Why isn't every type part of Eq in Haskell?

Question

Or rather, why isn\'t (==) usable on every data type? Why do we have to derive Eq ourseleves? In other languages, such as Python, C++, and surely o

Answer 1

I can't think of types that can't be compared.

let infiniteLoop = infiniteLoop

let iSolvedTheHaltingProblem f = f == infiniteLoop
-- Oops!