Given:
scala> trait Resource[A] { def f: String }
defined trait Resource
scala> case class Foo(x: String)
defined class Foo
And then an implicit:
scala> implicit def fooToResource(foo: Foo): Resource[Foo] =
new Resource[Foo] { def f = foo.x }
The following works:
scala> implicitly[Resource[Foo]](Foo("foo")).f
res2: String = foo
I defined a function:
scala> def f[A](x: A)(implicit ev: Resource[A]): String = ev.f
f: [A](x: A)(implicit ev: Resource[A])String
However, the following code fails to compile:
scala> f(Foo("foo"))
<console>:17: error: could not find implicit value for parameter ev: Resource[Foo]
f(Foo("foo"))
Secondly, then I tried:
scala> f2(Foo("bippy"))
<console>:17: error: could not find implicit value for parameter ev: Resource[Foo]
f2(Foo("bippy"))
^
Lastly, I attempted:
scala> def g(foo: Foo)(implicit ev: Resource[Foo]): String = ev.f
g: (foo: Foo)(implicit ev: Resource[Foo])String
scala> g(Foo("5"))
<console>:17: error: could not find implicit value for parameter ev: Resource[Foo]
g(Foo("5"))
^
However, it failed too. How can I fix f?
Ok with Peter Neyens' answer, this is not a typeclass, this is an implicit conversion, which you should avoid - there should have been some warning, asking that you import scala.language.implicitConversions.
As a complement, here is why the first implicitly works:
Implicitly is just:
def implicitly[T](implicit ev: T): T = e
When you write implicitly[T] without supplying a parameter, it will look for an implicit of type T in scope and return it. However, you call implicitly with a parameter (I believe there is no legitimate reason to do that, ever), so it would just return your parameter, Foo("foo"), an instance of Foo. Except that you explicitly stated that T should be Resource[Foo]. If you had written a type ascription, such as (Foo("foo"): Resource[Foo]), it would have worked the same way. implicitly is not relevant here.
The point is that Foo("foo") is not of the expected type Resource[Foo], but just a Foo. The compiler would reject that, except that at this point, the implicit conversion you defined above kicks in, and your Foo instance is transformed into a Resource[Foo]. Then, you can call f.
Next, you call your f(Foo("foo")). There is an implicit parameter, however this time, you don't supply it. So the compiler looks for one (while it did no such thing the first time), and as there is no such instance, fails.
The implicit def fooToResource is not a type class instance, but does return one if you supply a Foo, that's the reason the following line works :
implicitly[Resource[Foo]](Foo("foo")).f
A solution would be to change the Resource.f function to take a parameter of type A :
trait Resource[A] {
def f(a: A): String
}
You then could define a Resource type class instance for Foo as follows:
case class Foo(x: String)
implicit val fooResource = new Resource[Foo] {
def f(foo: Foo) = foo.x
}
We can rewrite f to use the changed Resource :
def f[A](a: A)(implicit resA: Resource[A]): String = resA.f(a)
Which does what (I think) you need :
f(Foo("hello world")) // String = hello world
Related
Given a made-up F type-class:
scala> trait F[A] {}
defined trait F
and this definition, which uses a context bound to require that the input A has a type-class instance of F:
scala> def f[A : F](x: A) = ???
f: [A](x: A)(implicit evidence$1: F[A])Nothing
I defined a Person and type-class instance:
scala> case class Person(name: String)
defined class Person
scala> implicit val person: F[Person] = new F[Person] {}
person: F[Person] = $anon$1#262b2c86
And the following compiles:
scala> f(Person("foo"))
scala.NotImplementedError: an implementation is missing
But, there's no String implementation, so it fails.
scala> f("foobar")
<console>:17: error: could not find implicit value for evidence parameter of type F[String]
f("foobar")
^
I then defined an F[String] using:
scala> implicit def fInstance(x: String) = new F[String] {}
fInstance: (x: String)F[String]
But, I can't run:
scala> f("foobar")
<console>:18: error: could not find implicit value for evidence parameter of type F[String]
f("foobar")
^
since I do not have an implicit F[String], but rather a String => F[String].
What's the proper way to use such an implicit def to meet the F[String] constraint, i.e. call the f function successfully with a type of String?
I got it to work via:
scala> implicit val x: F[String] = implicitly[String => F[String]].apply("foobar")
x: F[String] = $anon$1#7b7fdc8
scala> f("foobar")
scala.NotImplementedError: an implementation is missing
at scala.Predef$.$qmark$qmark$qmark(Predef.scala:230)
at .f(<console>:12)
... 33 elided
But I'm not sure if it's the right/clean way to do it.
You defined an implicit conversion. If you want to use a def to provide typeclass instances you just write the same as you'd write an implicit val but replace val with def.
implicit def fInstance = new F[String] {}
Normally you only use a def if you need type parameters, like here.
implicit def fInstance[A] = new F[List[A]] {}
Or
implicit def fInstance[A](implicit ev: F[A]) = new F[List[A]] {}
Your fInstance defines an implicit conversion, i.e. a way to turn a String into F[String]. For generating a typeclass instance, a method accepting implicit parameters can be used:
implicit def fInstance(implicit x: String) = new F[String] {}
it is typically used in FP libraries to derive one typeclass from another:
implicit def optionMonoid[A](implicit S: Semigroup[A]): Monoid[Option[A]] = ???
// or, which is the same
// implicit def optionMonoid[A: Semigroup]: Monoid[Option[A]] = ???
The idea is that F[String] can operate on any String in general, not being dependent on actual arguments provided into function. Of course, you can always provide instances explicitly:
f("foobar")(new F[String] { })
As a follow-up, the important part of typeclasses is that you can define them ad-hoc, i.e. not having access to definitions of F and String at all, and you are forced to scope implicits in Scala and import them, so it's totally ok.
Here is a simpler version of your definition (and you can remove implicit from fInstance):
implicit val singleFInstance: F[String] = fInstance("") // or fInstance("foobar"), etc.
Whether this is the right thing to do, very much depends on what F and f are supposed to mean.
But generally speaking: if F is really a type-class, fInstance(string) gives different results depending on the string (not just different instances, but different behavior), and f's signature is correct, then this is wrong and you should accept that calling f("foobar") isn't meaningful.
I know this can be done but I can't remember (or locate) the syntax for the life of me. How can I specify that a specific function must be present on a generic type in a function definition?
For instance, I know I can do this:
def blah[A](p: A)
What I really want is something like this (but my syntax is all wrong of course):
def blah[A(someFunction)](p: A)
I am not sure if using this is a good idea, but here is how it works:
def blah[A <: { def someFunction: String }](a: A) : Unit =
println(a.someFunction)
scala> class Test { def someFunction: String = "hello" }
defined class Test
scala> blah(new Test)
hello
scala> class Test2
defined class Test2
scala> blah(new Test2)
<console>:16: error: inferred type arguments [Test2] do not conform to method blah's type parameter bounds [A <: AnyRef{def someFunction: String}]
Note that this feature (so-called structural types) uses reflection, and has to be enabled using the language import import scala.language.reflectiveCalls
What you're looking for is called a "structural type".
Any example from here: https://twitter.github.io/scala_school/advanced-types.html
scala> def foo(x: { def get: Int }) = 123 + x.get
foo: (x: AnyRef{def get: Int})Int
scala> foo(new { def get = 10 })
res0: Int = 133
Note, however, that this is going to be slower than just declaring a trait that requiring that any argument implements that trait.
I have the following (simplified) code:
case class Value[T](value: T)
trait Absable[In,Out] {
def absoluteValue(in: In): Out
}
implicit class AbsValue[In, Out](in: Value[In]) {
def abs()(implicit ev: Absable[In, Out]): Value[Out] = Value(ev.absoluteValue(in.value))
}
implicit def AbsNumeric[A : Numeric] = new Absable[A, A] {
def absoluteValue(in: A) = implicitly[Numeric[A]].abs(in)
}
Now I want to use the abs function on a Value:
scala> Value(-3).abs()
res3: Value[Int] = Value(3)
scala> Value(-3).abs
<console>:14: error: could not find implicit value for parameter ev: Absable[Int,Nothing]
Value(-3).abs
^
I added an empty argument list in front of the implicit arguments to give callers more flexibility, but now when I omit the empty list at the call site the compiler can't find the implicit... So now instead of more flexibility callers get confusing compile errors.
I don't understand how leaving off the argument list can affect the type inference or implicit resolution.
I am using scala 2.11.6
Given the following:
scala> trait Foo { def get: String = "get" }
defined trait Foo
I implemented it and made an implicit:
scala> case class FooImpl(x: String) extends Foo {
| override def get = s"got $x"
| }
defined class FooImpl
scala> implicit val fooImpl = FooImpl("yo")
fooImpl: FooImpl = FooImpl(yo)
Lastly, I tried to write a method that implicitly resolves a Foo, returning get on that implicitly resolved class.
scala> def f[A: Foo](x: A) = x.get
<console>:11: error: Foo does not take type parameters
def f[A: Foo](x: A) = x.get
^
<console>:11: error: value get is not a member of type parameter A
def f[A: Foo](x: A) = x.get
^
But I got the above errors.
So I re-wrote it using the implicit keyword:
scala> def f(implicit x: Foo): String = x.get
f: (implicit x: Foo)String
scala> f
res0: String = got yo
Is it possible to re-write this example to not explicitly specify the implicit keyword?
Note - it's possible that I'm confusing this notation with TypeTag under the section, Using a Context bound of a Type Parameter.
You are using the wrong syntax, what you want is an upper bound:
scala> def f[A <: Foo](x: A) = x.get
f: [A <: Foo](x: A)String
Where you're saying that A is a subtype of Foo and that tells the compiler that A does have a method called get.
The syntax you're using (:) means that there's an implicit conversion from A to Foo[A], the problem is that Foo doesn't take a type parameter, you can also check it into the REPL where the column syntax is translated to an implicit parameter:
scala> trait Foo2[T]
defined trait Foo2
scala> def g[T: Foo2](x: Int): Int = x
g: [T](x: Int)(implicit evidence$1: Foo2[T])Int
I have seen a function named implicitly used in Scala examples. What is it, and how is it used?
Example here:
scala> sealed trait Foo[T] { def apply(list : List[T]) : Unit }; object Foo {
| implicit def stringImpl = new Foo[String] {
| def apply(list : List[String]) = println("String")
| }
| implicit def intImpl = new Foo[Int] {
| def apply(list : List[Int]) = println("Int")
| }
| } ; def foo[A : Foo](x : List[A]) = implicitly[Foo[A]].apply(x)
defined trait Foo
defined module Foo
foo: [A](x: List[A])(implicit evidence$1: Foo[A])Unit
scala> foo(1)
<console>:8: error: type mismatch;
found : Int(1)
required: List[?]
foo(1)
^
scala> foo(List(1,2,3))
Int
scala> foo(List("a","b","c"))
String
scala> foo(List(1.0))
<console>:8: error: could not find implicit value for evidence parameter of type
Foo[Double]
foo(List(1.0))
^
Note that we have to write implicitly[Foo[A]].apply(x) since the compiler thinks that implicitly[Foo[A]](x) means that we call implicitly with parameters.
Also see How to investigate objects/types/etc. from Scala REPL? and Where does Scala look for implicits?
implicitly is avaliable in Scala 2.8 and is defined in Predef as:
def implicitly[T](implicit e: T): T = e
It is commonly used to check if an implicit value of type T is available and return it if such is the case.
Simple example from retronym's presentation:
scala> implicit val a = "test" // define an implicit value of type String
a: java.lang.String = test
scala> val b = implicitly[String] // search for an implicit value of type String and assign it to b
b: String = test
scala> val c = implicitly[Int] // search for an implicit value of type Int and assign it to c
<console>:6: error: could not find implicit value for parameter e: Int
val c = implicitly[Int]
^
Here are a few reasons to use the delightfully simple method implicitly.
To understand/troubleshoot Implicit Views
An Implicit View can be triggered when the prefix of a selection (consider for example, the.prefix.selection(args) does not contain a member selection that is applicable to args (even after trying to convert args with Implicit Views). In this case, the compiler looks for implicit members, locally defined in the current or enclosing scopes, inherited, or imported, that are either Functions from the type of that the.prefix to a type with selection defined, or equivalent implicit methods.
scala> 1.min(2) // Int doesn't have min defined, where did that come from?
res21: Int = 1
scala> implicitly[Int => { def min(i: Int): Any }]
res22: (Int) => AnyRef{def min(i: Int): Any} = <function1>
scala> res22(1) //
res23: AnyRef{def min(i: Int): Int} = 1
scala> .getClass
res24: java.lang.Class[_] = class scala.runtime.RichInt
Implicit Views can also be triggered when an expression does not conform to the Expected Type, as below:
scala> 1: scala.runtime.RichInt
res25: scala.runtime.RichInt = 1
Here the compiler looks for this function:
scala> implicitly[Int => scala.runtime.RichInt]
res26: (Int) => scala.runtime.RichInt = <function1>
Accessing an Implicit Parameter Introduced by a Context Bound
Implicit parameters are arguably a more important feature of Scala than Implicit Views. They support the type class pattern. The standard library uses this in a few places -- see scala.Ordering and how it is used in SeqLike#sorted. Implicit Parameters are also used to pass Array manifests, and CanBuildFrom instances.
Scala 2.8 allows a shorthand syntax for implicit parameters, called Context Bounds. Briefly, a method with a type parameter A that requires an implicit parameter of type M[A]:
def foo[A](implicit ma: M[A])
can be rewritten as:
def foo[A: M]
But what's the point of passing the implicit parameter but not naming it? How can this be useful when implementing the method foo?
Often, the implicit parameter need not be referred to directly, it will be tunneled through as an implicit argument to another method that is called. If it is needed, you can still retain the terse method signature with the Context Bound, and call implicitly to materialize the value:
def foo[A: M] = {
val ma = implicitly[M[A]]
}
Passing a subset of implicit parameters explicitly
Suppose you are calling a method that pretty prints a person, using a type class based approach:
trait Show[T] { def show(t: T): String }
object Show {
implicit def IntShow: Show[Int] = new Show[Int] { def show(i: Int) = i.toString }
implicit def StringShow: Show[String] = new Show[String] { def show(s: String) = s }
def ShoutyStringShow: Show[String] = new Show[String] { def show(s: String) = s.toUpperCase }
}
case class Person(name: String, age: Int)
object Person {
implicit def PersonShow(implicit si: Show[Int], ss: Show[String]): Show[Person] = new Show[Person] {
def show(p: Person) = "Person(name=" + ss.show(p.name) + ", age=" + si.show(p.age) + ")"
}
}
val p = Person("bob", 25)
implicitly[Show[Person]].show(p)
What if we want to change the way that the name is output? We can explicitly call PersonShow, explicitly pass an alternative Show[String], but we want the compiler to pass the Show[Int].
Person.PersonShow(si = implicitly, ss = Show.ShoutyStringShow).show(p)
Starting Scala 3 implicitly has been replaced with improved summon which has the advantage of being able to return a more precise type than asked for
The summon method corresponds to implicitly in Scala 2. It is
precisely the same as the the method in Shapeless. The difference
between summon (or the) and implicitly is that summon can return a
more precise type than the type that was asked for.
For example given the following type
trait F[In]:
type Out
def f(v: Int): Out
given F[Int] with
type Out = String
def f(v: Int): String = v.toString
implicitly method would summon a term with erased type member Out
scala> implicitly[F[Int]]
val res5: F[Int] = given_F_Int$#7d0e5fbb
scala> implicitly[res5.Out =:= String]
1 |implicitly[res5.Out =:= String]
| ^
| Cannot prove that res5.Out =:= String.
scala> val x: res5.Out = ""
1 |val x: res5.Out = ""
| ^^
| Found: ("" : String)
| Required: res5.Out
In order to recover the type member we would have to refer to it explicitly which defeats the purpose of having the type member instead of type parameter
scala> implicitly[F[Int] { type Out = String }]
val res6: F[Int]{Out = String} = given_F_Int$#7d0e5fbb
scala> implicitly[res6.Out =:= String]
val res7: res6.Out =:= String = generalized constraint
However summon defined as
def summon[T](using inline x: T): x.type = x
does not suffer from this problem
scala> summon[F[Int]]
val res8: given_F_Int.type = given_F_Int$#7d0e5fbb
scala> summon[res8.Out =:= String]
val res9: String =:= String = generalized constraint
scala> val x: res8.Out = ""
val x: res8.Out = ""
where we see type member type Out = String did not get erased even though we only asked for F[Int] and not F[Int] { type Out = String }. This can prove particularly relevant when chaining dependently typed functions:
The type summoned by implicitly has no Out type member. For this
reason, we should avoid implicitly when working with dependently typed
functions.
A "teach you to fish" answer is to use the alphabetic member index currently available in the Scaladoc nightlies. The letters (and the #, for non-alphabetic names) at the top of the package / class pane are links to the index for member names beginning with that letter (across all classes). If you choose I, e.g., you'll find the implicitly entry with one occurrence, in Predef, which you can visit from the link there.