Why does this need both (Eq t, Num t) as type classes?

Question

So I am doing some exercised in Haskell and I thought the type of int2nat in

int2nat :: (Eq t, Num t) => t -> Nat

int2nat 0 = Zero
int2nat n = Succ (int2nat (n-1)

would be int2nat :: (Num t) => t -> Nat.

Can somebody explain why this is not the case and why the additional Eq t is correct instead?

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score 1 · Answer 1 · answered Jan 01 '22 at 19:37

1

Because you matched on the numeric literal 0 in the pattern. int2nat 0 = Zero desugars to something like int2nat n | n == fromInteger 0 = Zero.

answered Jan 01 '22 at 19:37

1 Answers1