[Haskell-beginners] list monad question

[Matthias Guedemann]

> How about a recursive function like this:
> 
>     alln 1 ls = map (:[]) ls
>     alln n ls = do
>         a <- ls
>         as <- alln (n-1) ls
>         return (a:as)
> 
> Note that `ls :: [t]` and `all (n-1) ls :: [[t]]` has different types but
> it's okay because both are in the list monad. 
> 
> Also, it can be done with list comprehensions:
> 
>     alln' 1 ls = [[a] | a<-ls] 
>     alln' n ls = [a:as | a<-ls, as<-alln' (n-1) ls]


Works great, thanks a lot

Matthias


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Dipl. Inf. Matthias Guedemann              / __\/ _\  /__\
Computer Systems in Engineering           / /   \ \  /_\
Otto-von-Guericke Universitaet Magdeburg / /___ _\ \//__
Tel.: 0391 / 67-19359                    \____/ \__/\__/
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