haskell - 在 Haskell 中，Integral typeclass 是否意味着 Show typeclass？

Question

我试图编译这段代码。

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)

那个代码没有编译，我得到的不是很长
错误消息说它不能推断(显示 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)

更改此行后它起作用了

isPalindrome :: Integral a => a -> Bool

到

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

所以我在想，既然 Integral 中的每个类型都在 Show 中，Haskell 编译器应该能够从 (Integral a) 推导出 (Show a)。

Answer 1

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

但并非 Integral 中的所有类型在 Show .在 Haskell98 中曾经是这种情况，因为

class Show n => Num n

但是这种父类(super class)关系阻止了大量有用的数字类型(“无限精度数字”、连续函数的全局结果等)。在现代 Haskell 中，类 Show和 Integral根本没有关系，因此编译器无法从另一个推断出一个。

然而，确实可以独立于实际 Show 显示任何整数类型。类(class);使用 showInt 为此功能。

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