On 28/05/07, Rodrigo Queiro <[EMAIL PROTECTED]> wrote:
After a little too long trying, I managed to get code from which the type
system will infer that type, without using 'undefined':
Yes, but you do it by writing a coerce :: a -> b, which is surely cheating.
--
-David House, [EMAIL PROTECTED
After a little too long trying, I managed to get code from which the type
system will infer that type, without using 'undefined':
*Main> :t map
map :: ((a -> a1) -> [a]) -> [a1]
import System.IO.Unsafe
import Data.IORef
import Prelude hiding (map)
import qualified Prelude as P
coerce a = unsafe
map :: ((a -> b) -> [a]) -> [b]
I am not following here: what do you mean? Clearly, this is not a
valid typing for map. Moreover, modulo undefinedness, there are no
functions with this typing.
map _ = []
Ah, well, and that one, of course... :-)
Cheers,
Stefan
On Sun, May 27, 2007 at 02:10:40PM +0200, Stefan Holdermans wrote:
> Andrew,
>
> >Which is beautifully symmetric. Alternatively, you can think about
> >how you actually use it:
> >
> > map :: ((a -> b) -> [a]) -> [b]
>
> I am not following here: what do you mean? Clearly, this is not a
> vali
On 5/27/07, Andrew Coppin <[EMAIL PROTECTED]> wrote:
map :: (a -> b) -> [a] -> [b]
map :: (a -> b) -> ([a] -> [b])
Which is beautifully symmetric. Alternatively, you can think about how
you actually use it:
map :: ((a -> b) -> [a]) -> [b]
No, now you're confusing things. The uncurried
Andrew,
Which is beautifully symmetric. Alternatively, you can think about
how you actually use it:
map :: ((a -> b) -> [a]) -> [b]
I am not following here: what do you mean? Clearly, this is not a
valid typing for map. Moreover, modulo undefinedness, there are no
functions with this t
On Sun, May 27, 2007 at 11:12:47AM +0100, Andrew Coppin wrote:
> So, clearly, (->) is not associative.
That's right.
> Now I'm left wondering why you can bracket the type for map in two
> different ways
Are you sure you can?
Best regards
Tomasz
___
Ha
As I lay in bed last night, a curios fact occurred to me. (Yes, I don't
get out very much...)
Consider the map function:
map :: (a -> b) -> [a] -> [b]
There are two ways you can think about this function. First, you can see
it as meaning
map :: (a -> b) -> ([a] -> [b])
Which is beautifu
