[Haskell-cafe] Can type be changed along a = chain?

2009-10-12 Thread michael rice 
The first of these works, but not the second. It would seem that the type 
cannot change along a = chain, but I may be missing something in the code.

Is the second example illegal? If so, is there a different way to change a 
String to an Int along the = chain?

Michael

===

import Data.Char

{-
transform :: IO ()
transform = putStrLn What is your word?
  getLine
 = \str - return ('Q':str)
 = \str - return ('Z':str)
 = \str - putStrLn $ Transformed word is  ++ show str
-}

transform :: IO ()
transform = putStrLn What is your digit string?
  getLine
 = \str - return ('9':str)
 = \str - return (read str :: Int)
 = \i - putStrLn $ The number is  ++ show i




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Re: [Haskell-cafe] Can type be changed along a = chain?

2009-10-12 Thread Niklas Broberg 
On Mon, Oct 12, 2009 at 6:37 PM, michael rice nowg...@yahoo.com wrote:

 transform :: IO ()
 transform = putStrLn What is your digit string?
   getLine
  = \str - return ('9':str)
  = \str - return (read str :: Int)
  = \i - putStrLn $ The number is  ++ show i

This code works perfectly for me. What problem are you seeing specifically?
Cheers,
/Niklas
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Re: [Haskell-cafe] Can type be changed along a = chain?

2009-10-12 Thread minh thu 
2009/10/12 michael rice nowg...@yahoo.com

  The first of these works, but not the second. It would seem that the type
 cannot change along a = chain, but I may be missing something in the code.

 Is the second example illegal? If so, is there a different way to change a
 String to an Int along the = chain?

 Michael

 ===

 import Data.Char

 {-
 transform :: IO ()
 transform = putStrLn What is your word?
   getLine
  = \str - return ('Q':str)
  = \str - return ('Z':str)
  = \str - putStrLn $ Transformed word is  ++ show str
 -}

 transform :: IO ()
 transform = putStrLn What is your digit string?
   getLine
  = \str - return ('9':str)
  = \str - return (read str :: Int)
  = \i - putStrLn $ The number is  ++ show i



Both seem good to me and my old ghci (6.6.1)...

Thu
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Re: [Haskell-cafe] Can type be changed along a = chain?

2009-10-12 Thread michael rice 
Dumb! I just figured out I was entering the input string in quotes.

So, I suppose the answer to my question is yes, type CAN be changed along a = 
chain. I was having trouble doing it in a different problem, created this small 
example to illustrate the problem, and then screwed it up putting quotes around 
my input string.

Thanks!

Michael

--- On Mon, 10/12/09, Niklas Broberg niklas.brob...@gmail.com wrote:

From: Niklas Broberg niklas.brob...@gmail.com
Subject: Re: [Haskell-cafe] Can type be changed along a = chain?
To: michael rice nowg...@yahoo.com
Cc: haskell-cafe@haskell.org
Date: Monday, October 12, 2009, 12:43 PM

On Mon, Oct 12, 2009 at 6:37 PM, michael rice nowg...@yahoo.com wrote:

transform :: IO ()
transform = putStrLn What is your digit
 string?
  getLine
 = \str - return ('9':str)
 = \str - return (read str :: Int)
 = \i - putStrLn $ The number is  ++ show i

This code works perfectly for me. What problem are you seeing specifically?
Cheers,/Niklas



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Re: [Haskell-cafe] Can type be changed along a = chain?

2009-10-12 Thread minh thu 
I hope you're not building some unneeded rules in your head. There is no
reason to believe there is something to be remembered about whether or not
types can change along a = chain. That chain has nothing special in
Haskell. = is just an operator, much like ++, ! or .

ghci :t (=)
(=) :: (Monad m) = m a - (a - m b) - m b

This says that, you provide an a and you get a b. Nothing says the a and b
have to be the same upon successive uses.

Likewise,

ghci :t (+)
(+) :: (Num a) = a - a - a

fromIntegral ((1 :: Int) + 2) + (3 :: Integer)
6

This shows clearly that the types are not the same along the + chain.


2009/10/12 michael rice nowg...@yahoo.com

  Dumb! I just figured out I was entering the input string in quotes.

 So, I suppose the answer to my question is yes, type CAN be changed along a
 = chain. I was having trouble doing it in a different problem, created
 this small example to illustrate the problem, and then screwed it up putting
 quotes around my input string.

 Thanks!

 Michael

 --- On *Mon, 10/12/09, Niklas Broberg niklas.brob...@gmail.com* wrote:


 From: Niklas Broberg niklas.brob...@gmail.com
 Subject: Re: [Haskell-cafe] Can type be changed along a = chain?
 To: michael rice nowg...@yahoo.com
 Cc: haskell-cafe@haskell.org
 Date: Monday, October 12, 2009, 12:43 PM

 On Mon, Oct 12, 2009 at 6:37 PM, michael rice 
 nowg...@yahoo.comhttp://mc/compose?to=nowg...@yahoo.com
  wrote:

  transform :: IO ()
 transform = putStrLn What is your digit string?
   getLine
  = \str - return ('9':str)
  = \str - return (read str :: Int)
  = \i - putStrLn $ The number is  ++ show i

 This code works perfectly for me. What problem are you seeing specifically?
 Cheers,
 /Niklas



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 Haskell-Cafe@haskell.org
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Re: [Haskell-cafe] Can type be changed along a = chain?

2009-10-12 Thread Ketil Malde 

minh thu not...@gmail.com writes:

 ghci :t (=)
 (=) :: (Monad m) = m a - (a - m b) - m b

 This says that, you provide an a and you get a b. Nothing says the a and b
 have to be the same upon successive uses.

(But note that the monad 'm' has to be the same all the way.  You can't
switch from, say, IO to ST in the middle of the computation.)

 Likewise,

Contrariwise, I'd say.

 ghci :t (+)
 (+) :: (Num a) = a - a - a

I.e. (+) requires the same type on both sides...

 fromIntegral ((1 :: Int) + 2) + (3 :: Integer)
 6

...but 'fromIntegral' converts it.  Did I miss some subtle point here?

-k
-- 
If I haven't seen further, it is by standing in the footprints of giants
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Re: [Haskell-cafe] Can type be changed along a = chain?

2009-10-12 Thread minh thu 
2009/10/12 Ketil Malde ke...@malde.org:

 minh thu not...@gmail.com writes:

 ghci :t (=)
 (=) :: (Monad m) = m a - (a - m b) - m b

 This says that, you provide an a and you get a b. Nothing says the a and b
 have to be the same upon successive uses.

 (But note that the monad 'm' has to be the same all the way.  You can't
 switch from, say, IO to ST in the middle of the computation.)

 Likewise,

 Contrariwise, I'd say.

 ghci :t (+)
 (+) :: (Num a) = a - a - a

 I.e. (+) requires the same type on both sides...

 fromIntegral ((1 :: Int) + 2) + (3 :: Integer)
 6

 ...but 'fromIntegral' converts it.  Did I miss some subtle point here?

I was talking about the types of the two + applications. They have a
different types, just like the = in the parent's message can have
different types. My fromIntegral has a role similar to the read s ::
Int of the original question.

Cheers,
Thu
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