I want to write a function that taking a string and return a list of char. Here is a function, but I think it is not do what I want ( I want to take a string and return a list of characters).
let rec string_to_char_list s =
match s with
| "" -> []
| n -> string_to_char_list n
Aside, but very important:
Your code is obviously wrong because you have a recursive call for which all the parameters are the exact same one you got in. It is going to induce an infinite sequence of calls with the same values in, thus looping forever (a stack overflow won't happen in tail-rec position).
The code that does what you want would be:
let explode s =
let rec exp i l =
if i < 0 then l else exp (i - 1) (s.[i] :: l) in
exp (String.length s - 1) []
Source:
http://caml.inria.fr/pub/old_caml_site/FAQ/FAQ_EXPERT-eng.html#strings
Alternatively, you can choose to use a library: batteries String.to_list or extlib String.explode
Try this:
let explode s = List.init (String.length s) (String.get s)
Nice and simple:
let rec list_car ch =
match ch with
| "" -> []
| ch -> String.get ch 0 :: list_car (String.sub ch 1 (String.length ch - 1));;
How about something like this:
let string_to_list str =
let rec loop i limit =
if i = limit then []
else (String.get str i) :: (loop (i + 1) limit)
in
loop 0 (String.length str);;
let list_to_string s =
let rec loop s n =
match s with
[] -> String.make n '?'
| car :: cdr ->
let result = loop cdr (n + 1) in
String.set result n car;
result
in
loop s 0;;
As of OCaml 4.07 (released 2018), this can be straightforwardly accomplished with sequences.
let string_to_char_list s =
s |> String.to_seq |> List.of_seq
Here is an Iterative version to get a char list from a string:
let string_to_list s =
let l = ref [] in
for i = 0 to String.length s - 1 do
l := (!l) # [s.[i]]
done;
!l;;
My code, suitable for modern OCaml:
let charlist_of_string s =
let rec trav l i =
if i = l then [] else s.[i]::trav l (i+1)
in
trav (String.length s) 0;;
let rec string_of_charlist l =
match l with
[] -> ""
| h::t -> String.make 1 h ^ string_of_charlist t;;
Related
I am trying to run the following code on a coding question website and it says there is a runtime error, but running it on the top-level ocaml seems to work fine. Could there be any source of error in the code? Thanks in advance
The question is to find the number of 'good segments' within the given list and a specific number. A good segment is defined as follows:
A and B are positive integers such that A < B.
x that satisfies A <= x <= B is not an element of the given list.
The following are the inputs.
n, which is the number of elements in the list that will be given.
a, b, c, ... which are the elements of the list.
t, which is the number that must be included in the segment.
The output should be a single number printed out.
Edited Code:
let rec drop_value l to_drop =
match l with
| [] -> []
| hd :: tl ->
let new_tl = drop_value tl to_drop in
if hd = to_drop then new_tl else hd :: new_tl
;;
let rec find_start li t cur_min =
match li with
| [] -> cur_min
| hd :: tl -> let new_min = abs (t - hd) in
if new_min = 0 then find_start tl t new_min
else if new_min < cur_min && t > hd then find_start tl t new_min
else find_start tl t cur_min
;;
let rec find_end li t cur_min =
match li with
| [] -> cur_min
| hd :: tl -> let new_min = abs (t - hd) in
if new_min = 0 then find_end tl t new_min
else if new_min < cur_min && t < hd then find_end tl t new_min
else find_end tl t cur_min
;;
let rec contains_value l value =
match l with
| [] -> false
| hd :: tl -> if hd = value then true else contains_value tl value
;;
let nums = ref [];;
let n = read_int () in
for i = 1 to n do
Scanf.scanf " %d" (fun a ->
nums := a :: !nums)
done;
Scanf.scanf " %d" (fun t ->
if contains_value !nums t then print_int 0
else let start = if List.length !nums = 1 then 1 else abs (find_start !nums t 1001 - t) in
let finish = find_end (drop_value !nums start) t 1001 + t in
if t > start && t < finish then (if start = 1 && List.length ! nums = 1 then print_int ((t - start + 1) * (finish - t) - 1) else print_int ((t - start) * (finish - t) - 1))
else let start = 1 in print_int ((t - start + 1) * (finish - t) - 1))
;;
eg.
5
4 8 13 24 30
10
should give
5
=> [9, 10], [9, 11], [9, 12], [10, 11], [10, 12]
You don't describe the exact input format that your code is going to get. This makes it pretty much impossible to debug your code.
When I compile and run your code (as m.ml) using the input you describe I see this:
$ ./m
5 4 8 13 24 30 10
Fatal error: exception Failure("int_of_string")
In fact no matter what format I try for the input I get the same result.
So that is probably what is happening at the website.
In my experience it always causes more harm than good to use scanf. Combining it with other input functions is probably going to make things worse.
If you describe the expected format of the input carefully, somebody on StackOverflow can recommend a way to get your numbers.
In the meantime here's a way to read all the numbers on one line:
let rec split_at list n =
if n = 0 then
([], list)
else
match list with
| [] -> ([], [])
| h :: t ->
let (a, b) = split_at t (n - 1) in (h :: a, b)
in
let (nums, t) =
let line = read_line () in
let nstrs = Str.split (Str.regexp "[ \t][ \t]*") line in
match List.map int_of_string nstrs with
| [] -> failwith "no numbers"
| n :: rest ->
if List.length rest <> n + 1 then
failwith "bad count"
else
let (nums, tlist) = split_at rest n in
(nums, List.hd tlist)
in
. . .
let count (l: char list) : bool =
let cnt = 0 in
let rec check l =
match l with
| [] -> false
| h::t -> if h = 'a' then
let cnt + 1
check t
else check t
in check []
;;
what causes the syntax error?
I want to add 1 if 'a' exists and -1 if 'b' exists
You should write it like this :
let rec count l =
match l with
| [] -> 0
| c::t -> if c = 'a' then 1 + count t else count t
But this isn't tail recursive. To make it tail recursive, you'll have to add an accumulator and write an auxiliary function:
let count l =
let rec aux cnt l =
match l with
| [] -> cnt
| c::t -> if c = 'a' then aux (cnt + 1) t else aux cnt t
in
aux 0 l
Now this is tail recursive but a little bit long to write, so you could just use List.fold_left and still have a tail recursive function:
let count l =
List.fold_left (fun cnt c -> if c = 'a' then cnt + 1 else cnt) 0 l
OCaml is a functional language, variables are immutable (their content might be mutable, but that is not the default).
Consequently,
let cnt + 1
is both a syntax error and a logical error: it is not possible to update cnt in such way.
As a first step, I would suggest to write a recursive version of count (and maybe rename it to exists_a) without using ifs :
let rec count l = match l with
| [] -> ...
| 'a' :: t -> ...
| _ :: t -> ...
I'm reading an input file of several lines. Each line has the following format:
Greeting "hello"
Greeting " Good morning"
Sit
Smile
Question "How are you?"
My current can read each line into a string list. Then I process it using this function which is supposed to break it into a string list list:
let rec process (l : string list) (acc : string list list) : string list list =
match l with
| [] -> acc
| hd :: tl -> String.split_on_char ' ' hd :: (process tl acc)
Which, unfortunately, does not work, since it also splits spaces inside quotation marks. Anyone think of a the right way to do this, possibly using map or fold_left, etc? This would be my expected output:
[["Greeting"; "/"hello/""];[Greeting; "/" Good morning"];["Sit"]]
and so on. Thank you!
You want a real (but very simple) lexical analysis. IMHO this is beyond what you can do with simple string splitting.
A scanner takes a stream of characters and returns the next token it sees. You can make a string into a stream by having an index that traverses the string.
Here is a scanner that is roughly what you would want:
let rec scan s offset =
let slen = String.length s in
if offset >= slen then
None
else if s.[offset] = ' ' then
scan s (offset + 1)
else if s.[offset] = '"' then
let rec qlook loff =
if loff >= slen then
(* Unterminated quotation *)
let tok = String.sub s offset (slen - offset) in
Some (tok, slen)
else if s.[loff] = '"' then
let tok = String.sub s offset (loff - offset + 1) in
Some (tok, loff + 1)
else qlook (loff + 1)
in
qlook (offset + 1)
else
let rec wlook loff =
if loff >= slen then
let tok = String.sub s offset (slen - offset) in
Some (tok, slen)
else if s.[loff] = ' ' || s.[loff] = '"' then
let tok = String.sub s offset (loff - offset) in
Some (tok, loff)
else
wlook (loff + 1)
in
wlook (offset + 1)
It handles a few cases that you didn't specify: what to do if there is an unclosed quotation. What to do with something like abc"def ghi".
The scanner returns None at the end of the string, or Some (token, offset), i.e., the next token and the offset to continue scanning.
A recursive function to break up a string would look something like this:
let split s =
let rec isplit accum offset =
match scan s offset with
| None -> List.rev accum
| Some (tok, offset') -> isplit (tok :: accum) offset'
in
isplit [] 0
This can be visualized with a state machine. You have 2 main states: looking for ' ' and looking for '"'. Processing strings is ugly and you can't pattern match it. So first thing I did is turn the string into a char list. Implementing the two states then becomes simple:
let split s =
let rec split_space acc word = function
| [] -> List.rev (List.rev word::acc)
| ' '::xs -> split_space (List.rev word::acc) [] xs
| '"'::xs -> find_quote acc ('"'::word) xs
| x::xs -> split_space acc (x::word) xs
and find_quote acc word = function
| [] -> List.rev (List.rev word::acc)
| '"'::xs -> split_space acc ('"'::word) xs
| x::xs -> find_quote acc (x::word) xs
in
split_space [] [] s
;;
# split ['a';'b';' ';'"';'c';' ';'d';'"';' ';'e'];;
- : char list list = [['a'; 'b']; ['"'; 'c'; ' '; 'd'; '"']; ['e']]
Now if you want to do it with strings that's left to you. The Idea would be the same. Or you can just turn the char list list into a string list at the end.
This is the code I have to make a palindrome function. I already created the listReverse and explode function before that I use to make the palindrome. Can someone help me finnish the palindrome function?
let rec listReverse l = match l with
|[] -> []
|head :: tail -> (listReverse tail) # [head]
(* explode : string -> char list
* (explode s) is the list of characters in the string s in the order in
* which they appear
* e.g. (explode "Hello") is ['H';'e';'l';'l';'o']
*)
let explode s =
let rec _exp i =
if i >= String.length s then [] else (s.[i])::(_exp (i+1)) in
_exp 0
let rec palindrome w =
let a = explode w in
let b = listReverse a in
if c :: d
else false
You should use the List.rev standard function to reverse lists. Ocaml being a free software, you should look at its implementation (file stdlib/list.ml)
Try to explain in plain English (not code) what you are trying to achieve when you write
if c :: d
else false
Also, note that
if foo = bar then true else false
should be simplified to
foo = bar
You can replace your if statement with this:
(* tells wheter its a palindrome or not; most is (List.length a)/2*)
let rec same l1 l2 cur most =
match l1, l2 with
| h1::t1, h2::t2 when h1 = h2 ->
if cur < most then same t1 t2 (cur+1) most
else true
| _ -> false in
same a b 0 ((List.length a)/2)
i'm trying to learn ocaml right now and wanted to start with a little program, generating all bit-combinations:
["0","0","0"]
["0","0","1"]
["0","1","0"]
... and so on
My idea is the following code:
let rec bitstr length list =
if length = 0 then
list
else begin
bitstr (length-1)("0"::list);
bitstr (length-1)("1"::list);
end;;
But i get the following error:
Warning S: this expression should have type unit.
val bitstr : int -> string list -> string list = <fun>
# bitstr 3 [];;
- : string list = ["1"; "1"; "1"]
I did not understand what to change, can you help me?
Best regards
Philipp
begin foo; bar end executes foo and throws the result away, then it executes bar. Since this makes only sense if foo has side-effects and no meaningful return value ocaml emits a warning if foo has a return value other than unit, since everything else is likely to be a programmer error (i.e. the programmer does not actually intend for the result to be discarded) - as is the case here.
In this case it really does make no sense to calculate the list with "0" and then throw it away. Presumably you want to concatenate the two lists instead. You can do this using the # operator:
let rec bitstr length list =
if length = 0 then
[list]
else
bitstr (length-1)("0"::list) # bitstr (length-1)("1"::list);;
Note that I also made the length = 0 case return [list] instead of just list so the result is a list of lists instead of a flat list.
Although sepp2k's answer is spot on, I would like to add the following alternative (which doesn't match the signature you proposed, but actually does what you want) :
let rec bitstr = function
0 -> [[]]
| n -> let f e = List.map (fun x -> e :: x) and l = bitstr (n-1) in
(f "0" l)#(f "1" l);;
The first difference is that you do not need to pass an empty list to call the function bitsr 2 returns [["0"; "0"]; ["0"; "1"]; ["1"; "0"]; ["1"; "1"]]. Second, it returns a list of ordered binary values. But more importantly, in my opinion, it is closer to the spirit of ocaml.
I like to get other ideas!
So here it is...
let rec gen_x acc e1 e2 n = match n with
| 0 -> acc
| n -> (
let l = List.map (fun x -> e1 :: x) acc in
let r = List.map (fun x -> e2 :: x) acc in
gen_x (l # r) e1 e2 (n - 1)
);;
let rec gen_string = gen_x [[]] "0" "1"
let rec gen_int = gen_x [[]] 0 1
gen_string 2
gen_int 2
Result:
[["0"; "0"]; ["0"; "1"]; ["1"; "0"]; ["1"; "1"]]
[[0; 0]; [0; 1]; [1; 0]; [1; 1]]