Re: [Haskell-cafe] How to use "bracket" properly ?

thanks for your quick answer. But 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); } will report: parse error on input `let' Daniel Peebles wrote:

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 wrote:

...

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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