Hi,
I haven't tried this with a recent version of Hugs, but MutVars
are not strict in the values they contain, i.e., writeVar will not
force the evaluation of its second argument. Replacing the last line
of mkSum with
strict (writeVar x) (val + i)
might improve matters.
Just a hunch,
--Sigbjorn
Jose Emilio Labra Gayo writes:
> Hi, I have sent this message to the hugs-users list but I don't know if
> it really works so I ask it again in this list.
>
> I need to work with ST and STArray to implement in Haskell an algorithm
> wich was written in C.
>
> I am using HUGS 1.4 beta (970719) in WIndows 95
>
> The next problem is very simplified but I don't know how to solve it in
> an efficient way:
>
> Suppose you want to implement the sum of the first n integers using a
> MutVar as a temporal store.
>
> The imperative algorithm could be: "for i=1 to n do x:=x+i; "
>
> I know that you could type "sum [1..n]" but suppose you want to mirror the
> imperative style and use the next code (I know it looks imperative :)
>
> -------------------------------------------------
> test::Int->Int
> test n = runST
> ( do
> x <- newVar 0
> s <- mkSums n x
> v <- readVar x
> return v
> )
>
> mkSums::Int -> MutVar s Int -> ST s ()
> mkSums n x = sequence [mkSum x i | i<- [1..n]]
>
> mkSum::MutVar s Int->Int->ST s ()
> mkSum x i = do
> val <- readVar x
> writeVar x (val + i)
> -------------------------------------------------
>
> Well, the problem is that it gets a "Control stack overflow" when n is
> large.
>
> I think the problem is with "mkSums" but I don't know if there is
> a different way to specify that I want to repeat a process n times
> in Haskell without creating the list [1..n] and folding it.
> any suggestions?
>
