Scala method types and methods as parameters
In the following code example, I do not understand why the function fun can be passed as an argument to the method addAction
. The method fun
is of t开发者_如何学Goype Unit
, while the method addAction
expects a function of type () => Unit
.
If fun
is of type () => Unit
, then why does the compiler complain that fun
is of type Unit
, when I try to add fun
to the actions list: actions = fun :: actions
?
package myscala
object MyScala {
def fun() { println("fun1 executed.") }
def addAction(a: () => Unit) {
actions = a :: actions
}
var actions: List[() => Unit] = List()
def main(args: Array[String]) {
// the following line would produce a compiler error (found: Unit, required: () => Unit), it's OK
// actions = fun :: actions
actions = (() => fun) :: actions // OK
// I would expect the same compiler error here (found: Unit, required: () => Unit), but it's OK why?
addAction(fun)
actions.foreach(_()) // prints twice "fun1 executed"
}
}
Take this as an introductory example:
def fun() { println("fun1 executed.") }
val a1 = fun
val a2: () => Unit = fun
Both lines compile and (thanks to type inference) they look equivalent. However a1
is of type Unit
while a2
is of type () => Unit
... How is this possible?
Since you are not explicitly providing type of a1
, compilers interprets fun
as a method fun
call of type Unit
, hence the type of a1
is the same as type of fun
. It also means that this line will print fun1 executed.
However, a2
has explicitly declared type of () => Unit
. The compiler helps you here and it understands that since the context requires a function of type () => Unit
and you provided a method matching this type, it shouldn't call that method, but treat it as first class function!
You are not doomed to specify type of a1
explicitly. Saying:
val a1 = fun _
Do you now understand where your problem is?
You need to write fun _
in the first case to avoid calling the method and performing eta-expansion instead.
This will work:
actions = (fun _) :: actions
If you don't do this, then fun
is evaluated.
For more details, see Section 6.7 (Method Values) of the Scala Language Reference.
As to why fun
is not evaluated in the second case, it is because type inference can clearly conclude that addAction
expects a function. By the way, the type of fun
is technically ()Unit
, not Unit
, that is, a method type, and not a value type. See Section 3.3.1 in the reference for more.
There is a difference between methods and functions. In your case actions
is a list of functions. When the compiler knows that a function is required (like in the case of addAction
) it can automatically convert a method fun
into a function. Now ::
is also a method, therefore the compiler also knows that it takes functions as parameters. But the problem is the syntactic sugar of the right-associative operator ::
. If you were to call it like a method: actions.::(fun)
it will compile (although I can't test it at the moment). When writing fun :: actions
the compiler thinks that fun
is an expression and therefore evaluates it and since it "returns" a Unit
you get your compiler error.
EDIT
Since I now have the possibility to test my hypothesis (which was wrong) here are your options:
// Usual syntax
actions.::[() => Unit](fun)
actions.::(fun: () => Unit)
actions.::(fun _)
// Operator syntax
(fun: () => Unit) :: actions
(fun _) :: actions
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