Re: Doing IO in foldr repeats lines?
Ian Lynagh wrote:
>
> > main :: IO()
> > main = do _ <- foldl foo (return 14) ["qq\n", "ww\n", "ee\n"]
> > putStr ""
>
> > foo :: IO Int -> String -> IO Int
> > foo io_l s = do l <- io_l
> > () <- putStr s
> > io_l
>
> prints (with both GHC and hugs):
>
> qq
> ww
> qq
> ee
> qq
> ww
> qq
>
> and I really don't understand why. Is the code re-evaluated every time
> foldl is expanded or something?
Nobody seems to have answered yet, so I try to explain it.
Look at your definition of foo: it actually duplicates its argument
action io_l. For the first application io_l is (return 14). Let's call
that io_l0. The resulting action is
io_l1 = do { l <- return 14; () <- putStr "qq"; return 14 }
which is passed at the next application. The result is
io_l2 = do { l <- do { l <- return 14; () <- putStr "qq"; return 14 }
; () <- putStr "ww"
; do { l <- return 14; () <- putStr "qq"; return 14 }
}
This can be reformulated as
io_l2'= do { l <- return 14
; () <- putStr "qq"
; l <- return 14
; () <- putStr "ww"
; l <- return 14
; () <- putStr "qq"
; return 14
}
And so on. Finally the complete action (io_l3) is executed by running
main and produces the output you observe.
Hope this helps,
- Andreas
--
Andreas Rossberg, [EMAIL PROTECTED]
:: be declarative. be functional. just be. ::
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Re: Doing IO in foldr repeats lines?
On Sat, Jan 20, 2001 at 05:08:48PM -0500, Scott Turner wrote: > At 21:17 2001-01-20 +, Ian Lynagh wrote: > > main = do _ <- foldl foo (return 14) ["qq\n", "ww\n", "ee\n"] > > putStr "" > > > > foo :: IO Int -> String -> IO Int > > foo io_l s = do l <- io_l > > () <- putStr s > > io_l > > It repeats lines > >and I really don't understand why. Is the code re-evaluated every time > >foldl is expanded or something? > > Since the result type of foo is IO Int, it is building an IO action, in > other words it is calculating a sequence of effects. The combined effects > are performed by main. Thanks, a private mail showing the evaluation highlighted my stupidity wonderfully :-) I now see what's going on and a return l instead of io_l as the last line is what I wanted. The mixture of io_l and return bar also explains why I was seeing really random looking output unlike the nice pattern of the more minimal case :-) Thanks Ian ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: Doing IO in foldr repeats lines?
At 21:17 2001-01-20 +, Ian Lynagh wrote: > main = do _ <- foldl foo (return 14) ["qq\n", "ww\n", "ee\n"] > putStr "" > > foo :: IO Int -> String -> IO Int > foo io_l s = do l <- io_l > () <- putStr s > io_l It repeats lines >and I really don't understand why. Is the code re-evaluated every time >foldl is expanded or something? Since the result type of foo is IO Int, it is building an IO action, in other words it is calculating a sequence of effects. The combined effects are performed by main. Repetition occurs due to the structure of foo. Notice that its action consists of first performing io_l, then putting a string, and then performing io_l _again_. If io_l prints the lines qq, ww, and qq, then the result will involve two such triplets. Try this. Remove either of the occurrences of io_l from the do block, and you will see quite different results. -- Scott Turner [EMAIL PROTECTED] http://www.billygoat.org/pkturner ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Doing IO in foldr repeats lines?
Hi all The following code: > module Main (main) where > import IO > main :: IO() > main = do _ <- foldl foo (return 14) ["qq\n", "ww\n", "ee\n"] > putStr "" > foo :: IO Int -> String -> IO Int > foo io_l s = do l <- io_l > () <- putStr s > io_l prints (with both GHC and hugs): qq ww qq ee qq ww qq and I really don't understand why. Is the code re-evaluated every time foldl is expanded or something? Thanks Ian, confused ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
