CODEWITHSUNDEEP
haskellfloating-pointinttypeclass
synapse
synapse

Reputation: 5728

Converting between integer types

It's a pretty stupid question but I'm kinda lost. Here's the function

f :: (Bool,Int) -> Int
f (True,n) = round (2 ** n)
f (False,n) = 0

And here's an error I'm getting

No instance for (Floating Int)
  arising from a use of `**'
Possible fix: add an instance declaration for (Floating Int)
In the first argument of `round', namely `(2 ** n)'
In the expression: round (2 ** n)
In an equation for `f': f (True, n) = round (2 ** n)

What should I add to make it work?

Upvotes: 2

Views: 700

Answers (1)

hammar
hammar

Reputation: 139830

(**) is floating-point exponentiation. You probably want to use (^) instead. 

f :: (Bool,Int) -> Int
f (True,n)  = 2^n
f (False,n) = 0

It's helpful to look at the types:

Prelude> :t (**)
(**) :: Floating a => a -> a -> a
Prelude> :t (^)
(^) :: (Num a, Integral b) => a -> b -> a

The error message is telling you that Int is not an instance of the Floating typeclass, and therefore you can't use (**) on it directly. You could convert to some floating-point type and back, but here it is better to simply use the integral version directly. Also note that (^) only requires the exponent to be integral. The base can be of any numeric type.

Upvotes: 7

Related Questions