Type Inference

sumof ::	(Num a) => [a] -> a
h ::	(Num a) => a
t ::	(Num a) => [a]

"So the return value is really (Num a) => a."

        sumof []    = 0
        sumof (h:t) = h + sumof t

"That fits with the other return value of 0."

"And everything else checks out okay."

• If you ask, it will say that the type is:

     sumof :: (Num a) => [a] -> a

• If we had put 0.0 instead of 0, it would have deduced:

     sumof :: (Fractional a) => [a] -> a

• (Fractional is a subclass of Num)

• Among other things, it supports division

• If we had put "Fred" we would have gotten a type error

• Because String is not an instance of Num

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Copyright © 1999,2008 Mark Dominus