Hi John, I'm trying to use the GHCI debugger on this code: 20 instance ArrowLoop SF where 21 loop (SF f) = SF $ \as -> 22 let (bs, cs) = unzip (f (zip as (stream cs))) in bs 23 where stream ~(x:xs) = x : stream xs 24 25 swap :: (a,b) -> (b,a) 26 swap (x,y) = (y,x) in order to watch the recursion of the `loop' function unfold. However, when I single step through the code, I never stop on line 22 (where I could, presumably, peek in at `bs' and `cs', in order to see them develop): *SF> :break swap Breakpoint 1 activated at SF.hs:26:1-18 *SF> runSF (loop (arr swap)) [1,2,3] Stopped at SF.hs:26:1-18 _result :: (b, a) = _ [SF.hs:26:1-18] *SF> :step Stopped at SF.hs:26:14-18 _result :: (b, a) = _ x :: a = _ y :: b = _ [SF.hs:26:14-18] *SF> : [1Stopped at SF.hs:23:34-42 _result :: [a] = _ xs :: [a] = _ [SF.hs:23:34-42] *SF> : Stopped at SF.hs:23:13-42 _result :: [a] = _ [SF.hs:23:13-42] *SF> : Stopped at SF.hs:23:30-42 _result :: [a] = _ x :: a = _ xs :: [a] = _ [SF.hs:23:30-42] *SF> : (Pattern repeats.) Do you have any advice? Thanks, -db On Mon, Oct 31, 2011 at 3:19 PM, John Lask wrote:

On 1/11/2011 1:35 AM, Captain Freako wrote:

you need to study ArrowLoop and understand that. In the code

Never mind, I misread the code, 'zip' and the lazy definition of stream should add the necessary laziness. -- ryan On Tue, Nov 1, 2011 at 3:36 PM, Ryan Ingram wrote:

Try

swap p = (snd p, fst p)

or, equivalently

swap ~(x,y) = (y,x)

-- ryan

On Tue, Nov 1, 2011 at 1:30 PM, Captain Freako wrote:

...

Hi John,

I'm trying to use the GHCI debugger on this code:

20 instance ArrowLoop SF where 21 loop (SF f) = SF $ \as -> 22 let (bs, cs) = unzip (f (zip as (stream cs))) in bs 23 where stream ~(x:xs) = x : stream xs 24 25 swap :: (a,b) -> (b,a) 26 swap (x,y) = (y,x)

in order to watch the recursion of the `loop' function unfold. However, when I single step through the code, I never stop on line 22 (where I could, presumably, peek in at `bs' and `cs', in order to see them develop):

*SF> :break swap Breakpoint 1 activated at SF.hs:26:1-18 *SF> runSF (loop (arr swap)) [1,2,3] Stopped at SF.hs:26:1-18 _result :: (b, a) = _ [SF.hs:26:1-18] *SF> :step Stopped at SF.hs:26:14-18 _result :: (b, a) = _ x :: a = _ y :: b = _ [SF.hs:26:14-18] *SF> : [1Stopped at SF.hs:23:34-42 _result :: [a] = _ xs :: [a] = _ [SF.hs:23:34-42] *SF> : Stopped at SF.hs:23:13-42 _result :: [a] = _ [SF.hs:23:13-42] *SF> : Stopped at SF.hs:23:30-42 _result :: [a] = _ x :: a = _ xs :: [a] = _ [SF.hs:23:30-42] *SF> : (Pattern repeats.)

Do you have any advice?

Thanks, -db

On Mon, Oct 31, 2011 at 3:19 PM, John Lask wrote:

...

On 1/11/2011 1:35 AM, Captain Freako wrote:

you need to study ArrowLoop and understand that. In the code

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

First, let's lay out our definitions: unzip [] = ([], []) unzip ((x,y):xys) = (x:xs, y:ys) where (xs,ys) = unzip xys zip [] _ = [] zip _ [] = [] zip (x:xs) (y:ys) = (x,y) : zip xs ys map _ [] = [] map f (x:xs) = f x : map f xs stream ~(a:as) = a : stream as -- equivalently stream xs = head xs : stream (tail xs) Now we want to evaluate this: runSF (loop (arr swap)) [1,2,3] Lets simplify some of the insides a bit: arr swap = SF $ map swap = SF $ map (\(x,y)->(y,x)) loop (arr swap) = SF $ \as -> let (bs,cs) = unzip (map swap (zip as (stream cs))) in bs runSF (loop (arr swap)) [1,2,3] = runSF (SF $ ...) [1,2,3] = (\as -> let (bs,cs) = unzip (map swap (zip as (stream cs))) in bs) [1,2,3] Here is our heap at this point; we are trying to evaluate bs: p = unzip (map swap (zip as (stream cs))) as = [1,2,3] bs = fst p cs = snd p snd p forces p, unzip forces its argument, map forces its second argument, and zip forces both its arguments. So now we have: p = unzip retmap retmap = map swap retzip retzip = zip as retstream retstream = stream cs as = [1,2,3] bs = fst p cs = snd p Evaluating further: retstream = head cs : retstream2 retstream2 = stream (tail cs) retzip = (1, head cs) : retzip2 retzip2 = zip [2,3] (stream (tail cs)) retmap = (head cs, 1) : retmap2 retmap2 = map swap retzip2 p = (head cs : xs1, 1 : ys1) (xs1,ys1) = unzip retmap2 bs = head cs : xs1 cs = 1 : ys1 bs = 1 : xs1 and we can now return the first cons cell of bs. This repeats until we get [1,2,3] back out; note that each value goes through both sides of the swap before coming out the 'front' again. -- ryan On Tue, Nov 1, 2011 at 1:30 PM, Captain Freako wrote:

Hi John,

I'm trying to use the GHCI debugger on this code:

20 instance ArrowLoop SF where 21 loop (SF f) = SF $ \as -> 22 let (bs, cs) = unzip (f (zip as (stream cs))) in bs 23 where stream ~(x:xs) = x : stream xs 24 25 swap :: (a,b) -> (b,a) 26 swap (x,y) = (y,x)

in order to watch the recursion of the `loop' function unfold. However, when I single step through the code, I never stop on line 22 (where I could, presumably, peek in at `bs' and `cs', in order to see them develop):

*SF> :break swap Breakpoint 1 activated at SF.hs:26:1-18 *SF> runSF (loop (arr swap)) [1,2,3] Stopped at SF.hs:26:1-18 _result :: (b, a) = _ [SF.hs:26:1-18] *SF> :step Stopped at SF.hs:26:14-18 _result :: (b, a) = _ x :: a = _ y :: b = _ [SF.hs:26:14-18] *SF> : [1Stopped at SF.hs:23:34-42 _result :: [a] = _ xs :: [a] = _ [SF.hs:23:34-42] *SF> : Stopped at SF.hs:23:13-42 _result :: [a] = _ [SF.hs:23:13-42] *SF> : Stopped at SF.hs:23:30-42 _result :: [a] = _ x :: a = _ xs :: [a] = _ [SF.hs:23:30-42] *SF> : (Pattern repeats.)

Do you have any advice?

Thanks, -db

On Mon, Oct 31, 2011 at 3:19 PM, John Lask wrote:

...

On 1/11/2011 1:35 AM, Captain Freako wrote:

you need to study ArrowLoop and understand that. In the code

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