Re: [Haskell-cafe] Newbie question related to list evaluation
Felipe Lessa wrote: As as side note, GHC's flag -Wall would have warned about creating a variable with a name already in scope. *makes a mental note* I've created bugs like this far too many times... ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie question related to list evaluation
AAh! Thanks a ton! Hemanth K On Jan 7, 2008 12:10 AM, Rodrigo Queiro <[EMAIL PROTECTED]> wrote: > You have used the name 'pos' twice, for both the parameter and the > returned value of the recursive call. The reason this results in an > infinite loop is that in code like > > let x = x + 1 > > Haskell treats both xs to be references to the same thing, so evaluates: > x > = x + 1 > = (x + 1) + 1 > = ((x + 1) + 1) + 1 > ... > > which results in the infinite loop. > > On 06/01/2008, Sai Hemanth K <[EMAIL PROTECTED]> wrote: > > Hi, > > > > I am new to functional and lazy programming languages ( that's correct, > my > > life has been pretty pathetic so far) and am not able to understand > GHC's > > behaviour for a particular function. Can someone help me please? > > > > I am trying to write a function which would compare two strings (from > > reverse) and return the position of first mismatch. This is part of the > > right-to-left scan of bayer-moore algorithm.. and it is essential for me > to > > do it from reverse. > > Since my goal is to learn haskell, I am not using Data.ByteString. > > > > My function is as follows: > > > > matchReverse :: String -> String ->Int->(Bool,Int) > > matchReverse [] [] pos = (True,pos) > > matchReverse _ [] pos = (False,pos) > > matchReverse [] _ pos = (False,pos) > > matchReverse (x:xs) (y:ys) pos = let (matched,pos) = matchReverse xs ys > (pos > > +1) > > in if matched then > > ((x==y),pos) > > else (False,pos) > > > > > > > > The behaviour I expected in four scenarios is as below: > > 1.matchReverse "kapilash" "kapilash" 0 --should return (True,0) > > 2.matchReverse "kapilash" "kapilast" 0 --should return (False,8) > > 3.matchReverse str1 str2 0 --should return > (False,0) > > 4.matchReverse str1 str1 0 --should return (True,0) > > > > where str1 and str2 are defined as below: > > str1 = replicate 1000 'a' > > str2 = 'b':(replicate 999 'a') > > > > what confounds me is that it is able to identify the first element of > the > > tuple in ALL the cases. > > Invoking fst on the each of the four calls instantly returns the > expected > > value.(even for the cases 3 and 4 where, there are thousand elements) > > But it seems to go into an infinite loop while calculating the 'snd' of > the > > tuple. Even for strings containing just one element each. > > can someone throw some light on this please? Why does it go into an > infinite > > loop? > > > > Many thanks > > Kapilash > > > > > > -- > > I drink I am thunk. > > ___ > > Haskell-Cafe mailing list > > Haskell-Cafe@haskell.org > > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > > > -- I drink I am thunk. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie question related to list evaluation
On Jan 6, 2008 4:40 PM, Jonathan Cast <[EMAIL PROTECTED]> wrote: > let is always recursive in Haskell, so this is a recursive definition > of pos. To break the recursion, use > > > matchReverse (x:xs) (y:ys) pos = let (matched, pos') = matchReverse > xs ys (pos + 1) > in if matched then ((x==y), pos') >else (False, pos') Actually, I think he meant matchReverse (x:xs) (y:ys) pos = let (matched, pos') = matchReverse xs ys (pos + 1) in if matched then ((x==y), pos) else (False, pos') As as side note, GHC's flag -Wall would have warned about creating a variable with a name already in scope. -- Felipe. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie question related to list evaluation
On 6 Jan 2008, at 10:34 AM, Sai Hemanth K wrote: Hi, I am new to functional and lazy programming languages ( that's correct, my life has been pretty pathetic so far) and am not able to understand GHC's behaviour for a particular function. Can someone help me please? I am trying to write a function which would compare two strings (from reverse) and return the position of first mismatch. This is part of the right-to-left scan of bayer-moore algorithm.. and it is essential for me to do it from reverse. Since my goal is to learn haskell, I am not using Data.ByteString. My function is as follows: matchReverse :: String -> String ->Int->(Bool,Int) matchReverse [] [] pos = (True,pos) matchReverse _ [] pos = (False,pos) matchReverse [] _ pos = (False,pos) matchReverse (x:xs) (y:ys) pos = let (matched,pos) = matchReverse xs ys (pos +1) in if matched then ((x==y),pos) else (False,pos) let is always recursive in Haskell, so this is a recursive definition of pos. To break the recursion, use matchReverse (x:xs) (y:ys) pos = let (matched, pos') = matchReverse xs ys (pos + 1) in if matched then ((x==y), pos') else (False, pos') The behaviour I expected in four scenarios is as below: 1.matchReverse "kapilash" "kapilash" 0 --should return (True,0) 2.matchReverse "kapilash" "kapilast" 0 --should return (False,8) 3.matchReverse str1 str2 0 --should return (False,0) 4.matchReverse str1 str1 0 --should return (True,0) where str1 and str2 are defined as below: str1 = replicate 1000 'a' str2 = 'b':(replicate 999 'a') what confounds me is that it is able to identify the first element of the tuple in ALL the cases. Invoking fst on the each of the four calls instantly returns the expected value.(even for the cases 3 and 4 where, there are thousand elements) But it seems to go into an infinite loop while calculating the 'snd' of the tuple. Even for strings containing just one element each. can someone throw some light on this please? Why does it go into an infinite loop? Many thanks Kapilash -- I drink I am thunk. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie question related to list evaluation
You have used the name 'pos' twice, for both the parameter and the returned value of the recursive call. The reason this results in an infinite loop is that in code like let x = x + 1 Haskell treats both xs to be references to the same thing, so evaluates: x = x + 1 = (x + 1) + 1 = ((x + 1) + 1) + 1 ... which results in the infinite loop. On 06/01/2008, Sai Hemanth K <[EMAIL PROTECTED]> wrote: > Hi, > > I am new to functional and lazy programming languages ( that's correct, my > life has been pretty pathetic so far) and am not able to understand GHC's > behaviour for a particular function. Can someone help me please? > > I am trying to write a function which would compare two strings (from > reverse) and return the position of first mismatch. This is part of the > right-to-left scan of bayer-moore algorithm.. and it is essential for me to > do it from reverse. > Since my goal is to learn haskell, I am not using Data.ByteString. > > My function is as follows: > > matchReverse :: String -> String ->Int->(Bool,Int) > matchReverse [] [] pos = (True,pos) > matchReverse _ [] pos = (False,pos) > matchReverse [] _ pos = (False,pos) > matchReverse (x:xs) (y:ys) pos = let (matched,pos) = matchReverse xs ys (pos > +1) > in if matched then > ((x==y),pos) > else (False,pos) > > > > The behaviour I expected in four scenarios is as below: > 1.matchReverse "kapilash" "kapilash" 0 --should return (True,0) > 2.matchReverse "kapilash" "kapilast" 0 --should return (False,8) > 3.matchReverse str1 str2 0 --should return (False,0) > 4.matchReverse str1 str1 0 --should return (True,0) > > where str1 and str2 are defined as below: > str1 = replicate 1000 'a' > str2 = 'b':(replicate 999 'a') > > what confounds me is that it is able to identify the first element of the > tuple in ALL the cases. > Invoking fst on the each of the four calls instantly returns the expected > value.(even for the cases 3 and 4 where, there are thousand elements) > But it seems to go into an infinite loop while calculating the 'snd' of the > tuple. Even for strings containing just one element each. > can someone throw some light on this please? Why does it go into an infinite > loop? > > Many thanks > Kapilash > > > -- > I drink I am thunk. > ___ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Newbie question related to list evaluation
Hi, I am new to functional and lazy programming languages ( that's correct, my life has been pretty pathetic so far) and am not able to understand GHC's behaviour for a particular function. Can someone help me please? I am trying to write a function which would compare two strings (from reverse) and return the position of first mismatch. This is part of the right-to-left scan of bayer-moore algorithm.. and it is essential for me to do it from reverse. Since my goal is to learn haskell, I am not using Data.ByteString. My function is as follows: matchReverse :: String -> String ->Int->(Bool,Int) matchReverse [] [] pos = (True,pos) matchReverse _ [] pos = (False,pos) matchReverse [] _ pos = (False,pos) matchReverse (x:xs) (y:ys) pos = let (matched,pos) = matchReverse xs ys (pos +1) in if matched then ((x==y),pos) else (False,pos) The behaviour I expected in four scenarios is as below: 1.matchReverse "kapilash" "kapilash" 0 --should return (True,0) 2.matchReverse "kapilash" "kapilast" 0 --should return (False,8) 3.matchReverse str1 str2 0 --should return (False,0) 4.matchReverse str1 str1 0 --should return (True,0) where str1 and str2 are defined as below: str1 = replicate 1000 'a' str2 = 'b':(replicate 999 'a') what confounds me is that it is able to identify the first element of the tuple in ALL the cases. Invoking fst on the each of the four calls instantly returns the expected value.(even for the cases 3 and 4 where, there are thousand elements) But it seems to go into an infinite loop while calculating the 'snd' of the tuple. Even for strings containing just one element each. can someone throw some light on this please? Why does it go into an infinite loop? Many thanks Kapilash -- I drink I am thunk. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
