Stable name equality

[Simon Peyton Jones]

|  > But is the reverse true?  That is, can we be sure that two pointer-
|  distinct stable name objects contain different indices?
|  
|  I believe so, yes. Each stable name table entry has a pointer to the
|  linked stable name object. Calling makeStableName# checks whether the
|  passed pointer already has a stable name, and, if so, returns the linked
|  stable name object.

Actually I'm confused.  Currently we have

data StableName a = StableName (StableName# a)

I believe (but there is no documentation to state) that the (StableName# a) 
 * Has kind (TYPE UnliftedRep)
 * Is the index of the entry in the Stable Name Table

But if it's the index, why isn't it an IntRep?  UnliftedRep is for pointers?

Moreover eqStableName# :: StableName# a -> StableName# b -> Bool
is directly implemented in the code generator (StgCmmPrim) by an equality
comparison.

If these things are correct, it would be great to write them down in a Note.

And if they are right, I'm now lost about what you question is.  Equality is
/already/ implemented by direct equality comparison, no?

Simon

|  -----Original Message-----
|  From: David Feuer <david at well-typed.com>
|  Sent: 20 August 2018 23:36
|  To: ghc-devs at haskell.org; Simon Peyton Jones <simonpj at microsoft.com>
|  Cc: David Feuer <david.feuer at gmail.com>; marlowsd at gmail.com
|  Subject: Re: Stable name equality
|  
|  On Monday, August 20, 2018 11:56:44 AM EDT Simon Peyton Jones via ghc-
|  devs wrote:
|  > I defer to SimonM but IIRC each stable name corresponds to an entry in
|  the stable-name table; the index in the table is the “stable” thing in a
|  stable name.  So I bet that comparison compares these indices.
|  >
|  > If two stable names are pointer-equal, they presumably contain the same
|  index into the table.
|  >
|  > But is the reverse true?  That is, can we be sure that two pointer-
|  distinct stable name objects contain different indices?
|  
|  I believe so, yes. Each stable name table entry has a pointer to the
|  linked stable name object. Calling makeStableName# checks whether the
|  passed pointer already has a stable name, and, if so, returns the linked
|  stable name object. The design seems a bit surprising to me, but it looks
|  like that's actually how it works, at least for now. Each call locks the
|  stable name table, so it shouldn't be possible to miss entry creation.