Trying to understand function application operator in Haskell

Question

I'm trying to wrap my head around the function application operator ($) in Haskell.

I'm working through the examples in Learn You a Haskell, and I thought I understood the following example:

Prelude> map ($ 3) [(+4), (*10), (^2), sqrt]
[7.0,30.0,9.0,1.7320508075688772]

I then tried the following variant, which also worked fine:

Prelude> map ($ 3) [(+4), (*10), (\x -> x^2), sqrt]
[7.0,30.0,9.0,1.7320508075688772]

Finally, I tried modifying the third function in the list as follows, which generates an error:

Prelude> map ($ 3) [(+4), (*10), (\x -> 2^x), sqrt] 
<interactive>:53:38:
    Ambiguous type variable `b0' in the constraints:
      (Floating b0)
        arising from a use of `sqrt' at <interactive>:53:38-41
      (Integral b0) arising from a use of `^' at <interactive>:53:33
      (Num b0) arising from the literal `3' at <interactive>:53:8
    Probable fix: add a type signature that fixes these type variable(s)
    In the expression: sqrt
    In the second argument of `map', namely
      `[(+ 4), (* 10), (\ x -> 2 ^ x), sqrt]'
    In the expression: map ($ 3) [(+ 4), (* 10), (\ x -> 2 ^ x), sqrt]
Prelude>

It seems if the final sqrt function is somehow begin associated with the previous list element, as the following variant works ok:

Prelude> map ($ 3) [(+4), (*10), (\x -> 2^x)]
[7,30,8]

Can someone enlighten me as to what's going on here?

one lesson to take from this, after trying something out at GHCi prompt, check its type, with Prelude> :t it. "It" is a special word, referring to the previous result, and :t asks to see a type. You could see that there are no decimal points in the numbers in your last example. Also, after entering :s +t at the prompt, the GHCi will report the type for every result it produces. — Will Ness
– Will Ness, Commented Mar 13, 2013 at 9:06

Daniel Fischer · Accepted Answer · 2013-03-12 13:54:47Z

17

The type of the used exponentiation operator is

(^) :: (Num a, Integral b) => a -> b -> a

so when you use \x -> 2^x, you get an Integral constraint for the 3. But sqrt imposes a Floating constraint. So the type of the 3 must satisfy

3 :: (Integral t, Floating t) => t

but there is no instance for both among the default type list, which is Integer and Double, so the defaulting fails, and you're left with an ambiguous type variable.

When you had \x -> x^2, there was only a Num constraint from the first functions, and Floating from sqrt, so the type was defaulted to Double.

You can make it work if you use

(**) :: Floating a => a -> a -> a

as your exponentiation operator, then the type can again be defaulted to Double.

answered Mar 12, 2013 at 13:54

Daniel Fischer

184k19 gold badges319 silver badges436 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

zurgl Over a year ago

Then he could fix it, replacing (^) by (**).

pmags Over a year ago

Thanks both Daniel and Zurgl -- that clarifies things considerably. Coming from Python I'm still wrapping my head around the implications of strict type checking.

Collectives™ on Stack Overflow

Trying to understand function application operator in Haskell

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related