In Haskell does Integral typeclass imply Show typeclass?

Question

I was trying to compile this code.

symmetric [] = True
symmetric [_] = True
symmetric l
    | (head l) == (last l) = symmetric (tail (init l))
    | otherwise = False

isPalindrome :: Integral a => a -> Bool
isPalindrome n = symmetric (show n)

That code did not compile and I got a not very long error message saying that it cannot deduce (Show a).

Could not deduce (Show a) arising from a use of ‘show’
from the context (Integral a)
  bound by the type signature for
             isPalindrome :: Integral a => a -> Bool
  at 4.hs:7:17-39
Possible fix:
  add (Show a) to the context of
    the type signature for isPalindrome :: Integral a => a -> Bool
In the first argument of ‘symmetric’, namely ‘(show n)’
In the expression: symmetric (show n)
In an equation for ‘isPalindrome’:
    isPalindrome n = symmetric (show n)

It worked after changing this line

isPalindrome :: Integral a => a -> Bool

to

isPalindrome :: (Show a, Integral a) => a -> Bool

So I was thinking since every type in Integral is in Show,a Haskell compiler should be able to deduce (Show a) from (Integral a).

Answer 1

So I was thinking since every type in Integral is in Show

But not every type in Integral is in Show . That used to be the case in Haskell98, due to

class Show n => Num n

But that superclass relation prevents an awful lot of useful number types (“infinite-precision numbers”, global results of continuous functions, etc.). In modern Haskell, the classes Show and Integral have no relation at all, hence the compiler can't infer one from the other.

It is, however, indeed possible to show any integral number type independently of the actual Show class; use the showInt function for this.

import Numeric (showInt)
isPalindrome :: Integral a => a -> Bool
isPalindrome n = symmetric $ showInt n []