I am trying to write a function that checks if a number is prime. I wrote this :
primeCheck :: Int -> Int -> Bool
primeCheck n i
| n == 2 = True
The problem is in converting from the Int for use in sqrt. Change the last line to
isPrime n = primeCheck n (floor (sqrt (fromIntegral n)))
Floating
is a typeclass, not a constructor or function. You do seem to have figured out that you need to convert the type of n
. The correct way to do this would be using fromIntegral
:
isPrime n = primeCheck n $ floor $ sqrt $ (fromIntegral n :: Double)
We can see why this works by following the type signatures of the functions.
From the type signature of isPrime
, we see n
has type Int
.
Since sqrt
expects some Floating
type (i.e. a type that is an instance of the typeclass Floating
), we can convert from an Int
to a Double
using fromIntegral
. Note that the signature of fromIntegral
is
(Integral a, Num b) => a -> b
Int
is an instance of Integral
(so the input type is okay) and Double
is an instance of Num
so the output type is okay.
Then we take the sqrt
to get a new Double
.
floor
expects an argument whose type is an instance of RealFrac
. Double
happens to be an instance of both Floating
and RealFrac
, so it will do the job (no conversion necessary).
floor
will convert the square root back to type Int
.
Note that since the output type of fromIntegral
is polymorphic, as is the input type of sqrt
and floor
, we have to specify the type of the conversion as Double
, otherwise the compiler won't know which Num
/Floating
/RealFrac
instance to convert to. You might see the error ambiguous type 'a' in ...
.
You can see the type signatures of many functions using Hoogle
EDIT
It turns out the explicit type signature for the fromIntegral
is not required. Thus
isPrime n = primeCheck n $ floor $ sqrt $ fromIntegral n
suffices. In my opinion, it would be clearer to just provide the explicit signature, but it is not necessary in this case. You can read more about it here.
Floating
and RealFrac
are type-classes, not functions.
This page gives a good overview of the standard Haskell numeric type-class hierarchy (section 6.3):
http://www.haskell.org/onlinereport/basic.html
In particular, this diagram of the type-classes is very helpful:
http://www.haskell.org/onlinereport/classes.gif
You clearly want to take the square root of an Int
by using the floating point sqrt
and floor
functions. The problem is that these functions have the following types:
sqrt :: Floating a => a -> a
floor :: (Integral b, RealFrac a) => a -> b
The obstacles are:
sqrt
takes a Floating
argument, and Int
is not in the Floating
type classfloor
requires a RealFrac
argument, but sqrt
produces only a Floating
You can solve the first with the fromIntegral
function:
fromIntegral :: (Integral a, Num b) => a -> b
which will allow you to convert an Int
(or any other integral type) into any Num
type (which includes Floating
.)
Next, looking at the chart of the numeric type-class hierarchy we see that Double
and Float
are members of both the Floating
and RealFrac
type-classes.
So by an explicit type coercion to Double
you can apply floor
:
isqrt :: Int -> Int
isqrt n = floor (sqrt (fromIntegral n) :: Double)
Now, as @lthread mentioned, you can get away with:
isqrt :: Int -> Int
isqrt n = (floor . sqrt . fromIntegral) n
and GHC will default the return type of sqrt
to Double
. It's perfectly fine to do this, but you should be aware that it is happening. For example, if you want to use this approach to take the integer square-root of an Integer
, you want to pass through an abritrary precision real type (like a Data.Number.BigFloat
) instead of a Double
.