CODEWITHSUNDEEP
haskelltypessqrt
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Reputation: 21817

Get sqrt from Int in Haskell

How can I get sqrt from Int.

I try so:

sqrt . fromInteger x

But get error with types compatibility.

Upvotes: 35

Views: 59569

Answers (3)

MarcoS
MarcoS

Reputation: 13564

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.

Upvotes: 50

augustss
augustss

Reputation: 23014

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.)

Upvotes: 55

Edward
Edward

Reputation: 1826

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.

Upvotes: 13

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