A third way is to put each datatype declaration in a separate module and
use qualified imports.
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a plausable workaround in haskell 98 is to use a typeclass like so
class EdgyNodelike a where
node :: a -> Int
edge0 :: a -> a
edge1 :: a -> a
instance EdgyNodelike BFSM where
node = bfsmnode
edge0 = bfsmedge0
edge1 = bfsmedge1
instance EdgyNodelik
G'day all.
On Tue, Jan 07, 2003 at 10:01:38PM -0600, Kim-Ee Yeoh wrote:
> Why can't field labels have the same name in different types?
Because it would generate two functions with the same name.
> I'm not an expert on programming languages, but doesn't it seem
> that Haskell, as a strongly-typ
Dear Haskellers,
Why can't field labels have the same name in different types?
Here's some actual code on finite state automata I'm working on:
> data BMC
> = BMC { node :: !Int,
> threshold :: !Float,
> edge0 :: BMC,
> edge1 :: BMC }
> data BFSM
Hal Daume III writes:
>You are correct. You can't say:
>
> let x = if ??? then () else (5::Integer)
I meant to ask if when you do this next
let x = if ??? then toDyn () else toDyn (5::Integer)
can you then dispatch on x's type.
In my sample code, I forgot the toDyns. eek!
Anyway, this nex
G'day all.
On Wed, Jan 08, 2003 at 08:42:20AM +1100, Thomas Conway wrote:
> I'm fairly new to Haskell, but worked on Mercury until recently.
> Mercury has a type "univ" which might be declared something like:
>
> data Univ = Univ a
The equivalent would be:
data Univ = forall a. Univ a
> I'm fairly new to Haskell, but worked on Mercury until recently.
> Mercury has a type "univ" which might be declared something like:
you would write this:
data Univ = forall a . Univ a
> mkUniv :: a -> Univ
right, 'mkUniv = Univ' is sufficient.
> getValue :: Univ -> Maybe a
yes, this is the
On Wed, 8 Jan 2003 07:19 am, Hal Daume III wrote:
> Another way to do it would be to define a "universal" type:
> > data Univ = UUnit () | UInteger Integer | UDouble Double | ...
>
I'm fairly new to Haskell, but worked on Mercury until recently.
Mercury has a type "univ" which might be declared so
You are correct. You can't say:
let x = if ??? then () else (5::Integer)
because then the compiler can't assign a type to x.
However, to achieve what you want, you might try something like this
(untested code follows):
> main = do
>l <- getLine
>let x = case maybeRead l of
>
Just want to make sure I understand.
There does not exist a Haskell implementation-specific extension which
provides the sort of dynamic type that would allow one to discriminate
at run time between built-in types; right? eg,
main=do
object<-readLn
case object of --or some such syntax
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