I’m talking about in the same run of racket, for immutable hash-eqs with the
same set of keys.
For example:
(define hsh1
(hasheq ‘k1 (delay 1) ‘k2 (delay 2) ‘k3 (delay 3)))
(define hsh2
(for/hasheq ([(k v) (in-hash hsh1)])
(values k (force v))))
Would this:
(for ([k (in-hash-keys hsh1)]
[v (in-hash-values hsh2)])
….)
Be the same as
(for ([(k v) (in-hash hsh2)])
….)
Are you saying I can’t count on that, or are you saying that I can’t count on
that between
separate runs of racket?
And could I for instance share the return values of hash-iterate-first and
hash-iterate-next
between the two hash tables within the same run of racket?
I think you’re saying no, but I’m still curious.
On Feb 21, 2015, at 8:41 AM, Robby Findler <ro...@eecs.northwestern.edu> wrote:
> I think the current implementation might guarantee the same order for
> the same hash table in a single run of racket, but you shouldn't reply
> on even that and the current implementation does not make that
> guarantee if you run the same program multiple times in racket. (And
> the precise details of what the current implementation guarantees
> depend on which kind of hash table you have and possibly the timing of
> gc and other mysterious things.)
>
> Robby
>
> On Sat, Feb 21, 2015 at 7:03 AM, Alexander D. Knauth
> <alexan...@knauth.org> wrote:
>> If I have two hash tables, both with the same set of keys, but with
>> different values for those keys, then can I count on the order being the
>> same for both hash-tables?
>>
>> I noticed this:
>> https://github.com/plt/racket/commit/abe1233734b3a8d46f96707cc3a2517340cb28a3
>> saying:
>>
>> make hash-table order invertible at build time
>>
>> For detecting and debugging accidental dependencies on hash-table
>> order, it might be helpful to invert the order at the lowest level.
>>
>> Does this mean I shouldn’t count on the order being the same? How does this
>> work?
>>
>>
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