This is getting off-topic, but...
I don't like this solution at all :). It is very dependent on read being
a true inverse of show, which isn't always the case. Perhaps more
importantly though, I don't like the fact that it will readily convert
between Int and Float, etc. For instance, in almost
Niall Dalton writes:
| Hi,
|
| Its been a while since I've been using Haskell seriously, so I might simply
| have overlooked the answer..
|
| Is there an arbitrary precision real number library available for Haskell?
| IIRC, atleast GHC uses the GMP library, but only for integers?
Hi.
Th
Hello!
The paper
http://research.microsoft.com/~simonpj/papers/hmap/
by Ralf Laemmel and Simon Peyton Jones gave a reference
implementation of a 'cast' function. Here's another implementation:
cast:: (Show a, Read b) => a -> Maybe b
cast = read_as . show
where
read_as s = ca
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Thanks! These examples and counterexamples help a lot
in understanding.
Markus
> The reason is that sum only works on integer lists, whereas the given
> type for fun requires that the first argument be a function that works
> on lists of any type, such as length.
> If sum were allowed as argumen
[EMAIL PROTECTED] wrote:
>
> I think I get it. But to be sure:
> The forall means: The type for a may not be the same
> throughout the whole function. It just has to follow
> the pattern specified inside the braces.
Not exactly. In:
> > fun::(forall a.[a]->Int)->[b]->[c]->Int
> > fun f x y = f x
Hi,
Its been a while since I've been using Haskell seriously, so I might simply
have overlooked the answer..
Is there an arbitrary precision real number library available for Haskell?
IIRC, atleast GHC uses the GMP library, but only for integers?
Thanks,
niall
--
T
> Interestingly, when I want hugs to show me the type
> of
> > fun::(forall a.[a]->Int)->[b]->[c]->Int
> it tells me: ERROR - Use of fun requires at least 1 argument
> Why that? At least I have explicitely specified the type.
Hmm, ghci behaves properly.
But using hugs I get the same error :-(.
N
I think I get it. But to be sure:
The forall means: The type for a may not be the same
throughout the whole function. It just has to follow
the pattern specified inside the braces.
Interestingly, when I want hugs to show me the type
of
> fun::(forall a.[a]->Int)->[b]->[c]->Int
it tells me: ERROR -
> concrete: What is the difference between
> (forall b. Term b => b -> b) -> a -> a
> and
> (Term b) => (b -> b) -> a -> a
> ?
One may want:
fun f x y = f x + f y
for instance:
fun length [True, False] [1,2]
Therefore, I would say, you need typing a la
fun::(forall a.[a]->Int)->[b]->[c]->
> http://research.microsoft.com/~simonpj/papers/hmap/
The technique you describe in that paper is exactly what I was
wanting for many times. I often stopped using algebraic data types
because of the immense amount of boilerplate I had to introduce.
Thanks!
But now I have a question regarding th
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