They're "in" different Monads. The first one does x <- [...], which
means that you're operating in the list Monad instance, and bracket
operates in the IO Monad. The second one uses let x = [...] which
doesn't have any effect on what Monad you're in, so the whole thing
can be in IO.
Note that when you do x <- [1..3]; y <- [4..6] you're going to get all
9 pairs of values from x and y, by the way.
Hope this helps,
Dan
On Mon, Oct 19, 2009 at 1:33 AM, zaxis
winSSQ count noRed noBlue = do { yesRed <- [1..33] \\ noRed; yesBlue <- [1..16] \\ noBlue; bracket (openFile "ssqNum.txt" WriteMode) (hClose) (\hd1 -> pickSSQ count yesRed yesBlue hd1); return () } will report: Couldn't match expected type `IO ()' against inferred type `[()]' In a stmt of a 'do' expression: bracket (openFile "ssqNum.txt" WriteMode) (hClose) (\ hd1 -> pickSSQ count yesRed yesBlue hd1)
However, the following works fine:
winSSQ count noRed noBlue = do let yesRed = [1..33] \\ noRed let yesBlue = [1..16] \\ noBlue bracket (openFile "ssqNum.txt" WriteMode) (hClose) (\hd1 -> pickSSQ count yesRed yesBlue hd1)
Why ? -- View this message in context: http://www.nabble.com/How-to-use-%22bracket%22-properly---tp25953522p2595352... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
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