Re: [Haskell-cafe] Make Show instance

2011-07-21 Thread Александр 
Hello, Willem Van Lint, Thank you for help, it works.
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Re: [Haskell-cafe] Make Show instance

2011-07-21 Thread Willem Van Lint 
Hi,

I think your main problem here is that you use Int in the pattern matching.
The point is that you are matching your value of type Tree k v to a pattern
such as:
- EmptyTree
- (Node (a, b) left right)

Patterns don't contain type information such as Int, but things like value
constructors and variables.
Note the difference between type constructors and value constructors (
http://book.realworldhaskell.org/read/defining-types-streamlining-functions.html
).
So you probably should use something like this:

data Tree k v = EmptyTree
| Node (k, v) (Tree k v) (Tree k v)

instance (Show k, Show v) => Show (Tree k v) where
 show EmptyTree =
 "Empty"
 show (Node (a, b) left right)  =
show left ++ "(" ++ show a ++ "," ++ show b ++ ")" ++ show right

I probably use (++) too much here though.
'(Show k, Show v) =>' tells us that the types k and v are of that typeclass
so that we can use show on values of this type.

I'm a beginner too though so I hope I was clear.



2011/7/21 Александр 

> Hello,  thank you for reply. I know that i can derive this. But i want to
> know how can i make it by hand.
>
> Thank you.
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Re: [Haskell-cafe] Make Show instance

2011-07-21 Thread Daniel Patterson 
The documentation for the Show typeclass has this very example: 
http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#t:Show

The summary? you need to define either showPrec or show, the latter of which is 
simpler, it is just a -> String.

So:

instance Show (Tree Int Int) where
show (Tree (Node (k,v)) left right) = ...

On Jul 21, 2011, at 10:55 AM, Александр wrote:

> Hello,  thank you for reply. I know that i can derive this. But i want to 
> know how can i make it by hand.
> 
> Thank you.
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> Haskell-Cafe mailing list
> Haskell-Cafe@haskell.org
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Re: [Haskell-cafe] Make Show instance

2011-07-21 Thread Александр 
Hello,  thank you for reply. I know that i can derive this. But i want to know 
how can i make it by hand.

Thank you.
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Re: [Haskell-cafe] Make Show instance

2011-07-21 Thread Steffen Schuldenzucker 


Hi.

On 07/21/2011 04:45 PM, Александр wrote:

Hello,

I have binary tree, with every leaf tuple - (k,v):

data Tree k v = EmptyTree
| Node (k, v) (Tree k v) (Tree k v)

How can i make Show Instance for (Tree Int Int) ?


The easiest way is automatic derivation:

data Tree k v = EmptyTree
  | Node (k, v) (Tree k v) (Tree k v)
  deriving (Eq, Ord, Show, Read)
  -- you normally want at least these four.

Then the compiler automatically declares the instances for you, given 
that 'k' and 'v' have them. For k = v = Int, this is the case.




I try:

instance Show (Tree k v) where
show EmptyTree = show "Empty"
show (Node (Int, Int) left right) = ..


I'm afraid to say that, but 'Int' doesn't make sense at this place. You 
would want


> show (Node (x, y) left right) = ...

instead. (That is, use any variables. Variables have to be lowercase.)

-- Steffen

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[Haskell-cafe] Make Show instance

2011-07-21 Thread Александр 
Hello,

I have binary tree, with every leaf tuple - (k,v):


data Tree k v = EmptyTree  
                | Node (k, v) (Tree k v) (Tree k v)

How can i make Show Instance for (Tree  Int Int) ?

I try:


instance Show (Tree k v) where
     show EmptyTree = show "Empty"
     show (Node (Int, Int) left right)  = ..

But get error:

 Not in scope: data constructor `Int'

I have a function:


fillTree :: Int -> Tree Int Int -> Tree Int Int
fillTree  100 tree = tree 
fillTree  x tree = let a = treeInsert (x, x) EmptyTree                   in 
fillTree (x + 1) a
If i execute it:


 No instance for (Show (Tree Int Int))
      arising from a use of `print'

How can i make Show instance for my Tree? 
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