Addressing parameter function's parameter in a scala method

Question

It's probably obvious, but it's making me crazy and I think my code is being bloated as a result (missing the obvious, not the making me crazy part).

I have a method as follows:

def fun(f: Int => Boolean, y: Int): Int

This method is going to be called like this fun(*some anon func*, y) and should increase and return y + 1 when some anon function (applied to some param) is greater than 0 and y - 1 otherwise. How to define fun? I'm trying something like

def fun(f: Int => Boolean, y: Int): Int = 
  if (f) y - 1
  else y + 1

Obviously, that doesn't work. I should put f(tmp), where tmp:Int but I'm now sure how to express that.

All examples I could find online always apply f on y, but this is just funny.

Update: here's a similar problem I have already solved, but am not happy with how did I do it:

For a given function forall(s: Set, p: Int => Boolean): Boolean, which returns true iff all the set elements x in S (def S: int => Boolean, indicator function for set S) satisfy p(x), create a function that will implement exists quantifier (exists function). In set theory, one can express this quantifier as (in terms of the problem described above) not all elements (not forall) of a set satisfy NOT p.

I did something like this:

def exists(s: Set, p: Int => Boolean): Boolean = {
  def negP(x: Int): Boolean = !p(x)
  !forall(s, negP)
}

My question is: how to do this without defining negP? I cannot state !p because Scala gives error for ! operator. I cannot use !p(x) because there is no x. Hence the problem above and my question.

TIA

Regarding anon function is greater than 0 -- f: Int => Boolean cannot be compared to 0: Int, so probably you should clarify your goals. — dveim
– dveim, Commented Nov 25, 2017 at 23:12
Your function f : what is the int parameter you want to call it with? If you already provide the value then it should not take a parameter — Ossip
– Ossip, Commented Nov 25, 2017 at 23:14
@dveim modified the question a bit. essentially, if I pass (x: Int) => x * x, then y would always go up, no matter what. However, when I pass (x: Int) => x - 2, it will get different results for x <= 2 and for x > 2. — hummingBird
– hummingBird, Commented Nov 25, 2017 at 23:24
It feels like you don't know what you want to do rather than not knowing how to do it — Dici
– Dici, Commented Nov 25, 2017 at 23:48

Raudbjorn · Accepted Answer · 2017-11-28 23:15:30Z

1

You could make it return a function that does that, like so:

def fun(f: Int => Int, y: Int): Int => Int = 
input => if (f(input) > 0) y - 1 else y + 1


  val test1 = fun(x => x*2, 1)
  println(test1(-1))  // "2"
  val test2 = fun(x => x - 100, 5)
  println(test2(101)) // "4"
  val test3 = fun(x => x, -10)
  println(test3(0))  // "-9"

EDIT: Note that I changed the signature of the input function, as I think it probably reflects the requirements better, changing it back to boolean should be trivial(see other answers). :)

EDIT(2): Seeing is you updated your question, I thought best to update my answer: You can always just inline a function, in your example, maybe you could do it like this:

def exists(s: Set, p: Int => Boolean): Boolean = !forall(s, x => !p(x))

edited Nov 28, 2017 at 23:15

answered Nov 25, 2017 at 23:53

Raudbjorn

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6 Comments

Dici Over a year ago

Although the OP didn't make what they wanted very clear, it seems like it could be the right answer

hummingBird Over a year ago

@Dici, you made it clear - you hate when people can't explain very well. What I did was giving an example and did my best to explain. Learning scala for the first time and also functional programming, certain concepts are giving me a headache since I don't know how to approach the problem. And to help - No, this doesn't resolve my answer. Thx for trying

hummingBird Over a year ago

@raudbjorn, unfortunately, this doesn't help because function signature has changed. Thx and sorry for issues in explaining. If it helps, I have Dici complaining about my issues in explaining to share some pain around :)

Dici Over a year ago

@hummingBird hum ? Hope it didn't come as aggressive, I wasn't particularly criticizing your question, just saying I was not sure what the exact requirements were although this answer seemed to meet them. Don't take things personally dude :p

hummingBird Over a year ago

@Dici i didn't in your first comment. i actually supported... just kinda minded when u started answering around to other ppl :P

|

jwvh · Accepted Answer · 2017-11-28 23:22:57Z

1

Your original problem can't be solved because you can't test the output of f() without invoking f(), and you can't invoke f() without an input argument to pass. I think the best you can hope for is something like this.

def fun(f: Int => Boolean, y: Int): Int => Int = (arg: Int) =>
  if (f(arg)) y - 1
  else        y + 1

Your "similar problem" looks to me like it should be a different question.

Your proposed solution can be solved via recursion, but using forall() it can be expressed more concisely.

def exists(s: Set[Int], p: Int => Boolean): Boolean =
  !forall(s, (x:Int) => !p(x))

edited Nov 28, 2017 at 23:22

answered Nov 28, 2017 at 22:34

jwvh

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2 Comments

hummingBird Over a year ago

Thx for you reply. It doesn't compile because I didn't expose foreach code. Essentially, that code uses recursion to iterate through all possible options and it was my goal - and task's goal to use foreach and some rules of first order logic to achieve the same goal. Just imagine that foreach is defined in a library and that it goes through all of the set elements and say if all of them satisfy the function p: Int => Boolean. Then exists can simply be resolved as NOT (FOREACH (P(X) == FALSE)) - not all of the elements give p(x) false

hummingBird Over a year ago

I agree ... I put it wrong and your updated answer does contain the solution I was looking for. I already accepted the answer though .. although both solutions give the same answer pretty much.

Ossip · Accepted Answer · 2017-11-25 23:18:45Z

0

It sounds like you don't actually want to pass a parameter to your anonymous function? Then try this:

def fun(f: () => Boolean, y: Int): Int = 
  if (f()) y - 1
  else y + 1

answered Nov 25, 2017 at 23:18

Ossip

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