[Haskell-beginners] Help in pattern matching

[Joe Fox]

Thanks Stepen

will try out factoring

On Mon, Mar 8, 2010 at 7:44 PM, Stephen Tetley <stephen.tetley at gmail.com>wrote:

> Hi Joe - I've cc'ed back to list...
>
> I'm supposing from what you've said that only DataConst1 & DataConst2
> have ids - so I've made a type synonym VId
>
>
> type VId = Int
>
> data TestData = DataConst1 VId String
>              | DataConst2 VId Int
>               | DataConst3 Sting String
>              | DataConst4 String Int
>
>
> If both DataConst1 & DataConst2 have VId's the you right - you have to
> pattern match for both:
>
> setVid_to_4 :: TestData -> TestData
> setVid_to_4 (DataConst1 _ y)     = DataConst1 4 y
> setVid_to_4 (DataConst2 _ y)     = DataConst2 4 y
> setVid_to_4 var                  = var
>
>
> If all the cases of TestData had a VId e.g.:
>
>
> data TestData = DataConst1 VId String
>              | DataConst2 VId Int
>              | DataConst3 VId Sting String
>              | DataConst4 Vid String Int
>
> ... then you could factor out VId to get these two data types
>
>
> data FactoredTD = FactoredTD VId TD
>
> data TD = DC1 String
>         | DC2 Int
>         | DC3 String String
>         | DC4 String Int
>
> setVid_to_4 :: FactoredTD -> FactoredTD
> setVid_to_4 (FactoredTD _ y)     = FactoredTD 4 y
>
>
> This factoring transformation is quite common.
>
> For your particular data type you could factor the other way:
>
>
> data TestDataX = DataConst1_or_2 VId (Either String Int)
>                | DataConst3 Sting String
>               | DataConst4 String Int
>
>
> setVid_to_4 :: TestDataX -> TestDataX
> setVid_to_4 (DataConst1_or_2 _ y) = DataConst1_or_2 4 y
> setVid_to_4 y                     = y
>
> However this factoring is quite horrible - I can't think of anywhere
> where I've seen it.
>
> Generally I wouldn't be too concerned about redundancy in pattern
> matching. If its easy to consistently name all the constructors in an
> algebraic data type, then that's a strong indication that you've put
> it into a good form.
>
> If you're still taxed by duplicate pattern matching you can then write
> projection and modification functions to do it once and once only:
>
>
> -- Has to return a Maybe as DC3 and DC4 have no ID
> getVId :: TestData -> Maybe VId
> getVid (DataConst1 iden _) = Just iden
> getVid (DataConst2 iden _) = Just iden
> getVid _                   = Nothing
>
>
> -- note - uses an update function that accesses the original value
> -- rather than a simple replace
> --
> updateVid :: (Vid -> Vid) -> TestData -> TestData
> updateVid f  (DataConst1 iden y) = DataConst1 (f iden) y
> updateVid f  (DataConst2 iden y) = DataConst2 (f iden) y
> updateVid f  var                 = var
>
> incrementVid :: TestData -> TestData
> incrementVid = updateVid (\x -> x+1)
>
> Best wishes
>
> Stephen
>
> On 8 March 2010 12:19, Joe Fox <fox.joe87 at gmail.com> wrote:
> > Hi Guys,
> >
> > First thanks for all yours inputs.
> >
> > Actually test is just a simple function , actually my actual function is
> > pretty complex and huge ...
> >
> > let me redefine Test function , i guess i took a bad example which
> created
> > lot of confusion
> >
> > the type of the test functions will be  something like this  , (i don't
> have
> > any ambiguity on type of the function, )
> >
> > test:: TestData -> TestData
> >
> > My question is like is there any way i can match multiple pattern in a
> > function statement .
> >
> > like this (I know this doesnt work , but just trying to explain whats on
> my
> > mind)
> >
> >
> > test var@(DataConst1 x y) == var@(DataConst2 x y) = var{id=4} -- let's
> say
> > id is the record holder for first record
> > test var@(_) = var --- any for other two its returns the same
> >
> > the working code would be like
> > test var@(DataConst1 x y) = var{id=4}
> > test var@(DataConst2 x y) = var{id=4} --- I want these two lines in one
> > line, because the logic is same
> > test var@(_) = var
> >
> > some thing of this sort.
> > I hope i didnt add to the confusion
> >
> >
> > Thanks
> > Joe
> >
>
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