Email: richard.ad...@lvvwd.com
Tel. (702) 856-3627
-Original Message-
From: Ivan Perez [mailto:ivanperezdoming...@gmail.com]
Sent: Friday, February 10, 2012 12:28 PM
To: j...@repetae.net
Cc: Richard Adams; haskell-cafe@haskell.org
Subject: Re: [Haskell-cafe] Cannot understand liftM2
To understand how liftM2 achieves the cartesian product, I think
one way is to find liftM2's implementation and (=) implementation
as part of []'s instantiation of the Monad class.
You can find the first in Control.Monad, and the second in
the standard prelude.
Lists are monads, and as John
Nice explanation. However, at
http://stackoverflow.com/questions/4119730/cartesian-product it was pointed
out that this
cartProd :: [a] - [b] - [(a, b)]
cartProd = liftM2 (,)
is equivalent to the cartesian product produced using a list comprehension:
cartProd xs ys = [(x,y) | x - xs, y - ys]
A good first step would be understanding how the other entry works:
cartProd :: [a] - [b] - [(a,b)]
cartProd xs ys = do
x - xs
y - ys
return (x,y)
It is about halfway between the two choices.
John
On Thu, Feb 9, 2012 at 9:37 AM, readams richard.ad...@lvvwd.com
Using the cool lambdabot pointless utility I found out that:
\x - snd(x) - fst(x)
is the same as:
liftM2 (-) snd fst
I like the elegance of this but I cannot reconcile it with its type. I
can't understand it.
I check the signature of liftM2 and I get:
Prelude :t liftM2
Prelude liftM2 ::
Hi Nicola,
Am Montag, den 11.12.2006, 17:15 +0100 schrieb Nicola Paolucci:
Using the cool lambdabot pointless utility I found out that:
\x - snd(x) - fst(x)
is the same as:
liftM2 (-) snd fst
I like the elegance of this but I cannot reconcile it with its type. I
can't understand
Quoth Nicola Paolucci, nevermore:
Prelude :t liftM2
Prelude liftM2 :: (Monad m) = (a1 - a2 - r) - m a1 - m a2 - m r
Can someone help me understand what's happening here ?
What does a Monad have to do with a simple subtraction ?
What is actually the m of my example ?
I'm honestly not sure
On 11/12/06, Nicola Paolucci [EMAIL PROTECTED] wrote:
Hi All,
I'm loving learning Haskell quite a bit.
It is stretching my brain but in a delightfull way.
I've googled, I've hoogled but I haven't found a clear explanation for
what exactly liftM2 does in the context below.
Using the cool
Hi Cale !
On 12/11/06, Cale Gibbard [EMAIL PROTECTED] wrote:
The monad instance which is being used here is the instance for ((-)
e) -- that is, functions from a fixed type e form a monad.
So in this case:
liftM2 :: (a1 - a2 - r) - (e - a1) - (e - a2) - (e - r)
I bet you can guess what this
Hi All, Hi Cale,
Can you tell me if I understood things right ? Please see below ...
On 12/11/06, Cale Gibbard [EMAIL PROTECTED] wrote:
The monad instance which is being used here is the instance for ((-)
e) -- that is, functions from a fixed type e form a monad.
So in this case:
liftM2 ::
The interpreter infers that m = (e -) because of the types of snd and fst.
When snd and fst are considered as monadic computations in the (e -)
monad, there types are:
Prelude :t fst
fst :: (a, b) - a
Prelude :t snd
snd :: (a, b) - b
Note that: (a, b) - a =~= m awhere m x = (a,b) - x
So
On Tuesday 12 December 2006 08:57, Nicola Paolucci wrote:
- How do I know - or how does the interpreter know - that the m of
this example is an instance of type ((-) e) ?
- Is it always like that for liftM2 ? Or is it like that only because
I used the function (-) ?
It's the snd that forces
On 12/11/06, Nicola Paolucci [EMAIL PROTECTED] wrote:
I am trying to understand this bit by bit I am sorry if this is either
very basic and easy stuff, or if all I wrote is completely wrong and I
did not understand anything. :D Feedback welcome.
Don't apologise - I, for one, am finding this
Hi Nicolas,
On 12/11/06, Nicolas Frisby [EMAIL PROTECTED] wrote:
The interpreter infers that m = (e -) because of the types of snd and fst.
When snd and fst are considered as monadic computations in the (e -)
monad, there types are:
Prelude :t fst
fst :: (a, b) - a
Prelude :t snd
snd :: (a,
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