Re: [Haskell-cafe] Re: Embedding newlines into a string?
Things to avoid - HaskellWiki - 7 Related Links: http://www.haskell.org/haskellwiki/Things_to_avoid#Related_Links The link was broken (it had an extra chunk of '- Haskell Wiki' ;) ) so I fixed it. For that matter, the Common Hugs Messages link is broken too but I can't seem to find the page it should point to. -- Ariel J. Birnbaum ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Ariel, --- Ariel J. Birnbaum [EMAIL PROTECTED] wrote: Things to avoid - HaskellWiki - 7 Related Links: http://www.haskell.org/haskellwiki/Things_to_avoid#Related_Links The link was broken (it had an extra chunk of '- Haskell Wiki' ;) ) so I fixed it. Thank you; sorry about the broken link. For that matter, the Common Hugs Messages link is broken too but I can't seem to find the page it should point to. I just fixed it. It was supposed to be an external link to the following Web page: Some common Hugs error messages http://www.cs.kent.ac.uk/people/staff/sjt/craft2e/errors/allErrors.html I discovered that link originally under the following subsection of HaskellWiki: Learning Haskell - 2 Material - 2.9 Reference http://www.haskell.org/haskellwiki/Learning_Haskell#Reference This time, I have checked my updated link to verify that it works. ;-) Benjamin L. Russell ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Common Misunderstandings - HaskellWiki http://www.haskell.org/haskellwiki/Common_Misunderstandings I didn't find this one... maybe it should be in a more prominent place? Things to avoid - HaskellWiki http://www.haskell.org/haskellwiki/Things_to_avoid I thought of this but it has more discussions about style than pitfalls... it seems to me more oriented to people who know Haskell well and want to write better. Maybe a link from here to the above would be a good idea. Thanks for adding this case to the wiki =) -- Ariel J. Birnbaum ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Ariel, --- Ariel J. Birnbaum [EMAIL PROTECTED] wrote: Common Misunderstandings - HaskellWiki http://www.haskell.org/haskellwiki/Common_Misunderstandings I didn't find this one... maybe it should be in a more prominent place? Things to avoid - HaskellWiki http://www.haskell.org/haskellwiki/Things_to_avoid I thought of this but it has more discussions about style than pitfalls... it seems to me more oriented to people who know Haskell well and want to write better. Maybe a link from here to the above would be a good idea. Added: Things to avoid - HaskellWiki - 7 Related Links: http://www.haskell.org/haskellwiki/Things_to_avoid#Related_Links Thanks for adding this case to the wiki =) My pleasure! Benjamin L. Russell ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
That's exactly what I was thinking about, but your hanoi_shower only handles list of exactly one action, but you have to handle longer lists, too. This could be done with explicit recursion This seems to be a common pitfall for Haskell newcomers: mistaking a single-element list pattern (such as [x]) for a pattern that iterates over every element in the list. I can't seem to find a page with a list of common pitfalls and mistakes... is there such a thing? -- Ariel J. Birnbaum ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Ariel, Check out the following HaskellWiki pages: Common Misunderstandings - HaskellWiki http://www.haskell.org/haskellwiki/Common_Misunderstandings Things to avoid - HaskellWiki http://www.haskell.org/haskellwiki/Things_to_avoid Hope these help Benjamin L. Russell --- Ariel J. Birnbaum [EMAIL PROTECTED] wrote: That's exactly what I was thinking about, but your hanoi_shower only handles list of exactly one action, but you have to handle longer lists, too. This could be done with explicit recursion This seems to be a common pitfall for Haskell newcomers: mistaking a single-element list pattern (such as [x]) for a pattern that iterates over every element in the list. I can't seem to find a page with a list of common pitfalls and mistakes... is there such a thing? -- Ariel J. Birnbaum ___ 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] Re: Embedding newlines into a string?
Ariel, In response to your comment, since there was apparently no section devoted to pitfalls of iterating over lists, I have added the section 1.4 Iterating Over a List in the following HaskellWiki page; viz: Common Misunderstandings - HaskellWiki http://www.haskell.org/haskellwiki/Common_Misunderstandings#Iterating_Over_a_List Hope this helps Benjamin L. Russell --- Benjamin L. Russell [EMAIL PROTECTED] wrote: Ariel, Check out the following HaskellWiki pages: Common Misunderstandings - HaskellWiki http://www.haskell.org/haskellwiki/Common_Misunderstandings Things to avoid - HaskellWiki http://www.haskell.org/haskellwiki/Things_to_avoid Hope these help Benjamin L. Russell --- Ariel J. Birnbaum [EMAIL PROTECTED] wrote: That's exactly what I was thinking about, but your hanoi_shower only handles list of exactly one action, but you have to handle longer lists, too. This could be done with explicit recursion This seems to be a common pitfall for Haskell newcomers: mistaking a single-element list pattern (such as [x]) for a pattern that iterates over every element in the list. I can't seem to find a page with a list of common pitfalls and mistakes... is there such a thing? -- Ariel J. Birnbaum ___ 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 mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Benjamin L. Russel wrote: hanoi_shower ((a, b) : moves) | null moves = ... | otherwise == ... Luke Palmer wrote: More idiomatic pedantry: the way you will see most Haskellers write this style of function is by pattern matching rather than guards: hanoi_shower [] = ... hanoi_shower ((a,b):moves) = ... These two versions are semantically different! Benjamin's versions works for lists of length 1 or more, Luke's version works for lists of length 0 or more. Luke's version looks like a typical Haskell solution, which would be expressed in lispy syntax like this: (define hanoi_shower (lambda (xs) (cond ((null xs) (...)) (true, (let ((a, (first (first xs))) (b, (rest (first xs))) (moves, (rest xs))) (...) The pattern matching in Haskell takes care of both the cond and the let, there's no need for guards or to actually call null or any selector functions. A nice exercise may be to implement the map function using primitive recursion. Tillmann ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Ok; I rewrote my recursive version of hanoi, preserving my semantics (i.e., working for lists of length 1 or more, rather than 0 or more, to start with) in a more Haskell-idiomatic manner; viz: hanoi_general_recursive.hs: hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1) hanoi_shower :: Show a = [(a, a)] - String hanoi_shower [(a, b)] = unlines [Move ++ show a ++ to ++ show b ++ .] hanoi_shower ((a, b):moves) = unlines [Move ++ show a ++ to ++ show b ++ .] ++ hanoi_shower moves (I wanted to start out with lists of length 1 as a base case before extending the base case to lists of length 0 because Luke Palmer had already solved it for 0, and I didn't want just to copy his solution--I can't learn anything if I just do that.) In WinHugs: Main :load hanoi_general_recursive.hs Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2)) Move 'a' to 'b'. Move 'a' to 'c'. Move 'b' to 'c'. Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 1)) Move 'a' to 'c'. Main putStr (hanoi_shower (hanoi 1 2 3 2)) Move 1 to 2. Move 1 to 3. Move 2 to 3. Main putStr (hanoi_shower (hanoi 1 2 3 1)) Move 1 to 3. Ok; it works now. Now that I have successfully created a recursive version that preserves my original semantics, it is time to extend the base case to handle lists of length 0. (Notice that I added a base case of n == 0 to hanoi itself as well, in addition to hanoi_shower; leaving this out in hanoi results in an error of ERROR - C stack overflow on an argument of n == 0 discs:) hanoi_general_recursive_base_0.hs: hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 0 = [] | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1) hanoi_shower :: Show a = [(a, a)] - String hanoi_shower [] = hanoi_shower ((a, b):moves) = unlines [Move ++ show a ++ to ++ show b ++ .] ++ hanoi_shower moves Now, let's sit back and watch the fun in WinHugs: Main :load hanoi_general_recursive_base_0.hs Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2)) Move 'a' to 'b'. Move 'a' to 'c'. Move 'b' to 'c'. Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 0)) Main putStr (hanoi_shower (hanoi 1 2 3 2)) Move 1 to 2. Move 1 to 3. Move 2 to 3. Main putStr (hanoi_shower (hanoi 1 2 3 0)) Great! Just for reference, here's the code for the other versions for comparison: hanoi_general_list_comprehension_unwords.hs [Note: This version adds an extra space before the final '.' on each line.]: hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1) hanoi_shower :: Show a = [(a, a)] - String hanoi_shower moves = unlines [unwords [Move, show a, to, show b, .] | (a, b) - moves] -- hanoi_general_list_comprehension_unlines.hs: hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1) hanoi_shower :: Show a = [(a, a)] - String hanoi_shower moves = unlines [Move ++ show a ++ to ++ show b ++ . | (a, b) - moves] -- hanoi_general_map_unlines.hs: hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1) hanoi_shower :: Show a = [(a, a)] - String hanoi_shower moves = unlines (map move moves) where move (a, b) = Move ++ show a ++ to ++ show b ++ . -- hanoi_general_recursive.hs [Note: This version only works for lists of length 1 or more.]: hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1) hanoi_shower :: Show a = [(a, a)] - String hanoi_shower [(a, b)] = unlines [Move ++ show a ++ to ++ show b ++ .] hanoi_shower ((a, b):moves) = unlines [Move ++ show a ++ to ++ show b ++ .] ++ hanoi_shower moves -- hanoi_general_recursive_base_0.hs [Note: This program is just the program contained in the file hanoi_general_recursive.hs, but extended to process lists of length 0 or more.]: hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 0 = [] | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1)
[Haskell-cafe] Re: Embedding newlines into a string?
Benjamin L. Russell wrote: but got stuck on outputting newlines as part of the string; quoting is done by the show function in Haskell, so you have to take care to avoid calling show. your code calls show at two positions: (1) when you insert the newline into the string (2) when you output the string with respect to (1): you use (show '\n') to create a newline-only string, which produces a machine-readable (!) textual representation of '\n'. try the difference between '\n' and show '\n' to see what I mean. instead of using (show '\n'), you should simply use \n to encode the string of length 1 containing a newline character. with respect to (2): the type of your top-level expression is String, which is automatically print'ed by the interpreter. but print x = putStrLn (show x), so there is another call to show at this point. to avoid this call, write an IO action yourself. try the difference between putStrLn (hanoi ...) and print (hanoi ...) to see what I mean. Last, but not least, I would like to point out a different aproach to multiline output which is often used by Haskell programmers: The worker functions in this aproach produces a list of strings, which is joined together with newlines by the unlines function. In your case: hanoi_helper :: ... - [String] | ... = [Move ++ ...] | otherwise = hanoi_helper ... ++ hanoi_helper ... hanoi n = hanoi_helper 'a' 'b' 'c' n and in the interpreter one of these: hanoi 2 -- outputs a list mapM_ putStrLn (hanoi 2) -- outputs each move in a new line putStrLn (unlines (hanoi 2)) -- same as previous line Tillmann ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Ok; much better. Here's my new type signature and definition: hanoi.hs: hanoi :: Int - IO () hanoi n = mapM_ putStrLn (hanoi_helper 'a' 'b' 'c' n) hanoi_helper :: Char - Char - Char - Int - [String] hanoi_helper source using dest n | n == 1 = [Move ++ show source ++ to ++ show dest ++ .] | otherwise = hanoi_helper source dest using (n-1) ++ hanoi_helper source using dest 1 ++ hanoi_helper using source dest (n-1) Then in WinHugs (Version Sep 2006): Hugs :load hanoi.hs Main hanoi 2 Move 'a' to 'b'. Move 'a' to 'c'. Move 'b' to 'c'. Great! One minor question: I tried out both of your following suggestions: mapM_ putStrLn (hanoi 2) -- outputs each move in a new line putStrLn (unlines (hanoi 2)) -- same as previous line and discovered that putStrLn with unlines (the lower option) in fact generates one extra blank line at the end. Just curious as to why Benjamin L. Russell --- Tillmann Rendel [EMAIL PROTECTED] wrote: Benjamin L. Russell wrote: but got stuck on outputting newlines as part of the string; quoting is done by the show function in Haskell, so you have to take care to avoid calling show. your code calls show at two positions: (1) when you insert the newline into the string (2) when you output the string with respect to (1): you use (show '\n') to create a newline-only string, which produces a machine-readable (!) textual representation of '\n'. try the difference between '\n' and show '\n' to see what I mean. instead of using (show '\n'), you should simply use \n to encode the string of length 1 containing a newline character. with respect to (2): the type of your top-level expression is String, which is automatically print'ed by the interpreter. but print x = putStrLn (show x), so there is another call to show at this point. to avoid this call, write an IO action yourself. try the difference between putStrLn (hanoi ...) and print (hanoi ...) to see what I mean. Last, but not least, I would like to point out a different aproach to multiline output which is often used by Haskell programmers: The worker functions in this aproach produces a list of strings, which is joined together with newlines by the unlines function. In your case: hanoi_helper :: ... - [String] | ... = [Move ++ ...] | otherwise = hanoi_helper ... ++ hanoi_helper ... hanoi n = hanoi_helper 'a' 'b' 'c' n and in the interpreter one of these: hanoi 2 -- outputs a list mapM_ putStrLn (hanoi 2) -- outputs each move in a new line putStrLn (unlines (hanoi 2)) -- same as previous line Tillmann ___ 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] Re: Embedding newlines into a string?
Hi mapM_ putStrLn (hanoi 2) -- outputs each move in a new line putStrLn (unlines (hanoi 2)) -- same as previous line putStr (unlines (hanoi 2)) is what you want. Unlines puts a trailing new line at the end of every line, including the final one. putStrLn puts an additional trailing new line, so you get 2 at the end. mapM_ putStrLn == putStr . unlines Thanks Neil ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Benjamin L. Russell wrote: Ok; much better. Here's my new type signature and definition: hanoi :: Int - IO () hanoi_helper :: Char - Char - Char - Int - [String] If you want, you can separate the algorithm and the output processing even more by providing three functions of these types: hanoi :: Int - [(Char, Char)] hanoi_helper :: Char - Char - Char - Int - [(Char, Char)] hanoi_shower :: [(Char, Char)] - String and at the interpreter level: putStr (hanoi_shower (hanoi 2)) added value: you can easily use the output of hanoi for automated processing (e.g. testing, controlling a robot, producing an animation, counting the number of steps). You can go one step further if you consider that towers don't have to be named by single characters, but can be named by everything: hanoi :: a - a - a - Int - [(a, a)] hanoi_helper :: a - a - a - Int - [(a, a)] hanoi_shower :: Show a = [(a, a)] - String now you can use putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2)) to get the same result as above, but you are also allowed to write putStr (hanoi_shower (hanoi 1 2 3 2)) if you want to use numeric tower names. Tillmann ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Hi mapM_ putStrLn == putStr . unlines I'm wondering which (==) you mean here ;) Expression equality, defined by: instance (Arbitrary a, Eq b) = Eq (a - b) where f == g = forall x :: a, f x == g x Using QuickCheck to generate the values, and an Eq over IO (), which can be defined using the IO test modelling thing at last years Haskell Workshop. There are answers to all these things, even if Haskell can't express all of them quite like this :) Thanks Neil ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Neil Mitchell wrote: Unlines puts a trailing new line at the end of every line, including the final one. putStrLn puts an additional trailing new line, so you get 2 at the end. Thanks for that clarification. mapM_ putStrLn == putStr . unlines I'm wondering which (==) you mean here ;) Tillmann ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Wow, that's very general. So you want to divide hanoi
into a main function, a helper function, and a display
function.
I tried it out, and got this far so far:
hanoi :: a - a - a - Int - [(a, a)]
hanoi a b c n = hanoi_helper a b c n
hanoi_helper :: a - a - a - Int - [(a, a)]
hanoi_helper source using dest n
| n == 1 = [(source, dest)]
| otherwise = hanoi_helper source dest using (n-1)
++ hanoi_helper source using dest 1
++ hanoi_helper using source
dest (n-1)
hanoi_shower :: Show a = [(a, a)] - String
hanoi_shower [(a, b)] = Move ++ show a ++ to ++
show b ++ .
However, when I tried to run the code in WinHugs, this
is what I got:
Hugs :load hanoi_general.hs
Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2))
Program error: pattern match failure: hanoi_shower
[('a','b'),('a','c')] ++ ([] ++ hanoi_helper 'b' 'a'
'c' (2 - 1))
There seems to be a bug in hanoi_shower.
I'm still trying to debug hanoi_shower, but I need to
stop for today and continue on this tomorrow.
Thanks for your help so far! Perhaps I can get this
general version fully working tomorrow.
Benjamin L. Russell
--- Tillmann Rendel [EMAIL PROTECTED] wrote:
Benjamin L. Russell wrote:
Ok; much better. Here's my new type signature and
definition:
hanoi :: Int - IO ()
hanoi_helper :: Char - Char - Char - Int -
[String]
If you want, you can separate the algorithm and the
output processing
even more by providing three functions of these
types:
hanoi :: Int - [(Char, Char)]
hanoi_helper :: Char - Char - Char - Int -
[(Char, Char)]
hanoi_shower :: [(Char, Char)] - String
and at the interpreter level:
putStr (hanoi_shower (hanoi 2))
added value: you can easily use the output of hanoi
for automated
processing (e.g. testing, controlling a robot,
producing an animation,
counting the number of steps).
You can go one step further if you consider that
towers don't have to be
named by single characters, but can be named by
everything:
hanoi :: a - a - a - Int - [(a, a)]
hanoi_helper :: a - a - a - Int - [(a, a)]
hanoi_shower :: Show a = [(a, a)] - String
now you can use
putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2))
to get the same result as above, but you are also
allowed to write
putStr (hanoi_shower (hanoi 1 2 3 2))
if you want to use numeric tower names.
Tillmann
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Re: [Haskell-cafe] Re: Embedding newlines into a string?
Yes, they are. That's what perhaps Neil Mitchell means by mapM_ putStrLn == putStr . unlines And whether the trailing newline is to be called the last blank line depends upon the convention; The string that is output in both the cases contains a single newline character. Are you calling that a blank line at end? And I actually meant to reply to haskell-cafe as well; omitted haskell-cafe by mistake. Anyway, you got the same answer from two people! This time I am including haskell-cafe in the recipients. Regards, Abhay On Mon, Apr 14, 2008 at 4:05 PM, Benjamin L. Russell [EMAIL PROTECTED] wrote: Abhay Parvate, Thank you; that answered my question. Then, the following two lines of code should be equivalent: In hanoi.hs: hanoi n = mapM_ putStrLn (hanoi_helper 'a' 'b' 'c' n) In hanoi_unlines.hs: hanoi n = putStr (unlines(hanoi_helper 'a' 'b' 'c' n)) I tested them both out on WinHugs (Version Sep 2006); they both generated one blank line at the end. Benjamin L. Russell --- Abhay Parvate [EMAIL PROTECTED] wrote: unlines puts newline after each string; putStrLn puts newline after the given string. As a result, the output contains two newlines in the end. You can use putStr instead, since the resultant string from 'unlines' will have a newline at the end. On Mon, Apr 14, 2008 at 2:12 PM, Benjamin L. Russell [EMAIL PROTECTED] wrote: Ok; much better. Here's my new type signature and definition: hanoi.hs: hanoi :: Int - IO () hanoi n = mapM_ putStrLn (hanoi_helper 'a' 'b' 'c' n) hanoi_helper :: Char - Char - Char - Int - [String] hanoi_helper source using dest n | n == 1 = [Move ++ show source ++ to ++ show dest ++ .] | otherwise = hanoi_helper source dest using (n-1) ++ hanoi_helper source using dest 1 ++ hanoi_helper using source dest (n-1) Then in WinHugs (Version Sep 2006): Hugs :load hanoi.hs Main hanoi 2 Move 'a' to 'b'. Move 'a' to 'c'. Move 'b' to 'c'. Great! One minor question: I tried out both of your following suggestions: mapM_ putStrLn (hanoi 2) -- outputs each move in a new line putStrLn (unlines (hanoi 2)) -- same as previous line and discovered that putStrLn with unlines (the lower option) in fact generates one extra blank line at the end. Just curious as to why Benjamin L. Russell --- Tillmann Rendel [EMAIL PROTECTED] wrote: Benjamin L. Russell wrote: but got stuck on outputting newlines as part of the string; quoting is done by the show function in Haskell, so you have to take care to avoid calling show. your code calls show at two positions: (1) when you insert the newline into the string (2) when you output the string with respect to (1): you use (show '\n') to create a newline-only string, which produces a machine-readable (!) textual representation of '\n'. try the difference between '\n' and show '\n' to see what I mean. instead of using (show '\n'), you should simply use \n to encode the string of length 1 containing a newline character. with respect to (2): the type of your top-level expression is String, which is automatically print'ed by the interpreter. but print x = putStrLn (show x), so there is another call to show at this point. to avoid this call, write an IO action yourself. try the difference between putStrLn (hanoi ...) and print (hanoi ...) to see what I mean. Last, but not least, I would like to point out a different aproach to multiline output which is often used by Haskell programmers: The worker functions in this aproach produces a list of strings, which is joined together with newlines by the unlines function. In your case: hanoi_helper :: ... - [String] | ... = [Move ++ ...] | otherwise = hanoi_helper ... ++ hanoi_helper ... hanoi n = hanoi_helper 'a' 'b' 'c' n and in the interpreter one of these: hanoi 2 -- outputs a list mapM_ putStrLn (hanoi 2) -- outputs each move in a new line putStrLn (unlines (hanoi 2)) -- same as previous line Tillmann ___ 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 mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
On Apr 14, 2008, at 7:51 , Benjamin L. Russell wrote: hanoi_shower :: Show a = [(a, a)] - String hanoi_shower [(a, b)] = Move ++ show a ++ to ++ show b ++ . You've just specified via pattern match that hanoi_shower always gets a 1-element list. Is that really what you intended? -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] [EMAIL PROTECTED] system administrator [openafs,heimdal,too many hats] [EMAIL PROTECTED] electrical and computer engineering, carnegie mellon universityKF8NH ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Hi hanoi_shower [] = ... hanoi_shower ((a, b) : moves) = ... or (preferably) with map hanoi_shower moves = unlines (map show_move moves) where show_move (a, b) = ... A nice list comprehension works wonders in these situations: hanoi_shower moves = unlines [Move ++ show a ++ to ++ show b ++ . | (a,b) - moves] I would personally remove the . from the end, as its a list of commands, not sentences - but that is personal choice. I'd also use unwords, as its slightly shorter: hanoi_shower moves = unlines [unwords [Move, show a, to, show b] | (a,b) - moves] Thanks Neil ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Benjamin L. Russell wrote: Wow, that's very general. So you want to divide hanoi into a main function, a helper function, and a display function. I tried it out, and got this far so far: [...] hanoi_shower :: Show a = [(a, a)] - String hanoi_shower [(a, b)] = Move ++ show a ++ to ++ show b ++ . That's exactly what I was thinking about, but your hanoi_shower only handles list of exactly one action, but you have to handle longer lists, too. This could be done with explicit recursion hanoi_shower [] = ... hanoi_shower ((a, b) : moves) = ... or (preferably) with map hanoi_shower moves = unlines (map show_move moves) where show_move (a, b) = ... Note the use of unlines again. I decided to use where to introduce a local binding to avoid cluttering the top-level scope with function names. Tillmann ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
hanoi :: a - a - a - Int - [(a, a)] hanoi a b c n = hanoi_helper a b c n Note that now hanoi is exactly the same function as hanoi_helper, so you may as well just get rid of hanoi_helper. =) -Brent ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
Now it works; viz (in response to Brent Yorgey's suggestion, I have merged hanoi and hanoi_helper): hanoi_general_list_comprehension_unwords.hs (based on Neil Mitchell's suggestion, except for the trailing '.'): hanoi :: a - a - a - Int - [(a, a)] hanoi source using dest n | n == 1 = [(source, dest)] | otherwise = hanoi source dest using (n-1) ++ hanoi source using dest 1 ++ hanoi using source dest (n-1) hanoi_shower :: Show a = [(a, a)] - String hanoi_shower moves = unlines [unwords [Move, show a, to, show b, .] | (a, b) - moves] Then, in WinHugs: Main :load hanoi_general_list_comprehension_unwords.hs Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2)) Move 'a' to 'b' . Move 'a' to 'c' . Move 'b' to 'c' . Main putStr (hanoi_shower (hanoi 1 2 3 2)) Move 1 to 2 . Move 1 to 3 . Move 2 to 3 . Notwithstanding Neil's advice on removing the trailing '.', which I appreciate, I would still prefer to retain it, because, in the interests of literate programming, I would like a sequence of English sentences as commands acceptable even to an English teacher. So, to be pedantic and remove the ' ' before the '.' at each line: hanoi_shower portion of hanoi_general_list_comprehension_unlines.hs (based on Neil Mitchell's suggestion): hanoi_shower :: Show a = [(a, a)] - String hanoi_shower moves = unlines [Move ++ show a ++ to ++ show b ++ . | (a, b) - moves] Then, in WinHugs: Main :load hanoi_general_list_comprehension_unlines.hs Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2)) Move 'a' to 'b'. Move 'a' to 'c'. Move 'b' to 'c'. Main putStr (hanoi_shower (hanoi 1 2 3 2)) Move 1 to 2. Move 1 to 3. Move 2 to 3. Splendid! Now, just for fun, let's see what other versions also work: hanoi_shower portion of hanoi_general_map_unlines.hs (based on Tillman Rendel's suggestion): hanoi_shower :: Show a = [(a, a)] - String hanoi_shower moves = unlines (map move moves) where move (a, b) = Move ++ show a ++ to ++ show b ++ . Then, in WinHugs: Main :load hanoi_general_map_unlines.hs Main putStr (hanoi_shower (hanoi 'a' 'b' 'c' 2)) Move 'a' to 'b'. Move 'a' to 'c'. Move 'b' to 'c'. Main putStr (hanoi_shower (hanoi 1 2 3 2)) Move 1 to 2. Move 1 to 3. Move 2 to 3. Wonderful! Thanks especially to Neil Mitchell and Tillman Rendel for their constructive suggestions. Nevertheless, I'm still working on the recursive version. So far, I've gotten this far: hanoi_shower portion of hanoi_general_recursive.hs (based on Tillman Rendel's suggestion): hanoi_shower :: Show a = [(a, a)] - String hanoi_shower ((a, b) : moves) | null moves = Move ++ show a ++ to ++ show b ++ . | otherwise == Move ++ show a ++ to ++ show b ++ . ++ hanoi_shower moves However, in WinHugs, I get the following error: Hugs :load hanoi_general_recursive.hs ERROR file:hanoi_general_recursive.hs:11 - Syntax error in declaration (unexpected `}', possibly due to bad layout) I haven't used recursion in Haskell much so far; I've only used it in Scheme, so I'm not used to it. I need to go to lunch now, so I'll work on this part later. Perhaps I can get it to work after lunch Benjamin L. Russell --- Neil Mitchell [EMAIL PROTECTED] wrote: Hi hanoi_shower [] = ... hanoi_shower ((a, b) : moves) = ... or (preferably) with map hanoi_shower moves = unlines (map show_move moves) where show_move (a, b) = ... A nice list comprehension works wonders in these situations: hanoi_shower moves = unlines [Move ++ show a ++ to ++ show b ++ . | (a,b) - moves] I would personally remove the . from the end, as its a list of commands, not sentences - but that is personal choice. I'd also use unwords, as its slightly shorter: hanoi_shower moves = unlines [unwords [Move, show a, to, show b] | (a,b) - moves] Thanks Neil ___ 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] Re: Embedding newlines into a string?
On Apr 14, 2008, at 23:45 , Benjamin L. Russell wrote: hanoi_shower :: Show a = [(a, a)] - String hanoi_shower ((a, b) : moves) | null moves = Move ++ show a ++ to ++ show b ++ . | otherwise == Move ++ show a ++ to ++ show b ++ . ++ hanoi_shower moves `==' after the `otherwise'? (I think the error involving `}' is a side effect of the single ugliest part of Haskell syntax, which specifies the parser inserting `}' as necessary to try to get a parse.) -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] [EMAIL PROTECTED] system administrator [openafs,heimdal,too many hats] [EMAIL PROTECTED] electrical and computer engineering, carnegie mellon universityKF8NH ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Embedding newlines into a string?
On Tue, Apr 15, 2008 at 3:45 AM, Benjamin L. Russell [EMAIL PROTECTED] wrote: hanoi_shower ((a, b) : moves) | null moves = Move ++ show a ++ to ++ show b ++ . | otherwise == Move ++ show a ++ to ++ show b ++ . ++ hanoi_shower moves More idiomatic pedantry: the way you will see most Haskellers write this style of function is by pattern matching rather than guards: hanoi_shower [] = hanoi_shower ((a,b):moves) = Move ++ show a ++ to ++ show b ++ .\n ++ hanoi_shower moves Luke ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
