From seq to _* in Scala - scala

Does anyone knows how we can convert from any Seq to _* in an automatic way? It's quite cumbersome to force the type every time we have a Seq and a method uses a parameter of type vararg.
def mean[T: Numeric](elems: T*): Double
...
elems = Seq(1.0, 2.0, 3.0)
mean(elems) // this doesn't compiles
mean(elems: _*) // this compiles but it is cumbersome

That's the only way. It's a reason why varargs are arguably best only used at the public interface of a library, and even then, particularly when you think the caller will be calling with literally specified elements instead of a collection. If a method will likely be called on a collection argument, varargs can backfire in its goal of reducing syntactic nois, as you've noticed.

If the method isn't generic, you can add an overload:
def mean(elems: Seq[Double]): Double = ...
def mean(elems: Double*)(implicit d: DummyImplicit): Double = mean(elems)
Alas, it doesn't work in this case:
scala> object X { def f[T: Numeric](x: T*) = x; def f[T: Numeric](x: Seq[T])(implicit d: DummyImplicit) = x }
defined module X
scala> X.f(Seq(1, 2))
<console>:9: error: ambiguous reference to overloaded definition, both method f in object X of type [T](x: Seq[T])(implicit evidence$2: Numeric[T], implicit d: DummyImplicit)Seq[T] and method f in object X of type [T](x: T*)(implicit evidence$1: Numeric[T])Seq[T] match argument types (Seq[Int])
X.f(Seq(1, 2))
^
because the compiler thinks T could be Int or Seq[Int], and stops before checking whether implicits are available for both (at least in Scala 2.10).

Related

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.

reducing an Array of Float using scala.math.max

I am confused by the following behavior - why does reducing an Array of Int work using math.max, but an Array of Float requires a wrapped function? I have memories that this was not an issue in 2.9, but I'm not completely certain about that.
$ scala -version
Scala code runner version 2.10.2 -- Copyright 2002-2013, LAMP/EPFL
$ scala
scala> import scala.math._
scala> Array(1, 2, 4).reduce(max)
res47: Int = 4
scala> Array(1f, 3f, 4f).reduce(max)
<console>:12: error: type mismatch;
found : (Int, Int) => Int
required: (AnyVal, AnyVal) => AnyVal
Array(1f, 3f, 4f).reduce(max)
^
scala> def fmax(a: Float, b: Float) = max(a, b)
fmax: (a: Float, b: Float)Float
scala> Array(1f, 3f, 4f).reduce(fmax)
res45: Float = 4.0
update : this does work
scala> Array(1f, 2f, 3f).reduce{(x,y) => math.max(x,y)}
res2: Float = 3.0
so then it is just reduce(math.max) which cannot be shorthanded?
The first thing to note is that math.max is overloaded, and if the compiler has no hint about the expected argument types, it just picks one of the overloads (I'm not clear yet on what rules govern which overload is picked, but it will become clear before the end of this post).
Apparently it favors the overload that takes Int parameters over the others. This can be seen in the repl:
scala> math.max _
res6: (Int, Int) => Int = <function2>
That method is most specific because the first of the following compiles (by virtue of numeric widening conversions) and the second does not:
scala> (math.max: (Float,Float)=>Float)(1,2)
res0: Float = 2.0
scala> (math.max: (Int,Int)=>Int)(1f,2f)
<console>:8: error: type mismatch;
found : Float(1.0)
required: Int
(math.max: (Int,Int)=>Int)(1f,2f)
^
The test is whether one function applies to the param types of the other, and that test includes any conversions.
Now, the question is: why can't the compiler infer the correct expected type? It certainly knows that the type of Array(1f, 3f, 4f) is Array[Float]
We can get a clue if we replace reduce with reduceLeft: then it compiles fine.
So surely this has to do with a difference in the signature of reduceLeft and reduce.
We can reproduce the error with the following code snippet:
case class MyCollection[A]() {
def reduce[B >: A](op: (B, B) => B): B = ???
def reduceLeft[B >: A](op: (B, A) => B): B = ???
}
MyCollection[Float]().reduce(max) // Fails to compile
MyCollection[Float]().reduceLeft(max) // Compiles fine
The signatures are subtly different.
In reduceLeft the second argument is forced to A (the collection's type), so type inference is trivial: if A==Float (which the compiler knows), then the compiler knows that the only valid overload of max is one that takes a Float as its second argument. The compiler only finds one ( max(Float,Float) ), and it happens that the other constraint (that B >: A) is trivially satisfied (as B == A == Float for this overload).
This is different for reduce: both the first and second arguments can be any (same) super-type of A (that is, of Float in our specific case). This is a much more lax constraint, and while it could be argued that in this case the compiler could see that there is only one possibility, the compiler is not smart enough here.
Whether the compiler is supposed to be able to handle this case (meaning that this is an inference bug) or not, I must say I don't know. Type inference is a tricky business in scala, and as far as I know the spec is intentionally vague about what can be inferred or not.
Since there are useful applications such as:
scala> Array(1f,2f,3f).reduce[Any](_.toString+","+_.toString)
res3: Any = 1.0,2.0,3.0
trying overload resolution against every possible substitution of the type parameter is expensive and could change the result depending on the expected type you wind up with; or would it have to issue an ambiguity error?
Using -Xlog-implicits -Yinfer-debug shows the difference between reduce(math.max), where overload resolution happens first, and the version where the param type is solved for first:
scala> Array(1f,2f,3f).reduce(math.max(_,_))
[solve types] solving for A1 in ?A1
inferExprInstance {
tree scala.this.Predef.floatArrayOps(scala.Array.apply(1.0, 2.0, 3.0)).reduce[A1]
tree.tpe (op: (A1, A1) => A1)A1
tparams type A1
pt ?
targs Float
tvars =?Float
}
It looks like this is a bug in the inferrer, cause with Int it infers types correctly:
private[this] val res2: Int = scala.this.Predef.intArrayOps(scala.Array.apply(1, 2, 4)).reduce[Int]({
((x: Int, y: Int) => scala.math.`package`.max(x, y))
});
but with Floats:
private[this] val res1: AnyVal = scala.this.Predef.floatArrayOps(scala.Array.apply(1.0, 3.0, 4.0)).reduce[AnyVal]({
((x: Int, y: Int) => scala.math.`package`.max(x, y))
});
If you explicitly annotate reduce with a Float type it should work:
Array(1f, 3f, 4f).reduce[Float](max)
private[this] val res3: Float = scala.this.Predef.floatArrayOps(scala.Array.apply(1.0, 3.0, 4.0)).reduce[Float]({
((x: Float, y: Float) => scala.math.`package`.max(x, y))
});
There is always scala.math.Ordering:
Array(1f, 2f, 3f).reduceOption(Ordering.Float.max)
It doesn't seem to be a bug. Consider the following code:
class C1 {}
object C1 {
implicit def c2toc1(x: C2): C1 = new C1
}
class C2 {}
class C3 {
def f(x: C1): Int = 1
def f(x: C2): Int = 2
}
(new C3).f _ //> ... .C2 => Int = <function1>
If I remove implicit conversion I will get an error "ambiguous reference". And because Int has an implicit conversion to Float Scala tries to find the most specific type for min, which is (Int, Int) => Int. The closest common superclass for Int and Float is AnyVal, that's why you see (AnyVal, AnyVal) => AnyVal.
The reason why (x, y) => min(x, y) works is probably because eta-expansion is done before type inference and reduce has to deal with (Int, Int) => Int which will be converted to (AnyVal, AnyVal) => AnyVal.
UPDATE: Meanwhile (new C3).f(_) will fail with "missing parameter type" error, which means f(_) depends on type inference and doesn't consider implicit conversions while f _ doesn't need parameter type and will expand to the most specific argument type if Scala can find one.

type parameter definition in scala [duplicate]

Sometime I stumble into the semi-mysterious notation of
def f[T](..) = new T[({type l[A]=SomeType[A,..]})#l] {..}
in Scala blog posts, which give it a "we used that type-lambda trick" handwave.
While I have some intutition about this (we gain an anonymous type parameter A without having to pollute the definition with it?), I found no clear source describing what the type lambda trick is, and what are its benefits. Is it just syntactic sugar, or does it open some new dimensions?
Type lambdas are vital quite a bit of the time when you are working with higher-kinded types.
Consider a simple example of defining a monad for the right projection of Either[A, B]. The monad typeclass looks like this:
trait Monad[M[_]] {
def point[A](a: A): M[A]
def bind[A, B](m: M[A])(f: A => M[B]): M[B]
}
Now, Either is a type constructor of two arguments, but to implement Monad, you need to give it a type constructor of one argument. The solution to this is to use a type lambda:
class EitherMonad[A] extends Monad[({type λ[α] = Either[A, α]})#λ] {
def point[B](b: B): Either[A, B]
def bind[B, C](m: Either[A, B])(f: B => Either[A, C]): Either[A, C]
}
This is an example of currying in the type system - you have curried the type of Either, such that when you want to create an instance of EitherMonad, you have to specify one of the types; the other of course is supplied at the time you call point or bind.
The type lambda trick exploits the fact that an empty block in a type position creates an anonymous structural type. We then use the # syntax to get a type member.
In some cases, you may need more sophisticated type lambdas that are a pain to write out inline. Here's an example from my code from today:
// types X and E are defined in an enclosing scope
private[iteratee] class FG[F[_[_], _], G[_]] {
type FGA[A] = F[G, A]
type IterateeM[A] = IterateeT[X, E, FGA, A]
}
This class exists exclusively so that I can use a name like FG[F, G]#IterateeM to refer to the type of the IterateeT monad specialized to some transformer version of a second monad which is specialized to some third monad. When you start to stack, these kinds of constructs become very necessary. I never instantiate an FG, of course; it's just there as a hack to let me express what I want in the type system.
The benefits are exactly the same as those conferred by anonymous functions.
def inc(a: Int) = a + 1; List(1, 2, 3).map(inc)
List(1, 2, 3).map(a => a + 1)
An example usage, with Scalaz 7. We want to use a Functor that can map a function over the second element in a Tuple2.
type IntTuple[+A]=(Int, A)
Functor[IntTuple].map((1, 2))(a => a + 1)) // (1, 3)
Functor[({type l[a] = (Int, a)})#l].map((1, 2))(a => a + 1)) // (1, 3)
Scalaz provides some implicit conversions that can infer the type argument to Functor, so we often avoid writing these altogether. The previous line can be rewritten as:
(1, 2).map(a => a + 1) // (1, 3)
If you use IntelliJ, you can enable Settings, Code Style, Scala, Folding, Type Lambdas. This then hides the crufty parts of the syntax, and presents the more palatable:
Functor[[a]=(Int, a)].map((1, 2))(a => a + 1)) // (1, 3)
A future version of Scala might directly support such a syntax.
To put things in context: This answer was originally posted in another thread. You are seeing it here because the two threads have been merged. The question statement in the said thread was as follows:
How to resolve this type definition: Pure[({type ?[a]=(R, a)})#?] ?
What are the reasons of using such construction?
Snipped comes from scalaz library:
trait Pure[P[_]] {
def pure[A](a: => A): P[A]
}
object Pure {
import Scalaz._
//...
implicit def Tuple2Pure[R: Zero]: Pure[({type ?[a]=(R, a)})#?] = new Pure[({type ?[a]=(R, a)})#?] {
def pure[A](a: => A) = (Ø, a)
}
//...
}
Answer:
trait Pure[P[_]] {
def pure[A](a: => A): P[A]
}
The one underscore in the boxes after P implies that it is a type constructor takes one type and returns another type. Examples of type constructors with this kind: List, Option.
Give List an Int, a concrete type, and it gives you List[Int], another concrete type. Give List a String and it gives you List[String]. Etc.
So, List, Option can be considered to be type level functions of arity 1. Formally we say, they have a kind * -> *. The asterisk denotes a type.
Now Tuple2[_, _] is a type constructor with kind (*, *) -> * i.e. you need to give it two types to get a new type.
Since their signatures do not match, you cannot substitute Tuple2 for P. What you need to do is partially apply Tuple2 on one of its arguments, which will give us a type constructor with kind * -> *, and we can substitue it for P.
Unfortunately Scala has no special syntax for partial application of type constructors, and so we have to resort to the monstrosity called type lambdas. (What you have in your example.) They are called that because they are analogous to lambda expressions that exist at value level.
The following example might help:
// VALUE LEVEL
// foo has signature: (String, String) => String
scala> def foo(x: String, y: String): String = x + " " + y
foo: (x: String, y: String)String
// world wants a parameter of type String => String
scala> def world(f: String => String): String = f("world")
world: (f: String => String)String
// So we use a lambda expression that partially applies foo on one parameter
// to yield a value of type String => String
scala> world(x => foo("hello", x))
res0: String = hello world
// TYPE LEVEL
// Foo has a kind (*, *) -> *
scala> type Foo[A, B] = Map[A, B]
defined type alias Foo
// World wants a parameter of kind * -> *
scala> type World[M[_]] = M[Int]
defined type alias World
// So we use a lambda lambda that partially applies Foo on one parameter
// to yield a type of kind * -> *
scala> type X[A] = World[({ type M[A] = Foo[String, A] })#M]
defined type alias X
// Test the equality of two types. (If this compiles, it means they're equal.)
scala> implicitly[X[Int] =:= Foo[String, Int]]
res2: =:=[X[Int],Foo[String,Int]] = <function1>
Edit:
More value level and type level parallels.
// VALUE LEVEL
// Instead of a lambda, you can define a named function beforehand...
scala> val g: String => String = x => foo("hello", x)
g: String => String = <function1>
// ...and use it.
scala> world(g)
res3: String = hello world
// TYPE LEVEL
// Same applies at type level too.
scala> type G[A] = Foo[String, A]
defined type alias G
scala> implicitly[X =:= Foo[String, Int]]
res5: =:=[X,Foo[String,Int]] = <function1>
scala> type T = World[G]
defined type alias T
scala> implicitly[T =:= Foo[String, Int]]
res6: =:=[T,Foo[String,Int]] = <function1>
In the case you have presented, the type parameter R is local to function Tuple2Pure and so you cannot simply define type PartialTuple2[A] = Tuple2[R, A], because there is simply no place where you can put that synonym.
To deal with such a case, I use the following trick that makes use of type members. (Hopefully the example is self-explanatory.)
scala> type Partial2[F[_, _], A] = {
| type Get[B] = F[A, B]
| }
defined type alias Partial2
scala> implicit def Tuple2Pure[R]: Pure[Partial2[Tuple2, R]#Get] = sys.error("")
Tuple2Pure: [R]=> Pure[[B](R, B)]

What are type lambdas in Scala and what are their benefits?

Sometime I stumble into the semi-mysterious notation of
def f[T](..) = new T[({type l[A]=SomeType[A,..]})#l] {..}
in Scala blog posts, which give it a "we used that type-lambda trick" handwave.
While I have some intutition about this (we gain an anonymous type parameter A without having to pollute the definition with it?), I found no clear source describing what the type lambda trick is, and what are its benefits. Is it just syntactic sugar, or does it open some new dimensions?
Type lambdas are vital quite a bit of the time when you are working with higher-kinded types.
Consider a simple example of defining a monad for the right projection of Either[A, B]. The monad typeclass looks like this:
trait Monad[M[_]] {
def point[A](a: A): M[A]
def bind[A, B](m: M[A])(f: A => M[B]): M[B]
}
Now, Either is a type constructor of two arguments, but to implement Monad, you need to give it a type constructor of one argument. The solution to this is to use a type lambda:
class EitherMonad[A] extends Monad[({type λ[α] = Either[A, α]})#λ] {
def point[B](b: B): Either[A, B]
def bind[B, C](m: Either[A, B])(f: B => Either[A, C]): Either[A, C]
}
This is an example of currying in the type system - you have curried the type of Either, such that when you want to create an instance of EitherMonad, you have to specify one of the types; the other of course is supplied at the time you call point or bind.
The type lambda trick exploits the fact that an empty block in a type position creates an anonymous structural type. We then use the # syntax to get a type member.
In some cases, you may need more sophisticated type lambdas that are a pain to write out inline. Here's an example from my code from today:
// types X and E are defined in an enclosing scope
private[iteratee] class FG[F[_[_], _], G[_]] {
type FGA[A] = F[G, A]
type IterateeM[A] = IterateeT[X, E, FGA, A]
}
This class exists exclusively so that I can use a name like FG[F, G]#IterateeM to refer to the type of the IterateeT monad specialized to some transformer version of a second monad which is specialized to some third monad. When you start to stack, these kinds of constructs become very necessary. I never instantiate an FG, of course; it's just there as a hack to let me express what I want in the type system.
The benefits are exactly the same as those conferred by anonymous functions.
def inc(a: Int) = a + 1; List(1, 2, 3).map(inc)
List(1, 2, 3).map(a => a + 1)
An example usage, with Scalaz 7. We want to use a Functor that can map a function over the second element in a Tuple2.
type IntTuple[+A]=(Int, A)
Functor[IntTuple].map((1, 2))(a => a + 1)) // (1, 3)
Functor[({type l[a] = (Int, a)})#l].map((1, 2))(a => a + 1)) // (1, 3)
Scalaz provides some implicit conversions that can infer the type argument to Functor, so we often avoid writing these altogether. The previous line can be rewritten as:
(1, 2).map(a => a + 1) // (1, 3)
If you use IntelliJ, you can enable Settings, Code Style, Scala, Folding, Type Lambdas. This then hides the crufty parts of the syntax, and presents the more palatable:
Functor[[a]=(Int, a)].map((1, 2))(a => a + 1)) // (1, 3)
A future version of Scala might directly support such a syntax.
To put things in context: This answer was originally posted in another thread. You are seeing it here because the two threads have been merged. The question statement in the said thread was as follows:
How to resolve this type definition: Pure[({type ?[a]=(R, a)})#?] ?
What are the reasons of using such construction?
Snipped comes from scalaz library:
trait Pure[P[_]] {
def pure[A](a: => A): P[A]
}
object Pure {
import Scalaz._
//...
implicit def Tuple2Pure[R: Zero]: Pure[({type ?[a]=(R, a)})#?] = new Pure[({type ?[a]=(R, a)})#?] {
def pure[A](a: => A) = (Ø, a)
}
//...
}
Answer:
trait Pure[P[_]] {
def pure[A](a: => A): P[A]
}
The one underscore in the boxes after P implies that it is a type constructor takes one type and returns another type. Examples of type constructors with this kind: List, Option.
Give List an Int, a concrete type, and it gives you List[Int], another concrete type. Give List a String and it gives you List[String]. Etc.
So, List, Option can be considered to be type level functions of arity 1. Formally we say, they have a kind * -> *. The asterisk denotes a type.
Now Tuple2[_, _] is a type constructor with kind (*, *) -> * i.e. you need to give it two types to get a new type.
Since their signatures do not match, you cannot substitute Tuple2 for P. What you need to do is partially apply Tuple2 on one of its arguments, which will give us a type constructor with kind * -> *, and we can substitue it for P.
Unfortunately Scala has no special syntax for partial application of type constructors, and so we have to resort to the monstrosity called type lambdas. (What you have in your example.) They are called that because they are analogous to lambda expressions that exist at value level.
The following example might help:
// VALUE LEVEL
// foo has signature: (String, String) => String
scala> def foo(x: String, y: String): String = x + " " + y
foo: (x: String, y: String)String
// world wants a parameter of type String => String
scala> def world(f: String => String): String = f("world")
world: (f: String => String)String
// So we use a lambda expression that partially applies foo on one parameter
// to yield a value of type String => String
scala> world(x => foo("hello", x))
res0: String = hello world
// TYPE LEVEL
// Foo has a kind (*, *) -> *
scala> type Foo[A, B] = Map[A, B]
defined type alias Foo
// World wants a parameter of kind * -> *
scala> type World[M[_]] = M[Int]
defined type alias World
// So we use a lambda lambda that partially applies Foo on one parameter
// to yield a type of kind * -> *
scala> type X[A] = World[({ type M[A] = Foo[String, A] })#M]
defined type alias X
// Test the equality of two types. (If this compiles, it means they're equal.)
scala> implicitly[X[Int] =:= Foo[String, Int]]
res2: =:=[X[Int],Foo[String,Int]] = <function1>
Edit:
More value level and type level parallels.
// VALUE LEVEL
// Instead of a lambda, you can define a named function beforehand...
scala> val g: String => String = x => foo("hello", x)
g: String => String = <function1>
// ...and use it.
scala> world(g)
res3: String = hello world
// TYPE LEVEL
// Same applies at type level too.
scala> type G[A] = Foo[String, A]
defined type alias G
scala> implicitly[X =:= Foo[String, Int]]
res5: =:=[X,Foo[String,Int]] = <function1>
scala> type T = World[G]
defined type alias T
scala> implicitly[T =:= Foo[String, Int]]
res6: =:=[T,Foo[String,Int]] = <function1>
In the case you have presented, the type parameter R is local to function Tuple2Pure and so you cannot simply define type PartialTuple2[A] = Tuple2[R, A], because there is simply no place where you can put that synonym.
To deal with such a case, I use the following trick that makes use of type members. (Hopefully the example is self-explanatory.)
scala> type Partial2[F[_, _], A] = {
| type Get[B] = F[A, B]
| }
defined type alias Partial2
scala> implicit def Tuple2Pure[R]: Pure[Partial2[Tuple2, R]#Get] = sys.error("")
Tuple2Pure: [R]=> Pure[[B](R, B)]

Passing functions for all applicable types around

I followed the advice found here to define a function called square, and then tried to pass it to a function called twice. The functions are defined like this:
def square[T](n: T)(implicit numeric: Numeric[T]): T = numeric.times(n, n)
def twice[T](f: (T) => T, a: T): T = f(f(a))
When calling twice(square, 2), the REPL spits out an error message:
scala> twice(square, 2)
<console>:8: error: could not find implicit value for parameter numeric: Numeric[T]
twice(square, 2)
^
Anyone?
I disagree with everyone here except Andrew Phillips. Well, everyone so far. :-) The problem is here:
def twice[T](f: (T) => T, a: T): T = f(f(a))
You expect, like newcomers to Scala often do, for Scala's compiler to take into account both parameters to twice to infer the correct types. Scala doesn't do that, though -- it only uses information from one parameter list to the next, but not from one parameter to the next. That mean the parameters f and a are analyzed independently, without having the advantage of knowing what the other is.
That means, for instance, that this works:
twice(square[Int], 2)
Now, if you break it down into two parameter lists, then it also works:
def twice[T](a: T)(f: (T) => T): T = f(f(a))
twice(2)(square)
So, basically, everything you were trying to do was correct and should work, except for the part that you expected one parameter to help figuring out the type of the other parameter (as you wrote it).
Here's a session from the Scala REPL.
Welcome to Scala version 2.8.0.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_20).
Type in expressions to have them evaluated.
Type :help for more information.
scala> def square[T : Numeric](n: T) = implicitly[Numeric[T]].times(n, n)
square: [T](n: T)(implicit evidence$1: Numeric[T])T
scala> def twice2[T](f: T => T)(a: T) = f(f(a))
twice2: [T](f: (T) => T)(a: T)T
scala> twice2(square)(3)
<console>:8: error: could not find implicit value for evidence parameter of type
Numeric[T]
twice2(square)(3)
^
scala> def twice3[T](a: T, f: T => T) = f(f(a))
twice3: [T](a: T,f: (T) => T)T
scala> twice3(3, square)
<console>:8: error: could not find implicit value for evidence parameter of type
Numeric[T]
twice3(3, square)
scala> def twice[T](a: T)(f: T => T) = f(f(a))
twice: [T](a: T)(f: (T) => T)T
scala> twice(3)(square)
res0: Int = 81
So evidently the type of "twice(3)" needs to be known before the implicit can be resolved. I guess that makes sense, but I'd still be glad if a Scala guru could comment on this one...
Another solution is to lift square into partially applied function:
scala> twice(square(_:Int),2)
res1: Int = 16
This way the implicit is applied to square as in:
scala> twice(square(_:Int)(implicitly[Numeric[Int]]),2)
res3: Int = 16
There is even another approach:
def twice[T:Numeric](f: (T) => T, a: T): T = f(f(a))
scala> twice[Int](square,2)
res1: Int = 16
But again, the type parameter don't get inferred.
Your problem is that square isn't a function (ie. a scala.Function1[T, T] aka (T) => T). Instead it's a type parametrized method with multiple argument lists one of which is implicit ... there's no syntax in Scala to define an exactly equivalent function.
Interestingly, your use of the Numeric type class means that the usual encodings of higher-ranked functions in Scala don't directly apply here, but we can adapt them to this case and get something like this,
trait HigherRankedNumericFunction {
def apply[T : Numeric](t : T) : T
}
val square = new HigherRankedNumericFunction {
def apply[T : Numeric](t : T) : T = implicitly[Numeric[T]].times(t, t)
}
This gives us a higher-ranked "function" with its type parameter context-bounded to Numeric,
scala> square(2)
res0: Int = 4
scala> square(2.0)
res1: Double = 4.0
scala> square("foo")
<console>:8: error: could not find implicit value for evidence parameter of type Numeric[java.lang.String]
square("foo")
We can now define twice in terms of HigherRankedNumericFunctions,
def twice[T : Numeric](f : HigherRankedNumericFunction, a : T) : T = f(f(a))
scala> twice(square, 2)
res2: Int = 16
scala> twice(square, 2.0)
res3: Double = 16.0
The obvious downside of this approach is that you lose out on the conciseness of Scala's monomorphic function literals.