Get sqrt from Int in Haskell

Question

How can I get sqrt from Int.

I try so:

sqrt . fromInteger x

But get error with types compatibility.

score 53 · Answer 1 · edited Feb 28 '14 at 00:00

53

Perhaps you want the result to be an Int as well?

isqrt :: Int -> Int
isqrt = floor . sqrt . fromIntegral

You may want to replace floor with ceiling or round. (BTW, this function has a more general type than the one I gave.)

edited Feb 28 '14 at 00:00

Uli Köhler

answered Jul 14 '11 at 15:23

augustss

3

ghc warns about this, since it doesn't know which exact type to use between `fromIntegral` and `floor` (could be `Double`,`Float`, etc.). To fix: `isqrt x = floor . sqrt $ (fromIntegral x :: Float)`, which is less elegant :( – Pontus Granström Aug 11 '14 at 21:32
I like it, anyway. No warns on GHC 8. – Manoel Vilela Jul 13 '17 at 22:07
ghc will warn if `-Wtype-defaults, -Werror=type-defaults`. – Abhijit Sarkar Dec 31 '22 at 08:49

score 50 · Accepted Answer · edited Jan 07 '21 at 19:58

50

Using fromIntegral:

Prelude> let x = 5::Int
Prelude> sqrt (fromIntegral  x)
2.23606797749979

both Int and Integer are instances of Integral:

fromIntegral :: (Integral a, Num b) => a -> b takes your Int (which is an instance of Integral) and "makes" it a Num.
sqrt :: (Floating a) => a -> a expects a Floating, and Floating inherit from Fractional, which inherits from Num, so you can safely pass to sqrt the result of fromIntegral

I think that the classes diagram in Haskell Wikibook is quite useful in this cases.

edited Jan 07 '21 at 19:58

erik

answered Jul 14 '11 at 14:58

MarcoS

2

Could `sqrt (fromIntegral x)` also be written as `sqrt $ fromIntegral x`? – rzetterberg Jul 14 '11 at 15:01
1

Yes indeed it could, since explicit application ($) does not bind as tightly as implicit application. – Edward Jul 14 '11 at 15:03
2

or even `sqrt . fromIntegral $ x` – pat Jul 14 '11 at 17:32
51

I prefer to write it as `((((($)))((sqrt))(fromIntegral (x))))`. – Thomas Eding Jul 15 '11 at 20:28

score 13 · Answer 3 · answered Jul 14 '11 at 15:02

Remember, application binds more tightly than any other operator. That includes composition. What you want is

sqrt $ fromIntegral x

Then

fromIntegral x

will be evaluated first, because implicit application (space) binds more tightly than explicit application ($).

Alternately, if you want to see how composition would work:

(sqrt .  fromIntegral) x

Parentheses make sure that the composition operator is evaluated first, and then the resulting function is the left side of the application.

3 Answers3