Re[2]: [Haskell-cafe] take the keywords from a string

Hello Sara, Sunday, June 18, 2006, 9:04:05 AM, you wrote: you should write stop alternatives first, because Haskell tries alternatives just sequentially, topmost first: listOfString :: [(String,String)] -> [String] listOfString [("","")] = [] listOfString [(s1,s2)] = s1: listOfString (lex s2)

I tried to write function lexList by using an intermediate function lisOfString as below:

module Lex where lexList :: String -> [String] lexList str = listOfString (lex str) lexList [] =[]

listOfString :: [(String,String)] -> [String] listOfString [(s1,s2)] = s1: listOfString (lex s2) listOfString [("","")] = []

When I try function lisOfString as below, it runs forever (non-stop) although I have the stop criteria for it ??

Lex>> lisOfString ["test1","test2(test3)"]

Thanks in advance.

S.

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On 6/17/06, Neil Mitchell wrote:

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Hi

On 6/18/06, Sara Kenedy wrote:

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Sorry, I am not clear at some point in your answer:

1) The function lex :: String -> [(String,String)] and filter :: (a -> Bool) -> [a] -> [a] So, I did not see how filter can use the list of tuple string of lex.

You can write a function lexList, of type String -> [String], by repeatedly calling lex - its not too hard. Once you have this the filter will work.

Thanks

Neil

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