In the OCaml repl if I run multiple phrases it will print the value of each.
# let add x y = x + y;;
val add : int -> int -> int = <fun>
# add 5 6;;
- : int = 11
However in the OCaml Playground it only prints the first one:
val add : int -> int -> int = <fun>
Why is that? Can I make it print everything?
Ah I figured it out. You need a newline after the last ;; otherwise it ignores it.
Related
So I need to write a function that will triple every number in a list of integers
Here is what I have so far:
let number = [1; 2; 3; 4];;
let rec print_list_int myList = match myList with
| [] -> print_endline "This is the end of the int list!"
| head::body ->
begin
print_int head * 3;
print_endline "";
print_list_int body *3
end
;;
print_list_int number;;
It doesn't seem to do anything useful, any ideas where I went wrong? Need it outputting but it also doesn't do that. Thanks in advance! :)
This expression:
print_int head * 3
is interpreted like this:
(print_int head) * 3
Because function calls (application) have high precedence. You need to parenthesize like this:
print_int (head * 3)
The similar case below is a different problem: (print_list_int body) * 3 doesn't make sense but print_list_int (body * 3) also doesn't make sense. You can't multiply a list by 3. However, you don't need to multiply at this call. The print_list_int function will (recursively) do the multiplying for you.
Update
If I make the changes I hinted at above, I see this in the OCaml toplevel:
val print_list_int : int list -> unit = <fun>
# print_list_int number;;
3
6
9
12
This is the end of the int list!
- : unit = ()
#
Note that the most elegant way to achieve what you're trying to do is to use List.iter. It applies a given function (returning unit) to each element of a List.
let print_triples = List.iter (fun x ->
print_endline (string_of_int (3*x))
);;
val print_triples : int list -> unit = <fun>
And here you go:
# print_triples [1;2;3;4;5];;
3
6
9
12
15
- : unit = ()
. Write a function that takes an integer list and return sum of all elements of the list. If the list is empty then return None.
This is my code now:
let rec sum (xs: int list) =
match xs with
| [] -> None
| [x] -> Some x
| hd::tl -> let m = (hd + (sum tl)) in
Some m
;;
The problem is that I can't seem to find a way to add up the last element without getting an error.
This is my error.
Error: This expression has type int but an expression was expected of type 'a option.
Your recursive call to sum does indeed return an int option. You know this because you're the author of the function, and you coded it up to return that type :-) You can either write a helper function that returns an int, or you can extract the int from the return value of sum, something like this:
let tlsum =
match sum tl with
| None -> (* figure this part out *)
| Some n -> (* figure this part out *)
You can define the addition of two int option.
let sum l =
let (+) a b =
match (a,b) with
| (None,x) | (x,None) -> x
| (Some x,Some y) -> Some (x+y)
in
let convert a = Some a in
let opt_l=List.map convert l in
List.fold_left (+) None opt_l
Test
# sum [];;
- : int option = None
# sum [1;2];;
- : int option = Some 3
That looks like an assignment so I'll be vague:
The easiest way to do that is probably to first define a function of type int list -> int that returns the "normal" sum (with 0 for the empty case). That function will be recursive and 0 will correspond to the base case.
Then write another function of type int list -> int option that checks whether its argument is empty or not and does the right thing based on that.
Trying to write the recursion directly probably is not a good idea since there are two cases when you will need to handle []: when it's the only element in the list, and when it's at the end of a nonempty list.
Can someone explain the syntax used for when you have nested functions?
For example I have a outer and an inner recursive function.
let rec func1 list = match list with
[] -> []
|(head::tail) ->
let rec func2 list2 = match list2 with
...
;;
I have spent all day trying to figure this out and I'm getting a ever tiring "Syntax error".
You don't show enough code for the error to be obvious.
Here is a working example:
# let f x =
let g y = y * 5 in
g (x + 1);;
val f : int -> int = <fun>
# f 14;;
- : int = 75
Update
Something that might help until you're used to OCaml syntax is to use lots of extra parentheses:
let rec f y x =
match x with
| h :: t -> (
let incr v = if h = y then 1 + v else v in
incr (f y t)
)
| _ -> (
0
)
It's particularly hard to nest one match inside another without doing this sort of thing. This may be your actual problem rather than nested functions.
Hello i resolved problem with ealier task.
Now if i have for example list = [ 2; 3; 2 ; 6 ] want to translate it like this [2;5;7;13].
I declared x as my first element and xs as my rest and used List.scan . Idea below
(fun x n -> x + n) 0
but this make something like this
val it : int list = [0; 2; 5; 7; 13]
How to rewrite it to make list looking like this [2;5;7;13] with using any starting parameter. When i delete 0 i get error message.
Another question how it's going to look like List.Fold i tried to write something similar but it can get only sum of this list ;( .
Here's how I would do this with a fold (with type annotations):
let orig = [2; 3; 2; 6]
let workingSum (origList:int list) : int list =
let foldFunc (listSoFar: int list) (item:int) : int list =
let nextValue =
match listSoFar with
| [] -> item
| head::_ -> head + item
nextValue::listSoFar
origList |> List.fold foldFunc [] |> List.rev
For help learning fold, here's how I would do this with a recursive function:
let workingSum' (origList: int list): int list =
let rec loop (listSoFar: int list) (origListRemaining:int list): int list =
match origListRemaining with
| [] -> listSoFar
| remainHead::remainTail ->
let nextValue =
match listSoFar with
| [] -> remainHead
| head::_ -> head + remainHead
loop (nextValue::listSoFar) remainTail
origList |> loop [] |> List.rev
Note that the signature of the inner loop function is really similar to the foldFunc of the previous example, with one major difference: instead of being passed in the next element, it's being passed in the remainder of the original list that hasn't been processed yet. I'm using a match expression to account for the two different possibilities of that remainder of the original list: either the list is empty (meaning we're done), or it's not (and we need to return a recursive call to the next step).
I'm trying to write a simple recursive function that look over list and return a pair of integer. This is easy to write in c/c++/java but i'm new to ocaml so somehow hard to find out the solution due to type conflict
it should goes like ..
let rec test p l = ... ;;
val separate : (’a -> bool) -> ’a list -> int * int = <fun>
test (fun x -> x mod 2 = 0) [-3; 5; 2; -6];;
- : int * int = (2, 2)
so the problem is how can i recursively return value on tuple ..
One problem here is that you are returning two different types: an int for an empty list, or a tuple otherwise. It needs to be one or the other.
Another problem is that you are trying to add 1 to test, but test is a function, not a value. You need to call test on something else for it to return a value, but even then it is supposed to return a tuple, which you can't add to an integer.
I can't figure out what you want the code to do, but if you update your question with that info I can help more.
One guess that I have is that you want to count the positive numbers in the list, in which case you could write it like this:
let rec test l =
match l with [] -> 0
| x::xs -> if x > 0 then 1 + (test xs)
else test xs;;
Update: since you've edited to clarify the problem, modify the above code as follows:
let test l =
let rec test_helper l pos nonpos =
match l with [] -> (pos, nonpos)
| x::xs -> if x > 0 then test_helper xs 1+pos, nonpos
else test_helper xs pos 1+nonpos
in test_helper l 0 0;;
Using the accumulators help a lot in this case. It also makes the function tail-recursive which is always good practice.
Been away from OCaml for a bit, but I think this will do the trick in regards to REALFREE's description in the comment
let rec test l =
match l with
[] -> (0,0)
| x::xs ->
if x > 0 then match (test xs) with (x,y) -> (x+1, y)
else match (test xs) with (x,y) -> (x, y+1);;
You can used the nested match statements to pull out pieces of the tuple to modify
EDIT:
I didn't know about the syntax Pascal Cuoq mentioned in his comment below, here's the code like that, it's neater and a little shorter:
let rec test l =
match l with
[] -> (0,0)
| x::xs ->
if x > 0 then let (x,y) = test xs in (x+1, y)
else let (x,y) = test xs in (x, y+1);;
But the accepted answer is still much better, especially with the tail recursion ;).