicit differs from function calls more than I expected.
Anyway, I still don't find a reason why the designers of scala decided to apply a different resolution logic for function and implicit overloading. (Edit: Later noticed why).
Let's see what happens in a real world example.
Suppose we are doing a Json parser that converts a Json string directly to scala Abstract data types, and we want it to support many standard collections.
The snippet in charge of parsing the iterable collections would be something like this:
trait Parser[+A] {
def parse(input: Input): ParseResult;
///// many combinators here
}
implicit def summonParser[T](implicit parserT: Parser[T]) = parserT;
/** #tparam IC iterator type constructor
* #tparam E element's type */
implicit def iterableParser[IC[E] <: Iterable[E], E](
implicit
parserE: Parser[E],
factory: IterableFactory[IC]
): Parser[IC[E]] = '[' ~> skipSpaces ~> (parserE <~ skipSpaces).repSepGen(coma <~ skipSpaces, factory.newBuilder[E]) <~ skipSpaces <~ ']';
Which requires a Parser[E] for the elements and a IterableFactory[IC] to construct the collection specified by the type parameters.
So, we have to put in implicit scope an instance of IterableFactory for every collection type we want to support.
implicit val iterableFactory: IterableFactory[Iterable] = Iterable
implicit val setFactory: IterableFactory[Set] = Set
implicit val listFactory: IterableFactory[List] = List
With the current implicit resolution logic implemented by the scala compiler, this snippet works fine for Set and List, but not for Iterable.
scala> def parserInt: Parser[Int] = ???
def parserInt: read.Parser[Int]
scala> Parser[List[Int]]
val res0: read.Parser[List[Int]] = read.Parser$$anonfun$pursue$3#3958db82
scala> Parser[Vector[Int]]
val res1: read.Parser[Vector[Int]] = read.Parser$$anonfun$pursue$3#648f48d3
scala> Parser[Iterable[Int]]
^
error: could not find implicit value for parameter parserT: read.Parser[Iterable[Int]]
And the reason is:
scala> implicitly[IterableFactory[Iterable]]
^
error: ambiguous implicit values:
both value listFactory in object IterableParser of type scala.collection.IterableFactory[List]
and value vectorFactory in object IterableParser of type scala.collection.IterableFactory[Vector]
match expected type scala.collection.IterableFactory[Iterable]
On the contrary, if the overloading resolution logic of implicits was like the one for function calls, this would work fine.
Edit 3: After many many coffees I noticed that, contrary to what I said above, there is no difference between the way the compiler decides which overloaded functions to call and which overloaded implicit to pick.
In the case of function call: from all the functions overloads such that the type of the argument is asignable to the type of the parameter, the compiler chooses the one such that the function's parameter type is assignable to all the others. If no function satisfies that, a compilation error is thrown.
In the case of implicit pick up: from all the implicit in scope such that the type of the implicit is asignable to the asked type, the compiler chooses the one such that the declared type is asignable to all the others. If no implicit satisfies that, an compilation error is thrown.
My mistake was that I didn't notice the inversion of the assignability.
Anyway, the resolution logic I proposed above (give me what I asked for) is not entirely wrong. It's solves the particular case I mentioned. But for most uses cases the logic implemented by the scala compiler (and, I suppose, all the other languages that support type classes) is better.
As explained in the Edit 3 section of the question, there are similitudes between the way the compiler decides which overloaded functions to call and which overloaded implicit to pick. In both cases the compiler does two steps:
Filters out all the alternatives that are not asignable.
From the remaining alternatives choses the most specific or complains if there is more than one.
In the case of the function call, the most specific alternative is the function with the most specific parameter type; and in the case of implicit pick is the instance with the most specific declared type.
But, if the logic in both cases were exactly the same, then why did the example of the question give different results? Because there is a difference: the assignability requirement that determines which alternatives pass the first step are oposite.
In the case of the function call, after the first step remain the functions whose parameter type is more generic than the argument type; and in the case of implicit pick, remain the instances whose declared type is more specific than the asked type.
The above words are enough to answers the question itself but don't give a solution to the problem that motivated it, which is: How to force the compiler to pick the implicit instance whose declared type is exactly the same than the summoned type? And the answer is: wrapping the implicit instances inside a non variant wrapper.

Scala unexplainable program behavior

For the following code:
object Test {
class MapOps(map: Map[String, Any]) {
def getValue[T](name: String): Option[T] = {
map.get(name).map{_.asInstanceOf[T]}
}
}
implicit def toMapOps(map: Map[String, Any]): MapOps = new MapOps(map)
def main(args: Array[String]): Unit = {
val m: Map[String, Any] = Map("1" -> 1, "2" -> "two")
val a = m.getValue[Int]("2").get.toString
println(s"1: $a")
val b = m.getValue[Int]("2").get
println(s"2: $b")
}
}
val a is computed without exception and the console prints 1: two,
but when computing val b, the java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer is thrown.
Besides, if I execute
val c = m.getValue[Int]("2").get.getClass.toString
println(s"1: $c")
The console prints "int".
Can someone explain why this code behaves like this?
This is certainly odd.
If you look at the following statement in the Scala REPL:
scala> val x = m.getValue[Int]("2")
x: Option[Int] = Some(two)
What I think is happening is this: the asInstanceOf[T] statement is simply flagging to the compiler that the result should be an Int, but no cast is required, because the object is still just referenced via a pointer. (And Int values are boxed inside of an Option/Some) .toString works because every object has a .toString method, which just operates on the value "two" to yield "two". However, when you attempt to assign the result to an Int variable, the compiler attempts to unbox the stored integer, and the result is a cast exception, because the value is a String and not a boxed Int.
Let's verify this step-by-step in the REPL:
$ scala
Welcome to Scala 2.12.1 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_151).
Type in expressions for evaluation. Or try :help.
scala> class MapOps(map: Map[String, Any]) {
| def getValue[T](name: String): Option[T] = {
| map.get(name).map{_.asInstanceOf[T]}
| }
| }
defined class MapOps
scala> import scala.language.implicitConversions
import scala.language.implicitConversions
scala> implicit def toMapOps(map: Map[String, Any]): MapOps = new MapOps(map)
toMapOps: (map: Map[String,Any])MapOps
scala> val a = m.getValue[Int]("2").get.toString
a: String = two
scala> println(s"1: $a")
1: two
So far so good. Note that no exceptions have been thrown so far, even though we have already used .asInstanceOf[T] and used get on the resulting value. What's significant is that we haven't attempted to do anything with the result of the get call (nominally a boxed Int that is actually the String value "two") except to invoke it's toString method. That works, because String values have toString methods.
Now let's perform the assignment to an Int variable:
scala> val b = m.getValue[Int]("2").get
java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer
at scala.runtime.BoxesRunTime.unboxToInt(BoxesRunTime.java:101)
... 29 elided
Now we get the exception! Note also the function in the stack trace that caused it: unboxToInt - it's clearly trying to convert the value stored in the Some to an Int and it fails because it's not a boxed Int but a String.
A big part of the problem is type erasure. Don't forget that a Some(Banana) and a Some(Bicycle) are - at runtime - both just Some instances with a pointer to some object. .asInstanceOf[T] cannot verify the type, because that information has been erased. However, the compiler is able to track what the type should be based upon what you've told it, but it can only detect the error when its assumptions are proven wrong.
Finally, with regard to the getClass call on the result. This is a bit of compiler sleight-of-hand. It's not actually calling a getClass function on the object, but - because it thinks it's dealing with an Int, which is a primitive - it simply substitutes an int class instance.
scala> m.getValue[Int]("2").get.getClass
res0: Class[Int] = int
To verify that the object actually is a String, you can cast it to an Any as follows:
scala> m.getValue[Int]("2").get.asInstanceOf[Any].getClass
res1: Class[_] = class java.lang.String
Further verification about the return value of get follows; note the lack of an exception when we assign the result of this method to a variable of type Any (so no casting is necessary), the fact that the valid Int with key "1" is actually stored under Any as a boxed Int (java.lang.Integer), and that this latter value can be successfully unboxed to a regular Int primitive:
scala> val x: Any = m.getValue[Int]("2").get
x: Any = two
scala> x.getClass
res2: Class[_] = class java.lang.String
scala> val y: Any = m.getValue[Int]("1").get
y: Any = 1
scala> y.getClass
res3: Class[_] = class java.lang.Integer
scala> val z = m.getValue[Int]("1").get
z: Int = 1
scala> z.getClass
res4: Class[Int] = int
Expanding on the last part of eje211's answer.
You told the compiler that a String was an Int, and now you're looking for a sensible explanation of the resulting behavior. That's understandable, but it's not really useful. Once you tell the compiler a lie, all bets are off. You can spend time investigating exactly when and where the compiler inserts checks that happen to discover your deceit, but your time would probably be better spent writing code that doesn't cause you to lie.
As the earlier answer pointed out, you can do that (avoid accidental lying) by using pattern matching. You'll need a ClassTag to make pattern matching work in cases like the above, but the end result will be code that is type-safe and correct.
It's an Int because you request an Int on this line :
val b = m.getValue[Int]("2").get
This calls this method:
def getValue[T](name: String): Option[T] = {
map.get(name).map{_.asInstanceOf[T]}
}
and applies it this way:
def getValue[Int](name: String): Option[Int] = {
map.get(name).map{_.asInstanceOf[Int]}
}
So if you ask for an Int, you get an Int.
In the case of "two", this is what happens in that case:
"two".asInstanceOf[Int]
That's what throws your exception.
A String is not an Int. You can't cast it that way. You can cast it this way:
"2".toInt
But that's different.
In general, using asInstanceOf[] is dangerous. Try pattern matching instead. If you must use, it's up you to make sure that the casts you attempt are valid. You're basically telling the compiler to bypass its own type-checks, particularly when you cast from Any.
It works when you add .toString because then, you change the type back to String, which is what it really was in the first place. The lie of what type the data was is corrected.

Scala asInstanceOf generic type does not fail in Try

Trying to cast a String to a Double obviously should fail:
scala> Try("abc".asInstanceOf[Double])
res11: scala.util.Try[Double] = Failure(java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Double)
However, if I define the above as a function:
scala> def convertAsTypeOf[T](anyValue: Any): Try[T] = Try(anyValue.asInstanceOf[T])
convertAsTypeOf: [T](anyValue: Any)scala.util.Try[T]
Strangely it returns Success when I try to cast a String to a Double:
scala> convertAsTypeOf[Double]("abc")
res10: scala.util.Try[Double] = Success(abc)
If I try to get the value out of Success, I get the following exception:
scala> convertAsTypeOf[Double]("abc").get
java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Double
at scala.runtime.BoxesRunTime.unboxToDouble(BoxesRunTime.java:119)
Why does this happen? Is there any way to fix this and make asInstanceOf[T] when it's in a generic function?
Because of type erasure, the type T is unknown at runtime, so asInstanceOf can't actually check the cast that it does. In fact, it compiles as a no-op. However, when the .get is finally done a cast is made to Double, and because the type is known here we can get a ClassCastException.
If you change convertAsTypeOf to use ClassTag:
def convertAsTypeOf[T](anyValue: Any)(implicit tag: ClassTag[T]): Try[T] =
Try(tag.runtimeClass.cast(anyValue).asInstanceOf[T])
You will then get errors when you expect:
scala> convertAsTypeOf[Double]("abc")
res1: scala.util.Try[Double] = Failure(java.lang.ClassCastException: Cannot cast java.lang.String to double)
The ClassTag represents the type T at runtime, allowing the value to be checked against it.
The reason this happens is type erasure, which makes asInstanceOf[T] for generic T useless. I don't know why the compiler doesn't issue any warning.
You can force the check to happen by using a ClassTag. Something like this:
import reflect.ClassTag
def convertAsTypeOf[T : ClassTag](anyValue: Any): Try[T] = Try {
anyValue match { case t: T => t }
}
This works because the compiler will use the implicit ClassTag in the case match if it is available (and generate a warning if it is not).

Why won't Scala use implicit conversion here?

I'm trying to call this set method documented here, in the Java library jOOQ, with signature:
<T> ... set(Field<T> field, T value)
This Scala line is a problem:
.set(table.MODIFIED_BY, userId)
MODIFIED_BY is a Field<Integer> representing the table column. userId is Int. Predef has an implicit conversion from Int to Integer, so why doesn't it use it? I get this:
type mismatch; found: org.jooq.TableField[gen.tables.records.DocRecord,Integer]
required: org.jooq.Field[Any]
Note: Integer <: Any
(and org.jooq.TableField[gen.tables.records.DocRecord,Integer] <:
org.jooq.Field[Integer]), but Java-defined trait Field is invariant in type T.
You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
Update - About Vinicius's Example
Rather than try to explain this in comments, here is a demonstration that there is no implicit conversion being called when you use a type with covariant parameter, like List[+T]. Let's say I put this code in a file, compile, and run it...
case class Foo(str: String)
object StackOver1 extends App {
implicit def str2Foo(s: String): Foo = {
println("In str2Foo.")
new Foo(s)
}
def test[T](xs: List[T], x: T): List[T] = {
println("test " + x.getClass)
xs
}
val foo1 = new Foo("foo1")
test(List(foo1), "abc")
}
You'll see that it calls test, but never the implicit conversion from String "abc" to Foo. Instead it's picking a T for test[T] that is a common base class between String and Foo. When you use Int and Integer it picks Any, but it's confusing because the runtime representation of the Int in the list is Integer. So it looks like it used the implicit conversion, but it didn't. You can verify by opening a Scala prompt...
scala> :type StackOver1.test(List(new java.lang.Integer(1)), 2)
List[Any]
I don't know anything aboutjOOQ, but I think the issue is that Scala does not understand java generics very well. Try:
scala> def test[T](a : java.util.ArrayList[T], b: T) = { println(a,b) }
scala> val a = new java.util.ArrayList[Integer]()
scala> val b = 12
scala> test(a,b)
<console>:11: error: type mismatch;
found : java.util.ArrayList[Integer]
required: java.util.ArrayList[Any]
Note: Integer <: Any, but Java-defined class ArrayList is invariant in type E.
You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
test(a,b)
Sounds familiar??
And to fix, just inform the type T to call the method: test[Integer](a,b) works fine.
EDIT:
There a few things involved here:
Erasure -> When compiled the type of the generic will disappear by erasure. The compiler will use Object which Scala, will treat as Any. However a ArrayList[Integer] is not an ArrayList[Any], even though Integer is any. The same way that TableField[gen.tables.records.DocRecord,Integer] is not a Field[Any].
Type inference mechanism -> it will figure out what type T should be and to do that it will use the intersection dominator of the types passed (in our case the first common ancestor). Page 36 of Scala Language Spec, which in our examples above will lead use to Any.
Implicit conversion -> it is the last step and would be called if there was some type to be converted to another one, but since the type of the arguments were determined to be the first common ancestor, there is no need to convert and we will never have a implicit conversion if we don't force the type T.
A example to show how the common ancestor is used to determine T:
scala> def test[T](a: T, b: T): T = a
scala> class Foo
scala> class Boo extends Foo
scala> test(new Boo,new Foo)
res2: Foo = Boo#139c2a6
scala> test(new Boo,new Boo)
res3: Boo = Boo#141c803
scala> class Coo extends Foo
scala> test(new Boo,new Coo)
res4: Foo = Boo#aafc83
scala> test(new Boo,"qsasad")
res5: Object = Boo#16989d8
Summing up, the implicit method does not get called, because the type inference mechanism, determines the types before getting the argument and since it uses the common ancestor, there is no need for a implicit conversion.
Your code produces an error due to erasure mechanism which disappear with the type information that would be important to determine the correct type of the argument.
#RobN, thanks for questioning my answer, I learned a lot with the process.

How to obtain the raw datatype of a parameter of a field that is specialized in Scala via reflection?

I have a class with a field that is specialized and is using a raw datatype. For instance a Tuple2[Int, String]:
scala> class TupleReflection(val tuple: Tuple2[Int, String])
defined class TupleReflection
scala> val refl = new TupleReflection((5, "hello"))
refl: TupleReflection = TupleReflection#1a597ec8
I want to use reflection now to find out the type parameters of the Tuple2 inside my 'refl' instance. (I cheat a little using 'head' to get the field because I know it's the only one.)
scala> val field = refl.getClass.getDeclaredFields.head
field: java.lang.reflect.Field = private final scala.Tuple2 TupleReflection.tuple
Now that I have the field I can query the generic types.
scala> field.getGenericType
res41: java.lang.reflect.Type = scala.Tuple2<java.lang.Object, java.lang.String>
The problem now is that the first type is Object. Is there a way to know, via reflection alone, the real type (Int) of that parameter?
Update:
I'm using this in a context of automatic serialization within my own API. Given a class marked with #Serializable I can serialize it. To do so I must build a tree of the fields and types of the class recursively using reflection so I can do a deep serialization.
If I'm working directly with a #Specialized class it works because the types are explicit and known at compile time at the invocation site. If a field in the hierarchy is #specialized I have no way to tell via reflection. Querying the fields or methods declared in a class doesn't yield the correct value. The type is present in runtime but only on the instance held in the field, not on the declaration of the field itself. Thus if the instance is null and can't do a "getClass" I can't know the correct types by reflection alone.
The Problem is you are using a Java reflection API which doesn't workaround the JVM's "Type Erasure" issue, because of that there's no way to find out the actual generic types with it.
Fortunately the coming 2.10 version of Scala implements new reflection API, which solves the type erasure issue. But since 2.10 hasn't been released yet, the API isn't standardized nor documented yet. Your best bet is to dig into it with some tool like debugger and ask more specific questions that come up here.
In Scala 2.10-M5 you can access the API like follows:
scala> reflect.runtime.universe.typeOf[(Int, String)]
res0: reflect.runtime.universe.Type = (Int, String)
scala> reflect.runtime.universe.typeOf[(Int, String)].typeArguments
res1: List[reflect.runtime.universe.Type] = List(Int, String)
scala> reflect.runtime.universe.typeOf[(Int, String)].typeArguments.head
res2: reflect.runtime.universe.Type = Int
Update #1
The following function shows how you can get a type of an instance:
scala> import reflect.runtime.universe._
import reflect.runtime.universe._
scala> def typeOf[T : TypeTag](x : T) = reflect.runtime.universe.typeOf[T]
typeOf: [T](x: T)(implicit evidence$1: reflect.runtime.universe.TypeTag[T])reflect.runtime.universe.Type
scala> typeOf(Seq((1,"sldf"),(20,"sldkfjew")))
res0: reflect.runtime.universe.Type = Seq[(Int, String)]
In fact it's all based around the [T : TypeTag] part which tells the compiler to magically create an implicit instance of the reflection to the type passed.
Update #2
scala> class TupleReflection(val tuple: Tuple2[Int, String])
defined class TupleReflection
scala> import reflect.runtime.universe._
import reflect.runtime.universe._
scala> typeOf[TupleReflection].member(newTermName("tuple")).typeSignature
res6: reflect.runtime.universe.Type = => (scala.Int, String)
scala> typeOf[TupleReflection].members
res7: Iterable[reflect.runtime.universe.Symbol] = List(constructor TupleReflection, value tuple, value tuple, method $asInstanceOf, method $isInstanceOf, method synchronized, method ##, method !=, method ==, method ne, method eq, constructor Object, method notifyAll, method notify, method clone, method getClass, method hashCode, method toString, method equals, method wait, method wait, method wait, method finalize, method asInstanceOf, method isInstanceOf, method !=, method ==)
scala> typeOf[TupleReflection].members.view.filter(_.isValue).filter(!_.isMethod).toList
res16: List[reflect.runtime.universe.Symbol] = List(value tuple)
I found a way but it's not pretty.
class TupleReflection(#(specializedFor #field)(Array(classOf[Int], classOf[String]) val tuple: Tuple2[Int, String])
I created the annotation specializedFor with runtime retention policy. It receives an Array of Class[_]. That way I can find in runtime exclusively with reflection the types of the Tuple2 field.
It's unsafe because I can't test that the Array contains the same types as the Tuple2.
In my API I have to first check if the annotation is present and if it is force the genericTypes to be those.