There are some discussions here about this, but I have some specific questions I wasn't able to find an answer for. So, by call-by-name, I mean =>T type, and by 0-arity function I mean () => T
I understand (I think) the conceptual difference, but probably I am missing something as I still have lots of questions:
Why do we have the concept of =>T at all if we could always use () => T?
Is there any syntax/functional limitations of each? For now I found only that => cannnot be used as a class field. Is this the only limitation?
Is the generated code always the same for both?
Should I always prefer =>T? And why?
Is it correct to call =>T a type? It looks for me like it does not have any type representation in scala.
1) It's just more handy to use it, especially inside DSLs:
def printAndGet[T](f: => T) = {
val res = f
println(res + " printed")
res
}
scala> :paste
// Entering paste mode (ctrl-D to finish)
val k = printAndGet {
val a = 5
5 * a
}
// Exiting paste mode, now interpreting.
25 printed
k: Int = 25
2) => T can only be a parameter of method or function. And actually => T and () => T aren't interchangable:
scala> def aaa(f: => String) = f
aaa: (f: => String)String
scala> val a: Function1[() => String, String] = aaa _
<console>:8: error: type mismatch;
found : (=> String) => String
required: (() => String) => String
val a: Function1[() => String, String] = aaa _
^
Thanks to #som-snytt, fоund this one:
scala> object O { def f(i: Int) = i; def f(i: => Int) = i + 1 }
defined object O
scala> O.f(5)
res12: Int = 5
scala> O.f(5: (=> Int))
<console>:1: error: no by-name parameter type allowed here
O.f(5: (=> Int))
^
Even this which should work if it compiles - but it doesn't (scala 2.11.2, 2.11.5 REPL just crashes):
scala> val k: (=> Int) => Int = O.f _
k: (=> Int) => Int = <function1>
scala> k(5) //should be 6
res18: Int = 5 //WTF?
Last one seems like a bug
3) Not exactly, if you want the same, just convert => T into () => T:
scala> def aaa(f: => String) = {f _}
aaa: (f: => String)() => String
Bytecode may also differ. For instance, compiler will more likely inline code from => T without generating lambda for it. So, the key difference is that () => T is actually an object (first class citizen), => T isn't.
4) see 1, but sometimes you may need to ensure that user knows that computation might be delayed - () => T is better then.
5) It's part of a type signature, just look at eta-expansion of:
scala> def aaa(f: => String) = {f}
aaa: (f: => String)String
scala> aaa _ //convert method into a function
res7: (=> String) => String = <function1>
scala> val a: ( => String) => String = aaa _
a: (=> String) => String = <function1>
However scala doesn't recognize it as independent type:
scala> val a: Function1[( => String), String] = aaa _
<console>:1: error: no by-name parameter type allowed here
val a: Function1[( => String), String] = aaa _
^
Related
def readf(value_name: String, operator:operand, compared_value:Int f:()=>Unit) {
I want to look up an Int in a HashMap based on the value_name String passed in. Taking that value I want to compare it
value operator compared_value
if that statement is true I want to execute f.
How do I pass in a comparator and evaluate this logic? Preferably without writing my own classes.
The result from the HashMap can be as an Option which you can then filter with the operand/compared_value. Then it should be just a map on the resulting option. It would look something like:
yourMap.get(value_name).filter(_ operator compared_value).map(f)
This does assume that operator is (Int, Int) => Boolean as asked in the comment
It's kind of hard to help with the editorial issues your question has, but I'll try. Assuming you have a myMap: Map[String, Int] in scope (which your question does not show) and assuming your operator is really of type (Int, Int) => Boolean, you could do:
scala> val myMap = Map("test" -> 8)
myMap: scala.collection.immutable.Map[String,Int] = Map(test -> 8)
scala> :paste
// Entering paste mode (ctrl-D to finish)
def readf(value_name: String, operator: (Int, Int) => Boolean, compared_value:Int, f:()=>Unit) {
myMap.get(value_name).filter(operator(_, compared_value)).foreach(_ => f())
}
// Exiting paste mode, now interpreting.
readf: (value_name: String, operator: Int => (Int => Boolean), compared_value: Int, f: () => Unit)Unit
scala> val eqComparator = ((_: Int) == (_: Int))
eqComparator: (Int, Int) => Boolean = <function2>
scala> val f = () => println("success")
f: () => Unit = <function0>
scala> readf("test", eqComparator, 8, f)
success
scala> readf("test", eqComparator, 5, f)
scala> readf("other", eqComparator, 8, f)
(Edited to remove the curried operator. It's probably clearer this way.)
Is it possible to do something like the following?
def takeCurriedFnAsArg(f: (Int)(implicit MyClass) => Result)
Yes, it is possible.
When you have the second curried parameter marked as implicit, the function seems to be not of type
Int => (MyClass => Result) => ResultOfFunction
which it would be if the curried higher order function parameter was a regular parameter; instead, it looks like this:
Int => ResultOfFunction
Here's a quick example:
scala> def curriedFn(i : Int)(implicit func : String => Int) : Boolean = (i + func("test!")) % 2 == 0
curriedFn: (i: Int)(implicit func: String => Int)Boolean
scala> implicit val fn : String => Int = s => s.length
fn: String => Int = <function1>
scala> curriedFn _
res4: Int => Boolean = <function1>
As you can see, the implicit parameter got 'eliminated'. Why and how? That's a question for someone more knowledgeable than me. If I had to guess, I'd say the compiler directly substitutes the parameter with the implicit value, but that might very well be false.
Anyway, digressions aside, here's an example very relevant to your situation:
scala> def foo(func : Int => Boolean) = if(func(3)) "True!" else "False!"
foo: (func: Int => Boolean)String
scala> foo(curriedFn)
res2: String = True!
Now if the second function parameter wasn't implicit:
scala> def curriedNonImplicit(i : Int)(fn : String => Int) : Boolean = (i + fn("test!")) % 2 == 0
curriedNonImplicit: (i: Int)(fn: String => Int)Boolean
scala> curriedNonImplicit _
res5: Int => ((String => Int) => Boolean) = <function1>
As you can see, the type of the function is a bit different. That means that the solution will look different too:
scala> def baz(func : Int => (String => Int) => Boolean) = if(func(3)(s => s.length)) "True!" else "False!"
baz: (func: Int => ((String => Int) => Boolean))String
scala> baz(curriedNonImplicit)
res6: String = True!
You have to specify the function directly inside the method, as it wasn't implicitly provided before.
I am trying to define a Map literal with key: String, value: (Any)=>String. I tried the following, but get a syntax error:
def foo(x: Int): String = /...
def bar(x: Boolean): String = /...
val m = Map[String, (Any) => String]("hello" -> foo, "goodbye" -> bar)
Funny that no one actually gave a type that would work. Here's one such:
def foo(x: Int): String = x.toString
def bar(x: Boolean): String = x.toString
val m = Map[String, (Nothing) => String]("hello" -> foo, "goodbye" -> bar)
The reason why it works this way is because Function1 is contra-variant on the input, so (Nothing) => String is a superclass of (Int) => String. It is also co-variant on the output, so (Nothing) => Any would be a superclass to any other Function1.
Of course, you can't use it like that. Without manifests, you can't even uncover what the original type of Function1 is. You could try something like this, though:
def f[T : Manifest](v: T) = v -> manifest[T]
val m = Map[String, ((Nothing) => String, Manifest[_])]("hello" -> f(foo), "goodbye" -> f(bar))
val IntManifest = manifest[Int]
val BooleanManifest = manifest[Boolean]
val StringManifest = manifest[String]
m("hello")._2.typeArguments match {
case List(IntManifest, StringManifest) =>
m("hello")._1.asInstanceOf[(Int) => String](5)
case List(BooleanManifest, StringManifest) =>
m("hello")._1.asInstanceOf[(Boolean) => String](true)
case _ => "Unknown function type"
}
Int => String is not a subclass of Any => String, rather, the contrary. You can't put (replace) an Int => String function when a code expects Any => String, since that code can apply the function with, say, "hi".
#Ben suggestion works, but how is it useful? you can't invoke the function once you get it from the Map.
If you really want to do this, maybe define foo as a partial function:
val foo: PartialFunction[Any, String] = {case i: Int => ....}
Obviously, this will fail at runtime if you pass it a string, but you can always test if the function is ok for use with your parameter by using isDefinedAt. (another alternative may be manifests, but I don't see the value here)
If I let the compiler infer it I seem to get an illegal type:
scala> val m = Map("hello" -> foo _, "goodbye" -> bar _)
m: scala.collection.immutable.Map[java.lang.String,(Boolean with Int) => String] =
Map((hello,<function1>), (goodbye,<function1>))
scala> m("hello")(8)
<console>:9: error: type mismatch;
found : Int(8)
required: Boolean with Int
m("hello")(8)
scala> var q = new Boolean with Int
<console>:5: error: illegal inheritance from final class Boolean
var q = new Boolean with Int
Anyway, what you want is not the type Any but a generic of "any type" which is _:
scala> val mm = Map[String, (_) => String]("hello" -> foo _, "goodbye" -> bar _)
mm: scala.collection.immutable.Map[String,Function1[_, String]] =
Map((hello,<function1>), (goodbye,<function1>))
I just posted a question about how to invoke such functions because I don't actually know.
Trait Function1 is contravariant for parameter, so def foo(x: Int): String is not a (Any) => String. So the following would work:
scala> def baz(x: Any): String = "baz"
baz: (x: Any)String
scala> val m2 = Map[String, (String) => String]("hello" -> baz)
m2: scala.collection.immutable.Map[String,(String) => String] = Map((hello,<function1>))
This is how I did it to fulfill a similar requirement.
object MapToMethods {
private def increment(x: Int): Int = x+1
private def decrement(x: Int): Int = x-1
val m: Map[String, Int => Int] =Map("increment" -> increment, "decrement" ->decrement)
println(m("increment")(2)) //prints 3
println(m("decrement")(3)) //prints 2
}
I know this question has come up many times in different ways. But it is still not clear to me. Is there a way to achieve the following.
def foo(a:Int, b:Int) = {}
foo(a,b) //right way to invoke foo
foo(getParams) // is there a way to get this working without explicitly unpacking the tuple??
def getParams = {
//Some calculations
(a,b) //where a & b are Int
}
It's a two step procedure. First turn foo into a function, then call tupled on it to make it a function of a tuple.
(foo _).tupled(getParams)
#dave-griffith is dead on.
You can also call:
Function.tupled(foo _)
If you want to wander into "way more information than I asked for" territory, there are also methods built into partially applied functions (and on Function) for currying. A few input/output examples:
scala> def foo(x: Int, y: Double) = x * y
foo: (x: Int,y: Double)Double
scala> foo _
res0: (Int, Double) => Double = <function2>
scala> foo _ tupled
res1: ((Int, Double)) => Double = <function1>
scala> foo _ curried
res2: (Int) => (Double) => Double = <function1>
scala> Function.tupled(foo _)
res3: ((Int, Double)) => Double = <function1>
// Function.curried is deprecated
scala> Function.curried(foo _)
warning: there were deprecation warnings; re-run with -deprecation for details
res6: (Int) => (Double) => Double = <function1>
Wherein the curried version is invoked with multiple argument lists:
scala> val c = foo _ curried
c: (Int) => (Double) => Double = <function1>
scala> c(5)
res13: (Double) => Double = <function1>
scala> c(5)(10)
res14: Double = 50.0
Finally, you can also uncurry/untuple if needed. Function has builtins for this:
scala> val f = foo _ tupled
f: ((Int, Double)) => Double = <function1>
scala> val c = foo _ curried
c: (Int) => (Double) => Double = <function1>
scala> Function.uncurried(c)
res9: (Int, Double) => Double = <function2>
scala> Function.untupled(f)
res12: (Int, Double) => Double = <function2>
Function.tupled(foo _)(getParams) or the one suggested by Dave.
EDIT:
To respond to your comment:
What if foo happens to be the
constructor of some class?
In that case, this trick won't work.
You can write a factory method in the companion object of your class and then obtain the tupled version of its apply method using one of the aforementioned techniques.
scala> class Person(firstName: String, lastName: String) {
| override def toString = firstName + " " + lastName
| }
defined class Person
scala> object Person {
| def apply(firstName: String, lastName: String) = new Person(firstName, lastName)
| }
defined module Person
scala> (Person.apply _).tupled(("Rahul", "G"))
res17: Person = Rahul G
With case classes you get a companion object with an apply method for free, and thus this technique is more convenient to use with case classes.
scala> case class Person(firstName: String, lastName: String)
defined class Person
scala> Person.tupled(("Rahul", "G"))
res18: Person = Person(Rahul,G)
I know that's a lot of code duplication but alas... we don't have macros (yet)! ;)
I appreciate some of the other answers which were closer to what you asked for, but I found it easier for a current project to add another function which converts tuple parameters into the split parameters:
def originalFunc(a: A, b: B): C = ...
def wrapperFunc(ab: (A, B)): C = (originalFunc _).tupled(ab)
Now, you can implement foo and make it take a param of the Tuple2 class like so.
def foo(t: Tuple2[Int, Int]) = {
println("Hello " + t._1 + t._2)
"Makes no sense but ok!"
}
def getParams = {
//Some calculations
val a = 1;
val b = 2;
(a, b) //where a & b are Int
}
// So you can do this!
foo(getParams)
// With that said, you can also do this!
foo(1, 3)
Using XScalaWT, this compiled under Scala 2.7:
class NodeView(parent: Composite) extends Composite(parent) {
var nodeName: Label = null
this.contains(
label(
nodeName = _
)
)
}
With 2.8.0 RC1, I get this error:
type mismatch; found : main.scala.NodeView required: org.eclipse.swt.widgets.Label
The types are:
label(setups: (Label => Unit)*)(parent: Composite) : Label
contains(setups: (W => Unit)*) : W
So it looks like _ now binds to the outer function instead of inner.
Is this change intentional?
UPDATE: Here is a minimized example:
Scala 2.7.7:
scala> var i = 0
i: Int = 0
scala> def conv(f: Int => Unit) = if (_:Boolean) f(1) else f(0)
conv: ((Int) => Unit)(Boolean) => Unit
scala> def foo(g: Boolean => Unit) { g(true) }
foo: ((Boolean) => Unit)Unit
scala> foo(conv(i = _))
scala> i
res4: Int = 1
Scala 2.8.0RC3:
scala> var i = 0
i: Int = 0
scala> def conv(f: Int => Unit) = if (_:Boolean) f(1) else f(0)
conv: (f: (Int) => Unit)(Boolean) => Unit
scala> def foo(g: Boolean => Unit) { g(true) }
foo: (g: (Boolean) => Unit)Unit
scala> foo(conv(i = _))
<console>:9: error: type mismatch;
found : Boolean
required: Int
foo(conv(i = _))
^
scala> foo(conv(j => i = j))
scala> i
res3: Int = 1
Interestingly enough, this works:
scala> foo(conv(println _))
1
Here is the answer I got from Lukas Rytz on the scala-user list:
Hi Alexey,
there has been a change in semantics
when we introduced named arguments. An
expression
foo(a = _)
used to be parsed as follows:
foo(x => a = x)
In 2.8, "a" is treated as a named
argument, i.e. the expression is
parsed as:
x => foo(a = x)
I will add a migration warning for
this change.
I noticed the same thing and asked on Dave's blog:
http://www.coconut-palm-software.com/the_new_visual_editor/doku.php?id=blog:simplifying_swt_with_scala#comment__930ba2f0a020203873d33decce01ebc2
No answer there yet though. Just as you're saying, it seems like the _ binds to the outer closure rather than the inner one.
Changing
nodeName = _
to
x => nodeName = x
fixes the problem.