←back to thread

# [derive(Clone)] Is Broken

(rgbcu.be)
128 points RGBCube | 1 comments | 04 Jul 25 20:31 UTC | HN request time: 0.33s | source
Show context
moomin ◴[08 Jul 25 07:04 UTC] No.44497757[source]▶
Haskell does this. If you derive Eq, it puts a condition on the generic parameter(s) requiring them to be Eq as well. Then if you use it with something that doesn’t implement Eq, your generic type doesn’t either.

It helps if you can express these preconditions in the first place, though.

replies(2): >>44497957 #>>44499255 #
1. evertedsphere ◴[08 Jul 25 12:14 UTC] No.44499255[source]▶
that's incorrect

haskell implements the "perfect derive" behaviour that the blog post is complaining about rust's lack of. the constraint is only propagated to the field types when using a haskell derive, not the parameters themselves (as in rust)

so the following compiles just fine:

    data Foo -- not Eq

    data Bar a = Bar 
      deriving Eq

    f :: Eq (Bar Foo) => ()
    f = ()