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Most (all?) functional programming languages implement structural equality automatically. Referential equality only makes sense when records can be mutated. Imm

• gizmo686 9y ago
  Haskell does not quite implement it automatically, but does support auto-generating it. Assuming all of of a type's component types support structural equality, a new type can support it by adding "deriving (Eq)" to the type declaration. Ocaml has both structural and referential equality tests automatically. Unfortunately (for C-style programmers), structural equality uses "=", while referential equality uses "==". From the Ocaml doc: >On mutable types such as references, arrays, byte sequences, records with mutable fields and objects with mutable instance variables, e1 == e2 is true if and only if physical modification of e1 also affects e2. On non-mutable types, the behavior of ( == ) is implementation-dependent; however, it is guaranteed that e1 == e2 implies compare e1 e2 = 0.
  • wyager 9y ago
    > Ocaml has both structural and referential equality tests automatically. Yep, and Ocaml equality is a giant pain. As for ==, referential equality is almost never what you want, so it’s (IMO) unwise to expose it except as an unsafe primitive. As for =, OCaml’s “polymorphic”/“structural” equality is usually wrong for any non-trivial data structure, so all the comparison-based standard library are functorized over comparison anyway. Use of polymorphic ‘=‘ is basically banned in the codebase I work on. Haskell’s Eq/Ord typeclasses are vastly easier to use and harder to mess up. I think this particular use case is a more or less unambiguous win for typeclasses.