Given this following code (which does not work):
fun func() =
val decimal = 0 (* the final result *)
val multiple = 0 (* keeps track of multiples, eg. In XXV, X would be a multiple *)
val current = 0 (* the digit currently being processed *)
val top = 0 (* value of the last element in the list *)
val last_add = 0 (* the last digit that wasn't a multiple, or subtraction operation *)
val last_sub = 0
val problem = 0 (* if value is 1 then there is a problem with the input *)
val myList = [1,2,3,4,5] (* the list has more values *)
val current = tl(myList) (* grab the last element from the list *)
val myList = tl(myList) (* remove the last element from the list *)
val top = tl(myList) (* grab the value at the end of the list *)
while (not(myList = [])) (* run while the list is not empty *)
if ( (not(myList = [])) andalso (current > top))
then
val decimal = decimal + current - top
val last_sub = top;
val myList = tl(myList)
else
if ( (myList = []) andalso (current = top))
then val decimal = decimal + current
val multiple = multiple + 1
else
if (last_sub = current)
then val problem = 1
else
val decimal = decimal + current
val multiple = 0
val last_add = current
This is only a partial code , which doesn't work at the moment , since the val is not
possible within an if statement .
I want to run in a while loop , how can I do that in ML ?
How can I assign and reassign values into variables that were previously declared in ML ?
The val keyword is not possible within the IF condition , so I cannot update the variables , any idea how to solve that ?
Regards
How can I assign and reassign values into variables that were
previously declared in ML ?
You cannot assign to variables after they are declared in ML.
The val keyword is not possible within the IF condition , so I cannot
update the variables , any idea how to solve that ?
Except at the top level, you usually use val and fun inside a let:
let
val x = blah blah
val y = blah blah
fun f x y = blah blah
in
some expression
end
However, note that this creates a new variable (which may hide any existing variable of the same name), which exists inside the scope of the body of the let. As said before, you cannot assign to an existing variable.
I want to run in a while loop , how can I do that in ML ?
You are almost there. The syntax is while condition do ( ... ). But a while loop is useless without mutable state.
If you want mutable state, you can use a mutable data structure. The language provides a simple "mutable cell" called ref: you create it by passing the initial value to the ref function, you get the current value with the ! operator, and you set a new value with the := operator. You also have to remember that if you want to run multiple imperative "statements", you must separate them with the ; operator, and possibly enclose the whole "block" of statements in parentheses due to precedence issues.
But using while loops and mutable state is really not the right way to go here. You are using a functional language, and it would be much better for you to re-write your algorithm to be purely functional. It's not hard to do. You would turn the body of your while loop into a tail-recursive helper function, and the "variables" that change between iterations of the loop would become arguments to this function. Instead of trying to "set" the values of these variables, it would simply recursively call itself with the new values for the next iteration. If it's tail-recursive, it's equivalent to iteration memory-wise.
Related
Why is this Ocaml statement giving me a syntax error?
let a = 0;; if a = 0 then let b = 0;;
Do if then else statements always have to return a value?
EDIT: Here is the code I am struggling with. I want to apply this function over a list with the map function. The function is supposed to look at each word in the list wordlist and add to the stringmap. If it has already been added to the string map then add 1 to its password.
module StringMap = Map.Make(String)
let wordcount = StringMap.empty
let findword testword =
let wordcount = (if (StringMap.mem testword wordcount)
then (StringMap.add testword ((StringMap.find testword wordcount)+1) wordcount)
else (StringMap.add testword 1 wordcount))
List.map findword wordlist
You can only have an if then without else if the then expression evaluates to unit () Otherwise, the expression will not type check. An if without an else is equivalent to writing if x then y else () which can only type check if y is unit.
Check this out for a reference.
(Terminology note: there are no statements in OCaml because everything is an expression, so the term "if statement" doesn't quite apply. I still understood what you meant, but I thought this was worth noting)
Yes, if is an expression in OCaml, not a statement. The best way to look at it is that there are no statements in OCaml. Everything is an expression. (Admittedly there are expressions that return (), which are similar to statements.)
You can only have if b then e if the type of e is unit (i.e., if it returns ()).
Note also that you can't just say let v = e, except at the top level of a module. At the top level it defines a global name in the module. In other cases you need to say let v = e1 in e2; the let defines a local symbol v for use in the expression e2.
One answer to the let b = problem - it works like this:
let a = 0
let b = if a = 0 then 0 else 1
(* or whatever value you need in the else branch *)
And then the Map problem: the manual says Map is applicative - that means Stringmap.add returns a new map. You must use a ref to store your map - see this ocaml toplevel protocol:
# module StringMap = Map.Make(String);;
# let mymap = ref StringMap.empty ;;
val mymap : '_a StringMap.t ref = {contents = <abstr>}
# mymap := StringMap.add "high" 1 !mymap;;
- : unit = ()
# StringMap.mem "high" !mymap;;
- : bool = true
# StringMap.mem "nono" !mymap;;
- : bool = false
# StringMap.find "high" !mymap;;
- : int = 1
# StringMap.find "nono" !mymap;;
Exception: Not_found.
I am trying to achieve the following: Finding the element at a specific index.
So if I had a list of [5; 2; 3; 6] and ask for the element at index 2, it would return 3.
let counter = 0;;
let increase_counter c = c + 1;;
let rec get_val x n = match x with
[] -> -1
| (h::t) ->
if (counter = n) then
h
else
increase_counter counter ; get_val t n
;;
But this code is giving me a bug saying that -1 is not of type 'unit'?
As Jeffrey Scofield said, you should write let counter = ref 0 to make counter mutable. Now, you can use the built in incr function to increment it (equivalent to counter := !counter + 1), and you'll get its value with !counter.
There is also a problem in your algorithm : if the counter is equal to n, you return the head of the list... you mean : if the head of the list is equal to n, you return the counter.
Your program is then :
let counter = ref 0;;
let rec get_val x n = match x with
[] -> -1
| (h::t) ->
if (h = n) then
!counter
else
begin incr counter ; get_val t n end
;;
Note that I've added begin and end around the else block so it can be interpreted as a sequence of instructions.
Your program now works, but it is not the best way to solve this problem with ocaml.
You should write something like
let get_val x n =
let rec get_val_aux x n counter = match x with
| [] -> -1
| h :: _ when h = n -> counter
| _ :: t -> get_val_aux t n (succ counter)
in
get_val_aux x n 0
;;
Here, we add a parameter to the get_val_aux function which we increment at each call. This function is nested within the get_val function to hide this additional parameter which is initialized with 0 on the first call.
Instead of using an if statement, we use the when condition to know when the element has been found, and add a new case to match the last case (not found). Note the use of the _ wildcard to avoid an unused variable.
The succ function (for successor) only adds 1 to its parameter. It is equivalent to counter + 1.
There are many problems with this code. If you ignore your immediate problem for a moment, you are treating OCaml variables like the variables of an imperative language. However, OCaml variables are immutable. This function
let increase_counter c = c + 1
Doesn't change the value of any variable. It just returns a number 1 bigger than what you give it.
The only error I get from the toplevel when I enter your code is for this expression:
increase_counter counter ; get_val t n
The compiler is warning you that the expression before ; is supposed to be executed for its side effects. I.e., it should almost always have type unit. Since (as I say) your function increase_counter returns an int, the compiler is warning you about this.
I would like to understand how sequential composition works much better than I do now in SML. I have to write a program that takes a list of integers and moves the integer at index zero to the last index in the list. ie. [4, 5, 6] -> [5, 6, 4].
The code I have right now is:
- fun cycle3 x =
= if length(x) = 1 then x
= else (List.drop(x, 1);
= x # [hd(x)]);
val cycle3 = fn : 'a list -> 'a list
The question lies in my else statement, what I want to happen is first concatenate the first term to the end, and then second drop the first term. It seems simple enough, I just don't understand how to perform multiple functions in a particular order using SML. My understanding was that the first function called has the scope of the second function that would have the scope of the third function.. etc etc.. What am I doing wrong here?
Most things in SML are immutable -- your function, rather than modifying the list, is building a new list. List.drop(x,1) evaluates to a new list consisting of all but the first element of x, but does not modify x.
To use your method, you would bind the result of List.drop(x,1) to a variable, as in the following:
fun cycle3 x = if length x = 1
then x
else let
val y = List.drop(x,1)
in
y # [hd(x)]
end
Alternately, a cleaner way of doing this same thing, that also handles the possibility of an empty list:
fun cycle3 [] = []
| cycle3 (x::xs) = xs # [x]
I'm new to SML and I'm attempting to get the index of an item in a list. I know that using List.nth will give me the value of an item at a index position, but I want the index value. There may even be a built in function that I'm not aware of. In my case, the list will not contain duplicates so if the item is in the list I get the index, if not it returns ~1. Here is the code I have so far. It works, but I don't think it is very clean:
val L=[1,2,3,4,5];
val m=length L-1;
fun Index(item, m, L)=if m<0 then ~1 else
if List.nth(L, m)=item then m else Index(item,m-1,L);
To elaborate on my previous comment, I suggest some changes for an implementation that fits better in the ML idiom:
fun index(item, xs) =
let
fun index'(m, nil) = NONE
| index'(m, x::xr) = if x = item then SOME m else index'(m + 1, xr)
in
index'(0, xs)
end
The individual changes are:
Have index return a value of type int option. NONE means the item is not in the list, SOME i means it is in the list, and the index of its first occurrence is i. This way, no special values (~1) need be used and the function's intended usage can be inferred from its type.
Hide the parameter m by renaming the function to index' and wrapping it into an outer function index that calls it with the appropriate arguments. The prime character (`) often indicates auxiliary values.
Use pattern matching on the list to get to the individual elements, eliminating the need for List.nth.
Also note that most commonly, function and variable names begin with a lowercase letter (index rather than Index), while capital letters are used for constructor constants (SOME) and the like.
I would like to propose a simpler and less efficient version of this index function. I agree that it is not as desirable to use exceptions rather than int option, and that it is not tail-recursive. But it is certainly easier to read and thus may serve as learning material:
fun index (x, []) = raise Subscript
| index (x, y::ys) =
if x = y then 0 else 1 + index (x, ys)
fun index(list,n)=
= if n=0 then hd(list) else index(tl(list),n-1);
val index = fn : 'a list * int -> 'a
index([1,2,3,4,5],2);
val it = 3 : int
index([1,2,3,4,5],0);
val it = 1 : int
I'm having a problem while trying to increment my value of x inside the inner foldl call. I make x equal to shiftValue that's passed in and attempt to increment it whenever I find a #" " or #"*" in the inner foldl call, but the value of x that gets returned is always the same as shiftvalue was when passed in.
The function takes in a tuple of (string, int) where the string will have leading spaces and asterisk chopped off that come before any other characters. Also any spaces or asterisk on the end not followed by any other characters will get chopped off. The int that is passed in is a shiftValue that tracks how many spaces the string was shifted over before getting passed into this function. Whenever I take off a leading space or asterisk I need to increment the shiftValue "x" by one.
The inner foldl call removes asterisks and spaces from the front. The outer foldl call removes them from the back. The asterisks and spaces get removed right, the x value just isn't getting updated.
(*Take string str and get rid of leading and following #"*"s and #" "s. For every
leading #"*" or #" " removed increment the shiftValue returned in the tuple*)
fun trimStarsOnNode (str, shiftValue) =
let
val x = shiftValue
in
((implode(rev (foldl (fn (cur, a) =>
if length a = 0 andalso cur = #"*" then a # []
else
if length a = 0 andalso cur = #" " then a # []
else a # [cur]) [] (rev (foldl (fn (cur, a) =>
if length a = 0 andalso cur = #"*" then (x = x + 1; a # [])
else
if length a = 0 andalso cur = #" " then (x = x + 1; a # [])
else a # [cur]) [] (explode str)))))), x)
end;
trimStarsOnNode ("***hello", 3); (* Should print out ("hello", 6) *) but prints out ("hello", 3)
Look at your x - in the beginning of your function, you do:
val x = shiftValue
Then, later, you try to do this:
x = x + 1
Remember, in SML, you can't change the value of a variable (actually, they're just called values in SML, for that reason). x = x + 1 just compares x and x + 1, so the value of the statement x = x + 1 is boolean false.
As Tayacan says, variables are not mutable in SML. If you want mutability you need to use reference types -- but usually, they are best avoided and it's preferable to stick to functional style.
It's also worth noting that your function is going to be very inefficient (O(n^2)), because of your use of list concatenation and length on every iteration. And it is incorrect, because it will also remove stars in the middle of the string (and then redundantly go over the whole list a second time). Finally, your solution is far too complicated.
FWIW, here is the shortest implementation I can think of, using the Substring library module and the function composition operator o:
fun isStarOrSpace c = (c = #"*" orelse c = #" ")
val trimStars =
let open Substring
in string o dropl isStarOrSpace o dropr isStarOrSpace o full end
This does not use your shiftValue because I don't understand what it's supposed to do. You can easily compute the number of removed characters by comparing the old and new string size. That is, your intended function (IIUC) could easily be expressed on top of mine as
fun trimStarsOnNode(s, shift) =
let val s' = trimStars s in (s', size s - size s' + shift) end
But to be honest, I don't understand what this version would be good for.
Edit: A version that returns the left drop count:
fun trimStars s =
let
open Substring
val ss = dropl isStarOrSpace (dropr isStarOrSpace (full s))
in
(string ss, #2(base ss))
end