I have found currying nicer than partial application or implicits for a certain scenario. However I have not found a way to curry for an argument group that is the first in a list of argument groups.
E.g. take an original function:
def a(x: Int)(y: Int) = x + y
It is easy to fix the first argument a la
def b = a(3)_
But how can you fix the second one, and only the second one, instead?
Type ascription works in this case:
scala> def a(a: Int)(b: Int) = a + b
a: (a: Int)(b: Int)Int
scala> a(_: Int)(3)
res12: Int => Int = <function1>
scala> res12(2)
res13: Int = 5
You could use the ordinary function for that:
def b(x: Int) = a(x)(3)
Related
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
I read, that in Scala you can overload any operator, not as in Groovy. But I saw no any example how to overload function call.
Is it possible / how to overload function call operator in Scala?
I know that "there is no operators in Scala", but despite this, I need to do what I would name that.
function call is spelled apply. Any object which has a method named apply overloads function call.
class Functionish(val int: Int, val str: String) {
def apply(i: Int): Int = i + int
def apply(s: String): String = s + str + s
}
val f = new Functionish(42, "hello")
println(f(4))
println(f("George, "))
prints
46
George, helloGeorge,
Of note: case class companion objects overload apply. This is why you can create a case class instance with
val instance = CaseClass(foo, bar, baz)
instead of
val instance = new CaseClass(foo, bar, baz)
Every "callable" object is called via apply method:
scala> def foo(i: Int) = i * i
foo: (i: Int)Int
scala> val f = foo _
f: Int => Int = <function1>
scala> f(42)
res2: Int = 1764
scala> f.apply(42)
res3: Int = 1764
Is that what are you looking for?
I'm fairly new to Scala and functional programming in general so I'm having a bit trouble wrapping my head around the concept of partially applied functions and function currying. There's also a pretty high chance that I'm gonna mix up some terminology, so all corrections are appreciated.
Note: I'm using the Scala Play framework but this is more of a Scala problem than a Play problem.
Given a function like this
def create(id: Long, userId: Long, label: String)
I get the userId and label as a (Int, String) tuple and the id as a Long. Basically what I'm trying to do is passing the id and the tuple to the function at the same time.
Now, I've read that passing a tuple to a function can be done something like this
scala> def f(a: Int, b: String) = 0
f: (a: Int, b: String)Int
scala> (f _).tupled((2, "Hello"))
res0: Int = 0
and that it is also possible to partially apply a function like this
scala> def f(a: Int, b: String) = 0
f: (a: Int, b: String)Int
scala> val ff = f(_: Int, "Hello")
ff: Int => Int = <function1>
scala> ff(2)
res1: Int = 0
So my initial idea was to combine these two concepts something like this
scala> def g(a: Long, b: Int, c: String) = 0
g: (a: Long, b: Int, c: String)Int
scala> val h = g(1, _: Int, _: String)
h: (Int, String) => Int = <function2>
scala> (h _).tupled((2, "Hello"))
However this results in an error
<console>:10: error: _ must follow method; cannot follow h.type
(h _).tupled((1, "Hello"))
^
So my question is first of all why doesn't this work because to me this makes sense. And secondly how would I go about achieving this effect?
Thanks for your help!
Just don't abstract it twice: don't do redundand underscore to the h as it's already a function after partial-applying:
scala> def create(a: Long, b: Int, c: String) = 0
create: (a: Long, b: Int, c: String)Int
scala> val h = create(1, _: Int, _: String)
h: (Int, String) => Int = <function2>
scala> h.tupled((1, "Hello"))
res0: Int = 0
More deeply, tupled is defined on functions (means object of Function, like Int => String), not on methods (like def f(i: Int): String) - so sometimes you need to convert a method to the function - it's called eta-expansion (or eta-abstraction more generally). When you do partial applying - eta-expansion is done automatically (you already have a (Int, String) => Int) - so you don't have to do it twice.
See The differences between underscore usage in these scala's methods, Difference between method and function in Scala for more information.
val list = List((1,2), (3,4))
list.map(tuple => {
val (a, b) = tuple
do_something(a,b)
})
// the previous can be shortened as follows
list.map{ case(a, b) =>
do_something(a,b)
}
// similarly, how can I shorten this (and avoid declaring the 'tuple' variable)?
def f(tuple: (Int, Int)) {
val (a, b) = tuple
do_something(a,b)
}
// here there two ways, but still not very short,
// and I could avoid declaring the 'tuple' variable
def f(tuple: (Int, Int)) {
tuple match {
case (a, b) => do_something(a,b)
}
}
def f(tuple: (Int, Int)): Unit = tuple match {
case (a, b) => do_something(a,b)
}
Use tupled
scala> def doSomething = (a: Int, b: Int) => a + b
doSomething: (Int, Int) => Int
scala> doSomething.tupled((1, 2))
res0: Int = 3
scala> def f(tuple: (Int, Int)) = doSomething.tupled(tuple)
f: (tuple: (Int, Int))Int
scala> f((1,2))
res1: Int = 3
scala> f(1,2) // this is due to scala auto-tupling
res2: Int = 3
tupled is defined for every FunctionN with N >= 2, and returns a function expecting the parameters wrapped in a tuple.
While this might look like a trivial suggestion, the f function, can be further simplified by just using _1 and _2 on a tuple.
def f(tuple: (Int, Int)): Unit =
do_something(tuple._1, tuple._2)
Obviously by doing this you're affecting readability (some meta-information about the meaning of the 1st and 2nd parameter of the tuple is removed) and should you wish to use elements of the tuple somewhere else in the f method you will need to extract them again.
Though for many uses this might be still the easiest, shortest and most intuitive alternative.
If I understand correctly you are trying to pass a tuple to a method with 2 args?
def f(tuple: (Int,Int)) = do_something(tuple._1, tuple._2)
by more readable, I mean giving variable names instead of using the _1 an _2 on the tuple
In this case, it's a good idea to use a case class instead of a tuple, especially since it only takes one line:
case class IntPair(a: Int, b: Int)
def f(pair: IntPair) = do_something(pair.a, pair.b)
If you get (Int, Int) from external code which can't be changed (or you don't want to change), you could add a method converting from a tuple to IntPair.
Another option: {(a: Int, b: Int) => a + b}.tupled.apply(tuple). Unfortunately, {case (a: Int, b: Int) => a + b}.apply(tuple) doesn't work.
Let's define f, a function that supports currying :
def f(a: Int)(b: Int) = a + b
This code doesn't compile
def g= f(1)
<console>:10: error: missing arguments for method f;
follow this method with `_' if you want to treat it as a partially applied function
def g= f(1)
I've found these 2 workarounds :
scala> def h = f(1) _
h: Int => Int
scala> def i : Int => Int = f(1)
i: Int => Int
But i don't understand why the inference engine need help in a trivial case like that ?
This is because def f(a: Int)(b: Int) is not a function but a method with multiple parameter lists. Scala doesn't unify both of them to a single concept, thus you have to distinguish between them as well.
To partially apply a method with multiple parameter lists (and therefore implicitly convert this method to a function) you have to add an underscore after the method, as the error message suggests. When you tell the compiler explicitly which signature you want to have, it can also implicitly convert a method to a function (as seen in your second workaround).
If you want to use currying in Scala it's best to create a function from the beginning:
scala> val f = (a: Int) => (b: Int) => a+b
f: Int => (Int => Int) = <function1>
scala> val g = f(1)
g: Int => Int = <function1>