Hi! do     x <- foo     bar' x 1 is equivalent to: foo >>= \ x -> bar' x 1 which is equivalent to flip bar' 1 =<< foo so, for the whole thing, you can say: do     flip bar' 1 =<< foo     flip bar' 2 =<< foo     flip bar' 3 =<< foo and if you define bar'' as, bar'' = flip bar' then it boils down to, do     bar'' 1 =<< foo     bar'' 2 =<< foo     bar'' 3 =<< foo To further get rid of the foo's, you might want to have a look at the Reader monad. But that would be overkill, I think. HTH, On 4 November 2010 16:19, John Smith wrote:

I have code which looks like this:

foo :: IO A

bar :: Bool -> A -> Int -> Bool -> IO ()

do  x <- foo  y <- foo  z <- foo

 bar True x 1 False  bar True y 2 False  bar True z 3 False

What is the best way to factor it, including eliminating the temporary x,y,z variables? I can get this down to

foo :: IO A

bar :: Bool -> A -> Int -> Bool -> IO ()

bar' a b = bar True a b False

do  x <- foo  y <- foo  z <- foo

 bar x 1  bar y 2  bar z 3

but I don't know what to do next. Is there some form of list comprehension or combination of map and lift which will eliminate the repetition?

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