Suppose I have functions that takes 2 Ints and returns 1 Int:
val sub : (Int, Int) => Int = (x,y) => x - y
val mul : (Int, Int) => Int = (x,y) => x * y
How will I create a method that accepts the above type. I tried:
def test(f:(Int,Int) => Int) : Unit = {
val a = f
println(a)
}
Which does not seem to work.
I guess by "Does not seem to work.", you mean you're trying to call the function and expected an output in the println.
The function works as designed. It's just not the way you probably hoped it will work...
So this for instance will not work:
scala> test(sub(5, 3))
<console>:14: error: type mismatch;
found : Int
required: (Int, Int) => Int
test(sub(5, 3))
But you can modify test() slightly:
scala> def test(f:(Int,Int) => Int, x: Int, y: Int) : Unit = {
| val a = f(x, y)
| println(a)
| }
test: (f: (Int, Int) => Int, x: Int, y: Int)Unit
And call it with the params:
scala> test(sub, 5, 3)
2
scala> test(mul, 5, 3)
15
Related
I have this scala code:
object C extends App{
def sum(a:Int,b:Int,c:Int)=a+b+c
//val d=sum
val d=sum _
println(d(1,2,3))
}
I have to write sum _,not sum
but In f#,I can write this:
let sum a b c=a+b+c
let d=sum
[<EntryPoint>]
let main argv =
printfn "%A" (d 1 2 3)
0 // return an integer exit code
why in scala I can't just write
val d=sum
Scala has
methods (by default def is a method) and
functions
And they are not same.
Taking java methods as example, you can not assign methods to a variable without evaluating the method. But you can do that with function and to achieve that in scala define sum as function.
scala> def sum: (Int, Int, Int) => Int = (a, b, c) => a + b + c
sum: (Int, Int, Int) => Int
scala> val sumVal = sum
sumVal: (Int, Int, Int) => Int = $$Lambda$912/0x0000000801667840#716f94c1
scala> sumVal(1, 2, 3)
res1: Int = 6
Longer version of defining a function is,
scala> def sum = new Function3[Int, Int, Int, Int] {
| def apply(a: Int, b: Int, c: Int): Int = a +b + c
| }
sum: (Int, Int, Int) => Int
scala> val sumVal = sum
sumVal: (Int, Int, Int) => Int = <function3>
scala> sumVal(1, 2, 3)
res2: Int = 6
I am having trouble understanding why I am unable to use a composed function and compose a new one. For eg: I have two functions f and g and I create a composed function composed1 from these. I tried to combine composed with fourth function lastOne and it fails.
I want to make two composed functions and for the second one Is there a way I can reuse the first composed function?
scala> def f(x: Int)(y: Int) = {
| x + y
| }
f: (x: Int)(y: Int)Int
scala> def g(a: Int)(b: Int) = {
| a + b
| }
g: (a: Int)(b: Int)Int
scala> def composed1(a: Int, b: Int) = {
| f(a) _ andThen g(b)
| }
composed1: (a: Int, b: Int)Int => Int
scala> composed1(2, 2)(5)
res1: Int = 9
scala> def lastOne(l: Int)(x: Int) = {
| l + x
| }
lastOne: (l: Int)(x: Int)Int
scala> def composed2(a: Int, b: Int, c: Int) = {
| composed1(a, b) _ andThen lastOne(c)
| }
<console>:14: error: _ must follow method; cannot follow Int => Int
composed1(a, b) _ andThen lastOne(c)
^
<console>:14: error: missing argument list for method lastOne
Unapplied methods are only converted to functions when a function type is expected.
You can make this conversion explicit by writing `lastOne _` or `lastOne(_)(_)` instead of `lastOne`.
composed1(a, b) _ andThen lastOne(c)
When I use all of them together it works
scala> def test(x: Int, y: Int, z: Int) = {
| f(x) _ andThen g(y) _ andThen lastOne(z)
| }
test: (x: Int, y: Int, z: Int)Int => Int
scala> test(2, 2, 4)(5)
res9: Int = 13
f()() and g()(), as you define them, are methods. Methods are not functions but methods can be promoted to functions via "eta expansion". One way to do that is using the underscore in place of a passed parameter.
andThen() is a method on the Function trait that takes a function as an argument and returns a new function. It looks as if you can also use a method as the passed argument but it is silently being promoted to Function status.
So composed1() looks like a method but it is actually a Function, because that's what andThen() returns, and you can't apply the underscore eta expansion to a Function. It only works on methods.
As an experiment, turn f()() into a Function that does the same thing...
def f :Int => Int => Int = (x: Int) => (y: Int) => x + y
...now composed1() won't compile.
So, now that we know that composed1() is a Function, how do we get what we want from composed2()? Simple. Skip the underscore.
def composed2(a: Int, b: Int, c: Int) =
composed1(a, b) andThen lastOne(c)
composed2(2, 2, 4)(5) //res0: Int = 13
What is the logical reason that the first form works and not the second?
scala> val d = (a: Int, b: Int) => a + b
d: (Int, Int) => Int = <function2>
scala> val d = (a: Int)(b: Int) => a + b
<console>:1: error: not a legal formal parameter.
Note: Tuples cannot be directly destructured in method or function parameters.
Either create a single parameter accepting the Tuple1,
or consider a pattern matching anonymous function: `{ case (param1, param1) => ... }
val d=(a:Int)(b:Int)=>a+b
Because multiple parameter lists aren't allowed on function declarations. If you want to curry a function, you do:
scala> val d: Int => Int => Int = a => b => a + b
d: Int => (Int => Int) = $$Lambda$1106/512934838#6ef4cbe1
scala> val f = d(3)
f: Int => Int = $$Lambda$1109/1933965693#7e2c6702
scala> f(4)
res6: Int = 7
You can also create a single parameter list and partially apply it:
scala> val d = (a: Int, b: Int) => a + b
d: (Int, Int) => Int = $$Lambda$1064/586164630#7c8874ef
scala> d(4, _: Int)
res2: Int => Int = $$Lambda$1079/2135563436#4a1a412e
We partially applied d with 4, and we got back a function, Int => Int, which means when we supply the next argument, we'll get the result:
scala> res2(3)
res3: Int = 7
We can also create a named method, and use eta-expansion to create a curried function out of it:
scala> def add(i: Int)(j: Int): Int = i + j
add: (i: Int)(j: Int)Int
scala> val curriedAdd = add _
curriedAdd: Int => (Int => Int) = $$Lambda$1115/287609100#f849027
scala> val onlyOneArgumentLeft = curriedAdd(1)
onlyOneArgumentLeft: Int => Int = $$Lambda$1116/1700143613#77e9dca8
scala> onlyOneArgumentLeft(2)
res8: Int = 3
Function currying is possible.
val curryFunc = (a: Int) => (b: Int) => a + b
curryFunc now has the type Int => (Int => Int)
The following:
val add = (a: Int, b: Int) => a + b
gets converted to:
object add extends Function2[Int, Int, Int] {
def apply(a: Int, b: Int) = a + b
}
while
val a1 = add(_: Int, 3)
gets converted to:
object a1 extends Function1[Int, Int] {
def apply(x: Int): Int = {
add(x, 3)
}
}
But when I do:
scala> val a2 = add _
a2: () => (Int, Int) => Int = <function0>
And then call a2, it throws an error:
scala> a2(1, 2)
<console>:11: error: too many arguments for method apply: ()(Int, Int) => Int in trait Function0
a2(1, 2)
^
Why is this? Why does the following work?
a2()(1, 2)
add is already a Function2[Int, Int, Int]. If you want a2 to have the same type, then a simple assignment will suffice.
scala> val a2 = add
a2: (Int, Int) => Int = <function2>
scala> a2(1, 2)
res3: Int = 3
What you're thinking of is eta-expansion of a method into a function. If we had:
def add(a: Int, b: Int): Int = a + b
Then, we would use add _ to get the eta-expansion to assign to a value.
scala> def a2 = add _
a2: (Int, Int) => Int
scala> a2(1, 2)
res4: Int = 3
But add is already a function, so the underscore has a different meaning. add is now a value and not a method. Since add is a value, it is like a parameter-less method that returns a Function2[Int, Int, Int]. And if we try to get the eta-expansion of that, we get () => Function2[Int, Int, Int].
Consider a simpler example where we have a simple val a = 1. a is essentially the same as a parameter-less method that returns 1 (def a = 1). If I try to obtain the eta-expansion, I will get () => Int.
scala> val a = 1
a: Int = 1
scala> val a2 = a _
a2: () => Int = <function0>
Dеar Scala,
scala> val f1: ((Int, Int)) => Int = { case (a, b) => a + b }
f1: ((Int, Int)) => Int = <function1>
scala> val f2: (Int, Int) => Int = { case (a, b) => a + b }
f2: (Int, Int) => Int = <function2>
huh?!
scala> f1(1, 2)
res2: Int = 3
Ok...
scala> def takesIntInt2Int(fun: (Int, Int) => Int) = fun(100, 200)
takesIntInt2Int: (fun: (Int, Int) => Int)Int
scala> def takesTuple2Int(fun: ((Int, Int)) => Int) = fun(100, 200)
takesTuple2Int: (fun: ((Int, Int)) => Int)Int
scala> takesIntInt2Int(f2)
res4: Int = 300
scala> takesIntInt2Int(f1)
<console>:10: error: type mismatch;
found : ((Int, Int)) => Int
required: (Int, Int) => Int
takesIntInt2Int(f1)
^
scala> takesTuple2Int(f1)
res6: Int = 300
scala> takesTuple2Int(f2)
<console>:10: error: type mismatch;
found : (Int, Int) => Int
required: ((Int, Int)) => Int
takesTuple2Int(f2)
Right. And now, look at this!
scala> takesTuple2Int { case (a, b, c) => a + b + c }
<console>:9: error: constructor cannot be instantiated to expected type;
found : (T1, T2, T3)
required: (Int, Int)
takesTuple2Int { case (a, b, c) => a + b + c }
^
scala> takesIntInt2Int { case (a, b, c) => a + b + c }
<console>:9: error: constructor cannot be instantiated to expected type;
found : (T1, T2, T3)
required: (Int, Int)
takesIntInt2Int { case (a, b, c) => a + b + c }
Like, srsly? o_O Both result in required: (Int, Int) error.
Why then use case at all in such anonymous functions?
See section 8.5 of the Scala reference (http://www.scala-lang.org/files/archive/nightly/pdfs/ScalaReference.pdf). The expression { case (a, b) => a + b } is interpreted differently based on the expected type. In your definition of f1 it created a PartialFunction[(Int, Int), Int] which was cast to a Function1[(Int, Int), Int], i.e. ((Int, Int)) => Int whereas in the definition of f2 it created a Function2[Int, Int, Int], i.e. (Int, Int) => Int.
These two interpretations relate to the two situations where you would commonly use case in an anonymous function.
One is for writing anonymous functions that accept tuples and work on their components, as you did with f1. An example would be the function you pass to the foreach or map method on a Map, e.g. Map(1 -> 2, 3 -> 4) map { case (k, v) => k + v }.
The second is for writing an anonymous function that does a match on its sole parameter. Your f2 is doing this, but not in any useful way. An example would be the anonymous function passed to collect, e.g. List(1, -2, 3) collect { case x if x > 0 => -x }.
Note that the two can be combined, that is functions like f1 can do complex matching as well. For example, Map(1 -> 2, 3 -> 4) collect { case (k, v) if k < 2 => v }.
Edit: res2 works because of tupling. If an application doesn't type check, the compiler will try wrapping the args in a tuple before failing.
But that is tried just for applications; it's not a general conversion, as you discovered. It will not try to upgrade a value Function2[A, B, C] to Function1[(A, B), C].