A solution with enums would severely suffer from the expression problem...
One would need to extent the enums every time one needs to support a new
function. Maybe could be solved with type classes, don't know.
On Mon, Apr 20, 2009 at 3:57 PM, Achim Schneider bars...@web.de wrote:
Lennart
Lennart Augustsson lenn...@augustsson.net wrote:
On Sun, Apr 19, 2009 at 10:43 PM, Peter Verswyvelen
bugf...@gmail.com wrote:
For example, suppose you have a predicate a - Bool, and a list of
these predicates [a - Bool], but you want to remove all functions
that are obviously equal in the
On Mon, Apr 20, 2009 at 7:57 AM, Achim Schneider bars...@web.de wrote:
Lennart Augustsson lenn...@augustsson.net wrote:
On Sun, Apr 19, 2009 at 10:43 PM, Peter Verswyvelen
bugf...@gmail.com wrote:
For example, suppose you have a predicate a - Bool, and a list of
these predicates [a -
Peter Verswyvelen bugf...@gmail.com wrote:
Sometimes I do miss the pragmatic C solution:- two function pointers
that are equal surely represent the same functions (although in C
nothing is really sure ;)
In haskell, they would, but C doesn't give you the same guarantee:
int evil = 0;
int
The problem is that you must note in the type signature the fact that
'a' must be a member of typeclass Eq in order to be able to use ==
count :: (Eq a) = a - [a] - Int
count x ys = length (filter (== x) ys)
JP
michael rice wrote:
Is there a general function to count list elements. I'm trying
