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Adventures in Computer Programming

1. jethr0 2007.01.16

   “head” and “tail” should be used sparingly, because “head” is a partially defined function (see “head []“).

   subst _    _  [       ] = []
   subst from to xs@(a:as) =
       if isPrefixOf from xs
           then to ++ drop (length from) xs
           else a : subst from to as
       where isPrefixOf as bs = and $ zipWith (==) as bs
   
2. jethr0 2007.01.16

   sorry, too smart (sic!) for my own good. my “isPrefixOf” is broken, instead use “Data.List.isPrefixOf”
3. Thomas David Baker 2007.01.17

   Thanks jethr0. I see you’ve moved the list to act on to the last of the arguments. Someone else on #haskell suggested that too, to improve composability.
4. Cale Gibbard 2007.01.20

   Yeah, the general rule is to always order parameters to functions in order of increasing likelihood of variation. In this case, for instance, it seems more likely that the same substitutions will be used on many different lists, rather than many different substitutions on the same initial list.
5. Joseph 2007.11.24

   The first function here is missing a check that can cause infinite recursion if an empty `find` list is given

    replace "abc" "" "A"
   

   it needs something like…

   replace _ [] list = list
   

   The second function given only replaces the first instance of the string, and if the `find` list is empty then it is equivalent to

   > to:xs
   

   the fourth line could be changed to…

   then to ++ subst from to (drop (length from) xs)
   

   but then it has the same problem as the first function, and needs another check

   substr [] _ list = list — or something like that
   

   I chose to write my replace as follows

   replace::(Eq a) => [a] -> [a] -> [a] -> [a]
   replace [] newSub list = join newSub list
   replace oldSub newSub list = _replace list where
   	_replace list@(h:ts) = if isPrefixOf oldSub list
   		then newSub ++ _replace (drop len list)
   		else h : _replace ts
   	_replace [] = []
   	len = length oldSub
   

   I decided that an empty match list would mean inserting the new element between all of the elements in the list, but maybe the new element should be inserted on the beginning of the list as well.

   join::[a] -> [a] -> [a]
   join glue [h] = [h]
   join glue (h:ts) = h : glue ++ join glue ts
   join _ [] = []
   

   another function that works nicely with replace

   strip::(Eq a) => [a] -> [a] -> [a]
   strip old = replace old []
   

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