On 6/11/08, Jonathan Cast [EMAIL PROTECTED] wrote:
This doesn't apply to f, though --- it has a function type, so the user
presumably already knows it won't be subject to updating, no?
Distinguishing functions from variables by the form of declaration,
rather than the type, seems somewhat
Ryan Ingram wrote:
sumns 0 = 0
sumns x = sumns (x-1) + n
Without the monomorphism restriction, computing n is a function call;
it is evaluated each time it is asked for.
I'm relatively new to Haskell, and since this topic already came up, I
was wondering if anyone could explain to me how
On Sat, 2008-06-14 at 17:19 +1000, Rafal Kolanski wrote:
Ryan Ingram wrote:
sumns 0 = 0
sumns x = sumns (x-1) + n
Without the monomorphism restriction, computing n is a function call;
it is evaluated each time it is asked for.
I'm relatively new to Haskell, and since this topic
On 6/11/08, Rex Page [EMAIL PROTECTED] wrote:
Please remind me, again, of the advantages of f being something different
from the formula defining it.
fibs !a !b = a : fibs b (a+b)
-- fibs :: Num a = a - a - [a]
n = head $ drop 100 $ fibs 1 1
-- n :: Integer (due to monomorphism
On Wed, 2008-06-11 at 20:24 -0700, Don Stewart wrote:
page:
Definition of f:
f = foldr (+) 0
Types:
0 :: (Num t) = t
foldr (+) 0 :: Num a = [a] - a
f :: [Integer] - Integer
Please remind me, again, of the advantages of f being something different
from the formula
Definition of f:
f = foldr (+) 0
Types:
0 :: (Num t) = t
foldr (+) 0 :: Num a = [a] - a
f :: [Integer] - Integer
Please remind me, again, of the advantages of f being something different from
the formula defining it.
- Rex Page
___
page:
Definition of f:
f = foldr (+) 0
Types:
0 :: (Num t) = t
foldr (+) 0 :: Num a = [a] - a
f :: [Integer] - Integer
Please remind me, again, of the advantages of f being something different
from the formula defining it.
Overloaded 'constants' take a dictionary as an
Neil Mitchell wrote:
http://haskell.org/hawiki/MonomorphismRestriction
Note to others (esp Cale): does this page not appear on the new wiki?
I did a very rough quick conversion:
http://www.haskell.org/haskellwiki/MonomorphismRestriction
The old wiki is locked, for obvious reasons. But
Hello,
I talked for a while with bd_ about this on #haskell, and I think maybe
I'm just being silly. But I can't get why:
lambda = \x - length (show x)
or
dot = length . show
is different from
pre x = length $ show x
I read about monomorphism restriction on the haskell 98 report, but I
