[racket] Different mutable vectors for each thread

I should have listened to the little voice in the back of my head that 
said I'd solve this myself in 15 minutes. That's exactly what I came up 
with.

Looks like we're both geniuses tonight!

Neil ⊥

On 07/14/2012 11:11 PM, Eric Dobson wrote:
> I think thread-cells are enough, you just need to do the allocation yourself.
>
> #lang racket
>
> (define get-thread-local-vector
>    (let ()
>      (define vector-cell (make-thread-cell #f))
>      (lambda ()
>        (or (thread-cell-ref vector-cell)
>            (let ((vec (vector 0 1)))
>              (thread-cell-set! vector-cell vec)
>              vec)))))
>
> (vector-set! (get-thread-local-vector) 0 'new-value)
> (get-thread-local-vector)
> (void (sync (thread (lambda () (print (get-thread-local-vector)) (newline)))))
>
> =>
>
> '#(new-value 1)
> '#(0 1)
>
>
>
> On Sat, Jul 14, 2012 at 10:50 PM, Neil Toronto <neil.toronto at gmail.com> wrote:
>> To head off wrong answers: It's not thread cells. By themselves, they're not
>> enough. Making them "preserved" also isn't the answer.
>>
>> I'd like a thread-cell-like thing that gives each thread its own mutable
>> vector. Here's why thread cells don't work:
>>
>> #lang racket
>>
>> (define v (make-thread-cell (vector 0 1)))
>> (sync (thread (λ () (vector-set! (thread-cell-ref v) 0 6))))
>> (thread-cell-ref v)
>>
>>
>> Output:
>>
>> #<thread>
>> '#(6 1)
>>
>>
>> Clearly, both the created thread and the runtime thread access the same
>> vector.
>>
>> Not only do I need this, but I need repeated access to the per-thread
>> vectors to be faster than allocating a new one. Ideally, after the first
>> access, they'd allocate nothing.
>>
>> Any ideas?
>>
>> Neil ⊥
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