- Beginners - Haskell.org

string instead of char
by John Moore 05 Oct '09

05 Oct '09

Hi, I am now writing a function that replaces vowels with the letter x. eg.put in a string "help" and out comes hxlp. I tried this: f x = case x of {'a' -> 'x';'e' -> 'x';'i' -> 'x';'o' -> 'x';'u' -> 'x';_ -> x} but this wont work on strings, only on the individual letters. Any direction would be very welcome. Yours john

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Advantage to installing Hugs with GHC?
by John Velman 05 Oct '09

05 Oct '09

My machine is an iMac running Leopard, 10.5.8. I've installed GHC from the haskell-platform-..(Leopard/x86) disk image. Ghci works fine as far as I can tell. Is there any particular advantage to also installing Hugs? Thanks, John Velman

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parse error
by John Moore 04 Oct '09

04 Oct '09

Hi, This is a function that returns true if divisible by 2. But it keeps telling me parse error on input | . What is wrong here. is_even n | mod n 2 == 0 | True Not sure if I need another statement for false John

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redirect stdout to /dev/null
by Hong Yang 04 Oct '09

04 Oct '09

On UNIX, we can easily redirect stdout to /dev/null in a Shell. How is this realized in Haskell on both UNIX and Windows? I used a library which gave me unwanted stdout display. I tried to find any function to hide the stdout output, but could not. Can someone help me out? Otherwise I have to modify the code in the library to comment out some print commands. Thanks, Hong

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main
by John Moore 03 Oct '09

03 Oct '09

Hi, I must be doing something wrong when I run a program it comes up on the screen as OK module loaded Main *Main> But it will not do what I program to do. What do I enter after *main> I tried putting in the arguments but ot just returns the arguments. For example: signum x | x < 0 = -1 | x == 0 = 0 | x > 0 = 1 Enter 5 it returns 5 What have I missed? Regards John

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$= operator
by Nathan Holden 03 Oct '09

03 Oct '09

I was reading Eleara examples out of Hackage, and I kept noticing lines like this: > t <- get GLFW.time > GLFW.time $= 0 Most of the rest of it made more-or-less sense (I mean, there IS a reason I'm still asking questions to beginners!) but I couldn't seem to find anything about it on my usual reference sources, and it's uses don't make that much sense to me.

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Help with Data.Time
by Hong Yang 02 Oct '09

02 Oct '09

I want to calculate the length of time between time ta and time tb. Time ta is a String looking like "N:NNAM" or "N:NNPM." I need tb to be 9:30AM on the current date. I think the flow should be 1) parse ta to utcTimeA :: UTCTime 2) get current date and form tb to utcTimeB :: UTCTime 3) call "diffUTCTime utcTimeA utcTimeB" So my program looks like as follows in a "do" expression: let utcTimeA = fromJust (parseTime defaultTimeLocale "%R%P" ta :: Maybe UTCTime) currentTime <- getCurrentTime let day = utctDay currentTime let utcTimeB = UTCTime day "09:30:00.0000000 UTC" let timeElapsed = diffUTCTime utcTimeA utcTimeB There are two difficulties I could not get around: 1) It seems that parseTime returns "Nothing". Why? 2) I knew "09:30:00.0000000 UTC" in the statement of "let utcTimeB ..." is incorrect, because it is a String instead of expected DiffTime type. After consulting Data.Time module (it just say "Data DiffTime"), I still cannot figure out how to get utcTimeB. Can someone help me? Thanks, Hong

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subtle dissimilarity between fromInteger and fromIntegral
by Hong Yang 30 Sep '09

30 Sep '09

Hi, Can someone explain the subtle dissimilarity between fromInteger and fromIntegral? Thanks, Hong

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Re: [Haskell-beginners] remove XML tags using Text.Regex.Posix
by Jan Jakubuv 30 Sep '09

30 Sep '09

Hi Robert, On Tue, Sep 29, 2009 at 12:25:07PM -0700, Robert Ziemba wrote: > I have been working with the regular expression package (Text.Regex.Posix). > My hope was to find a simple way to remove a pair of XML tags from a short > string. > > I have something like this "<tag>Data</tag>" and would like to extract > 'Data'. There is only one tag pair, no nesting, and I know exactly what the > tag is. > This is so simple that I would not recommend anything other than regular expressions. Use the following pattern: pat = "<tag>(.*)</tag>" It creates a group withing the matched string containing the data (it is done using parenthesis). Use `[[String]]` as a result type and you receive a list of matches where each match is described by a list of strings whose first member is the whole matched string (including <tag> and </tag>) and it is followed by values of groups (in our case we have just one group). Thus: *Main> "text<tag>data</tag>text" =~ pat :: [[String]] [["<tag>data</tag>","data"]] It is easy extract the data using `(!!)` and `head`: *Main> (!! 1) . head $ ("text<tag>7</tag>text" =~ pat :: [[String]]) "7" > My first attempt was this: > > "<tag>123</tag>" =~ "[^<tag>].+[^</tag>]"::String > > result: "123" > The problem with your pattern is that `[^<tag>]` doesn't mean what you think it does. Its meaning is “one character which is not `<`, `t`, `a`, or `>`” as Patrick already described in his mail. > Upon further experimenting I realized that it only works with more than 2 > digits in 'Data'. I occured to me that my thinking on how this regular > expression works was not correct - but I don't understand why it works at > all for 3 or more digits. > It doesn't work for all 3 or more digits: *Main> "<tag>tag</tag>" =~ "[^<tag>].+[^</tag>]" :: String "" Briefly, it doesn't work when the data contains one of characters `<`, `t`, `a`, `g`, `>`. Finally, consider using pat = "<tag>([^<]*)</tag>" which works with more tags in the same line as well. Sincerely, jan. -- Heriot-Watt University is a Scottish charity registered under charity number SC000278.

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remove XML tags using Text.Regex.Posix
by Robert Ziemba 30 Sep '09

30 Sep '09

I have been working with the regular expression package (Text.Regex.Posix). My hope was to find a simple way to remove a pair of XML tags from a short string. I have something like this "<tag>Data</tag>" and would like to extract 'Data'. There is only one tag pair, no nesting, and I know exactly what the tag is. My first attempt was this: "<tag>123</tag>" =~ "[^<tag>].+[^</tag>]"::String result: "123" Upon further experimenting I realized that it only works with more than 2 digits in 'Data'. I occured to me that my thinking on how this regular expression works was not correct - but I don't understand why it works at all for 3 or more digits. Can anyone help me understand this result and perhaps suggest another strategy? Thank you.

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