Thank you so much; this worked! :)
From: Kai Zhang [mailto:kai@kzhang.org]
Sent: Monday, December 1, 2014 08:59
To: Yasuaki Kudo
Cc: Haskell Cafe
Subject: Re: [Haskell-cafe] Typeclass Show Question
The type "a" in S a b must be an instance of "Show" in order to use "show". So you need:
instance Show a => Show (S a b) where
show (S x y) = show x
On Sun, Nov 30, 2014 at 3:55 PM, Yasuaki Kudo mailto:yasu@yasuaki.com > wrote:
Hi,
When I launch GHCI with the option –XGADTs and load the following program,
f = \x y -> (1, x,y)
data S a b where
S :: (Num a, Num b) => a -> b -> S a b
I encounter this dialog:
*Main> S 1 1
<interactive>:3:1:
No instance for (Show (S a0 b0))
arising from a use of `print'
Possible fix: add an instance declaration for (Show (S a0 b0))
In a stmt of an interactive GHCi command: print it
I tried to follow the recommendation but still could not figure out how. For example, after adding:
instance Show (S a b) where
show (S x y) = show(x)
I still get:
conmath.hs:10:24:
Could not deduce (Show a) arising from a use of `show'
from the context (Num a, Num b)
bound by a pattern with constructor
S :: forall a b. (Num a, Num b) => a -> b -> S a b,
in an equation for `show'
at conmath.hs:10:15-19
Possible fix:
add (Show a) to the context of
the data constructor `S'
or the instance declaration
In the expression: show (x)
In an equation for `show': show (S x y) = show (x)
In the instance declaration for `Show (S a b)'
Failed, modules loaded: none.
What should I do to show S 1 1 ?
Regards,
Yasu
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