haskell - Question about the argument type of sqrt and round

Question

following the signature:

sqrt :: Floating a => a -> a

why is (sqrt 2) legal? Isn't the number 2 a Integer which definitely doesn't satisfy Floating? Same Question about round, round (sqrt 2) is legal, sqrt returns type Floating but round needs ReadFrac.

round :: (RealFrac a, Integral b) => a -> b

Answer 1

The literal 2 is of type Num a => a. That is, it can be any numerical type. Specifically, Haskell will take the integer value 2 and call fromInteger on it (which is defined for any Num). So when you write

sqrt 2

Internally, what's happening is

sqrt (fromInteger 2) :: Floating a => a

And if you force the value, such as in GHCi, you'll get a Double since that's the default for Floating.

Likewise, the type of round (sqrt 2) is going to be Integral b => b and is going to require the sqrt 2 type to be RealFrac a => a. There exist types which are both RealFrac and Floating, so that's not a contradiction. In particular, GHC will happily default to Double here for the same reason as before. If you force the value to be printed, the entire result (Integral b => b) will default to Integer.

It's important to remember that all of these are universally quantified. Floating a => a doesn't mean "this is some floating type and that's all I know". It means "if you have any floating type, I can produce a value of that type". You get to choose which floating type to use, so if a constraint comes along later and says the value is also RealFrac a => a, that's fine because we've simply constrained ourselves to be both Floating and RealFrac. This is contrary to a language like Java, where if I have a value of some interface type, say Comparable, then all I can conclude is that it's some Comparable type, not that it works for all of them.