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Today's Topics:
1. Renato Leal has invited you to Dropbox (Dropbox)
2. sqrt root issues (Bryce Verdier)
3. Re: sqrt root issues (Sean Perry)
4. Re: sqrt root issues (Daniel Fischer)
5. More Deserialization Woes (Tom Hobbs)
6. Re: More Deserialization Woes (Benjamin Edwards)
----------------------------------------------------------------------
Message: 1
Date: Wed, 30 Jun 2010 19:03:07 +0000
From: Dropbox <no-re...@dropbox.com>
Subject: [Haskell-beginners] Renato Leal has invited you to Dropbox
To: beginners@haskell.org
Message-ID: <20100630190307.e690d307...@mailman.dropbox.com>
Content-Type: text/plain; charset="utf8"
We're excited to let you know that Renato Leal has invited you to Dropbox!
Renato Leal has been using Dropbox to sync and share files online and across
computers, and thought you might want it too.
Visit http://www.dropbox.com/link/20.R3oEQnNe2Y/NjIzNjAwNTgxNw to get started.
- The Dropbox Team
____________________________________________________
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Message: 2
Date: Wed, 30 Jun 2010 20:01:41 -0700
From: Bryce Verdier <bryceverd...@gmail.com>
Subject: [Haskell-beginners] sqrt root issues
To: haskellbeginners <beginners@haskell.org>
Message-ID: <4c2c0515.8010...@gmail.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Hi all,
I'm having a real hard time wrapping my head around a problem;
Here is a basic, no frills, function.
prime divisor divided
| divided == 1 = True
| divisor== divided= True
| mod divided divisor== 0 = False
| otherwise = next_prime
where next_prime = prime (divisor + 2) divided
and the calling function:
is_prime input = prime 3 input
I'm trying to get the square root of input as an Integer.
In GHCI:
:t (toInteger $ round $ sqrt 25)
(toInteger $ round $ sqrt 25) :: Integer
but if I change is_prime input = prime 3 (toInteger $ round $ sqrt input)
I get this error:
problem3.hs:19:33:
No instance for (RealFrac Integer)
arising from a use of `round' at problem3.hs:19:33-37
Possible fix: add an instance declaration for (RealFrac Integer)
In the first argument of `($)', namely `round'
In the second argument of `($)', namely `round $ sqrt input'
In the expression: toInteger $ round $ sqrt input
problem3.hs:19:41:
No instance for (Floating Integer)
arising from a use of `sqrt' at problem3.hs:19:41-50
Possible fix: add an instance declaration for (Floating Integer)
In the second argument of `($)', namely `sqrt input'
In the second argument of `($)', namely `round $ sqrt input'
In the expression: toInteger $ round $ sqrt input
Failed, modules loaded: none.
Can someone please help me out and tell me what I'm missing?
Thanks in advance,
Bryce
------------------------------
Message: 3
Date: Wed, 30 Jun 2010 21:08:08 -0700
From: Sean Perry <sha...@speakeasy.net>
Subject: Re: [Haskell-beginners] sqrt root issues
To: Bryce Verdier <bryceverd...@gmail.com>
Cc: haskellbeginners <beginners@haskell.org>
Message-ID: <1277957288.4698.125.ca...@turion>
Content-Type: text/plain
On Wed, 2010-06-30 at 20:01 -0700, Bryce Verdier wrote:
> Hi all,
>
> I'm having a real hard time wrapping my head around a problem;
>
> Here is a basic, no frills, function.
> prime divisor divided
> | divided == 1 = True
> | divisor== divided= True
> | mod divided divisor== 0 = False
> | otherwise = next_prime
> where next_prime = prime (divisor + 2) divided
>
> and the calling function:
> is_prime input = prime 3 input
>
> I'm trying to get the square root of input as an Integer.
>
> In GHCI:
> :t (toInteger $ round $ sqrt 25)
> (toInteger $ round $ sqrt 25) :: Integer
>
> but if I change is_prime input = prime 3 (toInteger $ round $ sqrt input)
>
> I get this error:
> problem3.hs:19:33:
> No instance for (RealFrac Integer)
> arising from a use of `round' at problem3.hs:19:33-37
> Possible fix: add an instance declaration for (RealFrac Integer)
> In the first argument of `($)', namely `round'
> In the second argument of `($)', namely `round $ sqrt input'
> In the expression: toInteger $ round $ sqrt input
>
> problem3.hs:19:41:
> No instance for (Floating Integer)
> arising from a use of `sqrt' at problem3.hs:19:41-50
> Possible fix: add an instance declaration for (Floating Integer)
> In the second argument of `($)', namely `sqrt input'
> In the second argument of `($)', namely `round $ sqrt input'
> In the expression: toInteger $ round $ sqrt input
> Failed, modules loaded: none.
>
> Can someone please help me out and tell me what I'm missing?
>
Prelude> let x = 25
Prelude> toInteger $ round $ sqrt x
<interactive>:1:12:
No instance for (RealFrac Integer)
arising from a use of `round' at <interactive>:1:12-16
Possible fix: add an instance declaration for (RealFrac Integer)
In the first argument of `($)', namely `round'
In the second argument of `($)', namely `round $ sqrt x'
In the expression: toInteger $ round $ sqrt x
<interactive>:1:20:
No instance for (Floating Integer)
arising from a use of `sqrt' at <interactive>:1:20-25
Possible fix: add an instance declaration for (Floating Integer)
In the second argument of `($)', namely `sqrt x'
In the second argument of `($)', namely `round $ sqrt x'
In the expression: toInteger $ round $ sqrt x
Prelude> :t x
x :: Integer
Prelude> :t 25
25 :: (Num t) => t
Left to its own devices ghci chooses "Integer" for the type of x but the
constant 25 is the more generic "Num". This is the source of your
problems.
is_prime2 :: (Floating a, RealFrac a) => a -> Bool
is_prime2 input = prime 3 (toInteger $ round $ sqrt input)
satisfies the compilation. But do you really need the call to toInteger?
The round() call by itself appears to be sufficient. If you just want
round then you do not need the type specifier at all.
You should look more closely at your definition of prime though. It is
easy to give it numbers that has it recursing for ever.
------------------------------
Message: 4
Date: Thu, 1 Jul 2010 11:49:23 +0200
From: Daniel Fischer <daniel.is.fisc...@web.de>
Subject: Re: [Haskell-beginners] sqrt root issues
To: beginners@haskell.org
Message-ID: <201007011149.23759.daniel.is.fisc...@web.de>
Content-Type: text/plain; charset="iso-8859-1"
On Thursday 01 July 2010 05:01:41, Bryce Verdier wrote:
> Hi all,
>
> I'm having a real hard time wrapping my head around a problem;
>
> Here is a basic, no frills, function.
> prime divisor divided
>
> | divided == 1 = True
> | divisor== divided= True
> | mod divided divisor== 0 = False
> | otherwise = next_prime
>
> where next_prime = prime (divisor + 2) divided
>
> and the calling function:
> is_prime input = prime 3 input
>
> I'm trying to get the square root of input as an Integer.
>
> In GHCI:
> :t (toInteger $ round $ sqrt 25)
>
> (toInteger $ round $ sqrt 25) :: Integer
>
> but if I change is_prime input = prime 3 (toInteger $ round $ sqrt
> input)
>
> I get this error:
> problem3.hs:19:33:
> No instance for (RealFrac Integer)
> arising from a use of `round' at problem3.hs:19:33-37
> Possible fix: add an instance declaration for (RealFrac Integer)
> In the first argument of `($)', namely `round'
> In the second argument of `($)', namely `round $ sqrt input'
> In the expression: toInteger $ round $ sqrt input
>
> problem3.hs:19:41:
> No instance for (Floating Integer)
> arising from a use of `sqrt' at problem3.hs:19:41-50
> Possible fix: add an instance declaration for (Floating Integer)
> In the second argument of `($)', namely `sqrt input'
> In the second argument of `($)', namely `round $ sqrt input'
> In the expression: toInteger $ round $ sqrt input
> Failed, modules loaded: none.
>
> Can someone please help me out and tell me what I'm missing?
The call to toInteger is unnecessary, round will return an Integer if you
want one.
What you are missing is a call to fromInetger or fromIntegral.
sqrt takes an argument whose type belongs to the class Floating.
The divisibilty test (mod) requires the type of input to be a member of
Integral.
There are no standard types which belong to both classes (and it wouldn't
really make sense to make a type an instance of Integral and Floating).
So you have to convert input to a type acceptable for sqrt.
is_prime input = prime 3 (round . sqrt . fromIntegral $ input)
With an integer-literal it works, because an integer literal n stands for
(fromInteger n), so the integer is automatically converted to whatever
number type is needed.
>
> Thanks in advance,
>
> Bryce
------------------------------
Message: 5
Date: Thu, 1 Jul 2010 11:55:47 +0100
From: Tom Hobbs <tvho...@googlemail.com>
Subject: [Haskell-beginners] More Deserialization Woes
To: beginners@haskell.org
Message-ID:
<aanlktim3x_a9wlokdc-0hihd-qnklrnv1kcmc-mfa...@mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"
Hi guys,
Thanks for all the previous help I'm having with serialization in Haskell, I
couldn't have gotten as far as I have without the help from this group. Many
thanks!
Now I've put you in a good mood, maybe you can help me with my latest
problem... :-)
I have the following code which reads strings from a handle;
import qualified Data.ByteString.Lazy.UTF8 as UTF
readNames 0 _ = []
readNames n h = do
length <- fmap (fromIntegral . runGet
getWord32be) $ L.hGet h 4
name <- L.hGet h length
(UTF.toString name) : readNames (n-1) h
Where n is the number of names remaining to read and h is the handle to the
stream that I'm reading from. It is my intention that this function would
return a [String] of all the names. I'm sure further explainations probably
aren't necessary, but just in case;
- "length" becomes equal to some number (represented by four bytes) which
describes the length of the string to read
- "name" becomes the string I'm trying to read
- Then I add that String to the list and go again
But the problem I'm getting is this;
Couldn't match expected type `[a]' against inferred type `IO b'
In a stmt of a 'do' expression:
length <- fmap (fromIntegral . runGet
getWord32be) $ L.hGet h 4
In the expression:
do { length <- fmap (fromIntegral . runGet
getWord32be)
$ L.hGet h 4;
name <- L.hGet h length;
(UTF.toString name) : readNames (n - 1) h }
In the definition of `readNames':
readNames n h
= do { length <- fmap (fromIntegral . runGet
getWord32be)
$ L.hGet h 4;
name <- L.hGet h length;
(UTF.toString name) : readNames (n - 1) h }
I (think I) understand what the message is saying, but not why it's
occurring. I read the documentation as "UTF8.tostring" essentially rips the
[Char] out of the ByteString and returns that, so where does the "IO b" come
into it? Is it complaining because the reads from the handle are IO
operations and it's expecting me to use the IO monad somewhere? What i want
to be able to do is do the IO stuff inside the monad, but end up with a
[String] which I can then use in a pure function.
RWH and Learn You A Good Haskell all cover some stream reading, but their
examples are to different from what I'm trying to do that I'm getting stuck.
Any pointers are very gratefully accepted.
Many thanks,
Tom
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Message: 6
Date: Thu, 1 Jul 2010 13:03:39 +0200
From: Benjamin Edwards <edwards.b...@gmail.com>
Subject: Re: [Haskell-beginners] More Deserialization Woes
To: Tom Hobbs <tvho...@googlemail.com>
Cc: beginners@haskell.org
Message-ID:
<aanlktinvkqfjusmtcwk6b4f_4vs4vv5vcfccxhqve...@mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"
It's the type signature of hGet that is your undoing. it returns IO
ByteString, and then you are trying to apply a pure function to that. You
need to lift the pure function into the IO monad.
On 1 July 2010 12:55, Tom Hobbs <tvho...@googlemail.com> wrote:
> Hi guys,
>
> Thanks for all the previous help I'm having with serialization in Haskell,
> I couldn't have gotten as far as I have without the help from this group.
> Many thanks!
>
> Now I've put you in a good mood, maybe you can help me with my latest
> problem... :-)
>
> I have the following code which reads strings from a handle;
>
> import qualified Data.ByteString.Lazy.UTF8 as UTF
>
> readNames 0 _ = []
> readNames n h = do
> length <- fmap (fromIntegral . runGet
> getWord32be) $ L.hGet h 4
> name <- L.hGet h length
> (UTF.toString name) : readNames (n-1) h
>
> Where n is the number of names remaining to read and h is the handle to the
> stream that I'm reading from. It is my intention that this function would
> return a [String] of all the names. I'm sure further explainations probably
> aren't necessary, but just in case;
>
> - "length" becomes equal to some number (represented by four bytes) which
> describes the length of the string to read
> - "name" becomes the string I'm trying to read
> - Then I add that String to the list and go again
>
> But the problem I'm getting is this;
>
> Couldn't match expected type `[a]' against inferred type `IO b'
> In a stmt of a 'do' expression:
> length <- fmap (fromIntegral . runGet
> getWord32be) $ L.hGet h 4
> In the expression:
> do { length <- fmap (fromIntegral . runGet
> getWord32be)
> $ L.hGet h 4;
> name <- L.hGet h length;
> (UTF.toString name) : readNames (n - 1) h }
>
> In the definition of `readNames':
> readNames n h
> = do { length <- fmap (fromIntegral . runGet
> getWord32be)
> $ L.hGet h 4;
> name <- L.hGet h length;
> (UTF.toString name) : readNames (n - 1) h }
>
> I (think I) understand what the message is saying, but not why it's
> occurring. I read the documentation as "UTF8.tostring" essentially rips the
> [Char] out of the ByteString and returns that, so where does the "IO b" come
> into it? Is it complaining because the reads from the handle are IO
> operations and it's expecting me to use the IO monad somewhere? What i want
> to be able to do is do the IO stuff inside the monad, but end up with a
> [String] which I can then use in a pure function.
>
> RWH and Learn You A Good Haskell all cover some stream reading, but their
> examples are to different from what I'm trying to do that I'm getting stuck.
>
> Any pointers are very gratefully accepted.
>
> Many thanks,
>
> Tom
>
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://www.haskell.org/mailman/listinfo/beginners
>
>
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