Get sqrt from Int in Haskell

问题

How can I get sqrt from Int.

I try so:

sqrt . fromInteger x

But get error with types compatibility.

回答1:

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.

回答2:

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.)

回答3:

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.

来源：https://stackoverflow.com/questions/6695267/get-sqrt-from-int-in-haskell