Types of a function

Question

I saw on a Haskell tutorial this function:

factorial :: (Integral a) => a -> a

My question is why should I use that instead of this:

factorial :: Int->Int

It's because with the first one we can use every type of numbers and in the second one we can only use an Int?

A detail: that type means you can use any type of integer numbers (that is, any type which is an instance of Integer). For instance, factorial 0.5 won't work. — duplode
– duplode, Commented Sep 22, 2015 at 21:39

2 revs, 2 users 67% · Accepted Answer · 2015-09-23 19:17:35Z

This answer is a mashup of the comments to the question.

Your interpretation is correct, except for one detail: that type means you can use any type of integer numbers (that is, any type which is an instance of Integer). For instance, factorial 0.5 won't work, as expected for a factorial function.

So if i put (Num a) instead of (Integral a) factorial 0.5 will work?
@paulo_ferreira, when you do that, and test it, what happens? Can you figure out why?

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