是什么导致Haskell这样的类型错误？

Question

I'm playing with Haskell to evaluate simple limits with tables of values. I have the following function defined:

f :: (Integral a) => a -> a
f x = div 1 $ subtract 6 x

and in GHCI, I let leftSide = [5.90, 5.91..5.99] and let rightSide = [6.10,6.09..6.01] , then:

GHCI> map f leftSide

Which barfs up this error:

<interactive>:50:5
  No instance for (Integral Double) arising from a use of `f'
  Possible fix: add an instance declaration for (Integral Double)
  In the first argument of `map', namely `f'
  In the expression: map f leftSide
  In an equation for `it': it = map f leftSide

Changing my type declaration for f to (Integral Double a) => a -> a makes the compiler complain about how `Integral' is applied to too many type arguments. What's going on here?

Answer 1

I think you just tried the wrong division - you really want (/) not div ...

your problem is that div which demands an Integral type (for example an Integer ):

Prelude> :t div
div :: Integral a => a -> a -> a

but then you use it with a Fractional ( 6.10 , ...) by map ing it over your leftSide and rightSide

now GHCi defaults this to Double - but Double is not an instance of Integral - this is exactly what Haskell is complaining about.

---

the thing you tried does not work and I guess you wanted to write (Integral a, Double a) => ... but in the end (if it would work - it does not since Double is type and not a type-class) it would be the same as saying f :: Double -> Double anyway - which would get you the error again (as there is no div for Double )

To make it short: use (/) instead of div and it should work:

f :: Fractional r => r -> r
f x = (1 /) $ subtract 6 x

here is your first example:

Prelude> let leftSide = [5.90, 5.91..5.99]
Prelude> map f leftSide
[-10.000000000000036,-11.111111111111128
,-12.49999999999999,-14.285714285714228
,-16.66666666666653,-19.999999999999716
,-24.999999999999424,-33.33333333333207
,-49.999999999996625,-99.99999999998437]

---

btw: you can make this shorter if you use point-free:

f = (1 /) . (6 -)

or a bit cleaner / more readable if you write it out

f x = 1 / (6 - x)