Question

Let’s take a function of type (Monad m) => a -> m a. For example:

ghci> let f x = Just (x+1)

I’d like to be able to apply it any number of times. The first thing I tried was

ghci> let times n f = foldr (>=>) return $ replicate n f

The problem is that it won’t work for large n:

ghci> 3 `times` f $ 1
Just 4
ghci> 1000000 `times` f $ 1
Just *** Exception: stack overflow

It doesn’t work also the other way:

ghci> let timesl n f = foldl' (<=<) return $ replicate n f
ghci> 3 `timesl` f $ 1
Just 4
ghci> 1000000 `timesl` f $ 1
Just *** Exception: stack overflow

Actually, what works is using ($!) strictness operator

ghci> let timesStrict n f = foldr1 ((>=>) . ($!)) $ replicate n f
ghci> 3 `timesStrict` f $ 1
Just 4
ghci> 10000000 `timesStrict` f $ 1
Just 10000001

Is there a nicer or more idiomatic solution? Or probably a stricter one? I still easily get stack overflows if f is a heavy-weight function.

UPD: I found that writing times in a pointful form does not solve the problem of composing heavy-weight monadic actions neither. This works for f x = Just (x+1) but fails in the real world:

times f 0 a = return a
times f i a = (f $! a) >>= times f (i - 1)

Answer 1

If you make f strict as in

f x = let y = x+1 in y `seq` Just y

or

-- remember to enable -XBangPatterns
f !x = Just (x+1)

and leave the rest alone, your code runs in constant space (albeit slowly) even with very large n:

ghci> times 4000000000 f 3
Just 4000000003