A few newbie questions about tracing/debugging and order of execution
Heya, I decided to play around with Haskell and see what it's like. I used a small problem to explore it. Basically, given two words, find the least number of 1 letter changes that will go from one word to the other (e.g. for "fig" and "dog" either fig -> fog -> dog or fig -> dig -> dog). I came up with a solution, but I have to say it was quite difficult to get any debugging information, and when I did, the result was fairly surprising! Here's the out put I got: dagger:~/stuff$ ./figt /usr/share/dic/twords fig dog Word: dog visited: ["bit","bin","bid","bib","bug","bog","beg","bag","fix","fit","fir","fin","fie","fib","fog","fag","wig","rig","pig","jig","gig","dig","big","fig"] Word: dig visited: ["big","fig"] Word: fig visited: [] fig -> dig -> dog The first thing I found surprising was that the debug output comes in the "reverse" order - I would have expected it to be the output for fig, then dig, then dog. Secondly, I'm not entirely sure why I only got output for the statements along the "solution" path. If I try a combination that has no solution, I get no output at all. I got the program working, and I'm quite intrigued, but I figure it's probably important to figure out why the above things happen... I'm using GHC 6.4.1, but I get the same thing if I use runhugs. Thanks! --------- figt.hs -------------- import System import System.IO import Data.Set import Debug.Trace type Word = String data Path = None | Path Word Path instance Show Path where showsPrec _ path = showsPath path where showsPath :: Path -> ShowS showsPath None = ("<None>"++) showsPath (Path w None) = (w ++) showsPath (Path w p) = (showsPath p) . ((" -> " ++ w)++) findPath :: Word -> Word -> Set String -> Path findPath orig target allWords = let gen = singleton orig in iteratePath [(Path orig None)] gen [] where iteratePath :: [Path] -> Set String -> [String] -> Path iteratePath [] _ _ = None iteratePath (None: _) _ _ = None iteratePath (fullPath@(Path w _):remain) generated visited = let allPossibilities = findPossibleWords w existant = Prelude.filter (\p -> member p allWords) allPossibilities new = Prelude.filter (\e -> not (member e generated)) existant marked = foldl (\x y -> insert y x) generated new thingum = trace ("Word: " ++ w ++ "\n\tvisited: " ++ (show visited) ++ "\n") fullPath todo = remain ++ Prelude.map (\u -> (Path u thingum)) new in if target == w then thingum else if length todo > 0 then iteratePath todo marked (w:visited) else None findPossibleWords :: String -> [String] findPossibleWords word = concatMap enumChars (splits) where enumChars (prefix, _:cs) = Prelude.map (\c -> prefix ++ [c] ++ cs) ['a'..'z'] splits = Prelude.map (\x -> splitAt x word) [0..((length word) -1)] findAllWordsSizeN :: String -> Int -> Set String findAllWordsSizeN content wordLength = fromDistinctAscList (Prelude.filter (\x -> wordLength == (length x)) (lines content)) main :: IO () main = do [wordsFile, orig, target] <- getArgs contents <- readFile wordsFile allWordsSizeN <- (return (findAllWordsSizeN contents (length orig))) putStrLn (show (findPath orig target allWordsSizeN ))
On Dec 28, 2005, at 6:10 AM, Hunter Kelly wrote:
Heya, I decided to play around with Haskell and see what it's like. I used a small problem to explore it. Basically, given two words, find the least number of 1 letter changes that will go from one word to the other (e.g. for "fig" and "dog" either fig -> fog -> dog or fig -> dig -> dog).
I came up with a solution, but I have to say it was quite difficult to get any debugging information, and when I did, the result was fairly surprising!
I see you are using Debug.Trace to generate your debug messages. The 'trace' function is a sort of strange one, because it breaks the usual rules that Haskell follows; it allows you to generate output in the middle of a pure computation. It works by generating output _when it is evaluated_. However, without the IO monad to make everything sequenced and well-behaved, it can be difficult to predict when that will occur. In the particular program you posted, the 'trace' thunk is not evaluated until after the recursive call has completed, which gives the reversed output. Furthermore, when there is no solution, the 'trace' thunk isn't evaluated at all (the magic of laziness!), so you never see that output. To make your traces show up where you expect, you need to make sure that your trace function gets forced earlier and on both success and failure paths. Since your function is written as a big let...in if ... block you can do something like this: ..... = let ... .... in trace traceString (if ... then else ) That way, the trace will be output before the 'if' is evaluated, so you will get output for both branches. I can't tell from a quick inspection if it will return the results in the order you expect, but I think it may. Rob Dockins Speak softly and drive a Sherman tank. Laugh hard; it's a long way to the bank. -- TMBG
Yes, thank you, that did the trick! It produced the output for all steps, and in the order I would expect. Are there other techniques that people use to get debugging output? H On 12/28/05, Robert Dockins <robdockins@fastmail.fm> wrote:
On Dec 28, 2005, at 6:10 AM, Hunter Kelly wrote:
Heya, I decided to play around with Haskell and see what it's like. I used a small problem to explore it. Basically, given two words, find the least number of 1 letter changes that will go from one word to the other (e.g. for "fig" and "dog" either fig -> fog -> dog or fig -> dig -> dog).
I came up with a solution, but I have to say it was quite difficult to get any debugging information, and when I did, the result was fairly surprising!
I see you are using Debug.Trace to generate your debug messages. The 'trace' function is a sort of strange one, because it breaks the usual rules that Haskell follows; it allows you to generate output in the middle of a pure computation. It works by generating output _when it is evaluated_. However, without the IO monad to make everything sequenced and well-behaved, it can be difficult to predict when that will occur. In the particular program you posted, the 'trace' thunk is not evaluated until after the recursive call has completed, which gives the reversed output. Furthermore, when there is no solution, the 'trace' thunk isn't evaluated at all (the magic of laziness!), so you never see that output.
To make your traces show up where you expect, you need to make sure that your trace function gets forced earlier and on both success and failure paths. Since your function is written as a big let...in if ... block you can do something like this:
..... = let ... .... in trace traceString (if ... then else )
That way, the trace will be output before the 'if' is evaluated, so you will get output for both branches. I can't tell from a quick inspection if it will return the results in the order you expect, but I think it may.
Rob Dockins
Speak softly and drive a Sherman tank. Laugh hard; it's a long way to the bank. -- TMBG
Hi, Mostly I just use ghci. If something isn't working (or even sometimes if it is), I break it down into smaller parts until I can understand and test each part separately. Usually once the functions are broken into small, easy to manage pieces, it becomes obvious where the error is. Sometimes though, in certain kinds of algorithms, you have some code which runs for quite a long time before producing an example of data on which it fails. For this purpose, Debug.Trace is quite handy for extracting a report of the input data where things are breaking, but I usually find that these cases are pretty rare, and Debug.Trace is still no substitute for decomposing your problem further. - Cale On 28/12/05, Hunter Kelly <retnuh@gmail.com> wrote:
Yes, thank you, that did the trick! It produced the output for all steps, and in the order I would expect.
Are there other techniques that people use to get debugging output?
H
On 12/28/05, Robert Dockins <robdockins@fastmail.fm> wrote:
On Dec 28, 2005, at 6:10 AM, Hunter Kelly wrote:
Heya, I decided to play around with Haskell and see what it's like. I used a small problem to explore it. Basically, given two words, find the least number of 1 letter changes that will go from one word to the other (e.g. for "fig" and "dog" either fig -> fog -> dog or fig -> dig -> dog).
I came up with a solution, but I have to say it was quite difficult to get any debugging information, and when I did, the result was fairly surprising!
I see you are using Debug.Trace to generate your debug messages. The 'trace' function is a sort of strange one, because it breaks the usual rules that Haskell follows; it allows you to generate output in the middle of a pure computation. It works by generating output _when it is evaluated_. However, without the IO monad to make everything sequenced and well-behaved, it can be difficult to predict when that will occur. In the particular program you posted, the 'trace' thunk is not evaluated until after the recursive call has completed, which gives the reversed output. Furthermore, when there is no solution, the 'trace' thunk isn't evaluated at all (the magic of laziness!), so you never see that output.
To make your traces show up where you expect, you need to make sure that your trace function gets forced earlier and on both success and failure paths. Since your function is written as a big let...in if ... block you can do something like this:
..... = let ... .... in trace traceString (if ... then else )
That way, the trace will be output before the 'if' is evaluated, so you will get output for both branches. I can't tell from a quick inspection if it will return the results in the order you expect, but I think it may.
Rob Dockins
Speak softly and drive a Sherman tank. Laugh hard; it's a long way to the bank. -- TMBG
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On Dec 28, 2005, at 12:07 PM, Hunter Kelly wrote:
Yes, thank you, that did the trick! It produced the output for all steps, and in the order I would expect.
Are there other techniques that people use to get debugging output?
Well, if you are writing code in the IO monad, obviously you can just insert 'putStrLn', just like in your favorite imperative language. For pure code, I generally try to write small functions and then compose them to do larger tasks. That way I can exercise each smaller function individually and see that I get the results I expect. It is also more likely that your functions will just be "obviously correct". If that isn't possible for some reason I go to Debug.Trace and try to find good places to sprinkle in the calls to 'trace'. I find that attaching them to control structures usually works best. If that doesn't cut it for you, there are also some interactive Haskell debuggers. (http://www.haskell.org/libraries/#tracing). Rob Dockins Speak softly and drive a Sherman tank. Laugh hard; it's a long way to the bank. -- TMBG
I agree with the suggestions of the other posters. Sometimes I'll also use "error <data to print>" to get a "one time" breakpoint :-) cheers "Hunter Kelly" <retnuh@gmail.com> wrote in message news:75a1c1310512280907w6907badey7d4187592ce7f3fd@mail.gmail.com... Yes, thank you, that did the trick! It produced the output for all steps, and in the order I would expect. Are there other techniques that people use to get debugging output? H On 12/28/05, Robert Dockins <robdockins@fastmail.fm> wrote:
On Dec 28, 2005, at 6:10 AM, Hunter Kelly wrote:
Heya, I decided to play around with Haskell and see what it's like. I used a small problem to explore it. Basically, given two words, find the least number of 1 letter changes that will go from one word to the other (e.g. for "fig" and "dog" either fig -> fog -> dog or fig -> dig -> dog).
I came up with a solution, but I have to say it was quite difficult to get any debugging information, and when I did, the result was fairly surprising!
I see you are using Debug.Trace to generate your debug messages. The 'trace' function is a sort of strange one, because it breaks the usual rules that Haskell follows; it allows you to generate output in the middle of a pure computation. It works by generating output _when it is evaluated_. However, without the IO monad to make everything sequenced and well-behaved, it can be difficult to predict when that will occur. In the particular program you posted, the 'trace' thunk is not evaluated until after the recursive call has completed, which gives the reversed output. Furthermore, when there is no solution, the 'trace' thunk isn't evaluated at all (the magic of laziness!), so you never see that output.
To make your traces show up where you expect, you need to make sure that your trace function gets forced earlier and on both success and failure paths. Since your function is written as a big let...in if ... block you can do something like this:
..... = let ... .... in trace traceString (if ... then else )
That way, the trace will be output before the 'if' is evaluated, so you will get output for both branches. I can't tell from a quick inspection if it will return the results in the order you expect, but I think it may.
Rob Dockins
Speak softly and drive a Sherman tank. Laugh hard; it's a long way to the bank. -- TMBG
participants (4)
-
Cale Gibbard -
Hunter Kelly -
Robert Dockins -
Srinivas Nedunuri