[m-rev.] for review: threadsafe version_arrays in Java

[Peter Wang]

On 19 April 2010 12:00, Paul Bone <pbone at csse.unimelb.edu.au> wrote:
>> @@ -678,6 +731,23 @@ ML_va_get(ML_va_ptr VA, MR_Integer I, MR_Word *Xptr)
>>  }
>>
>>  int
>> +ML_va_set_dolock(ML_va_ptr VA0, MR_Integer I, MR_Word X, ML_va_ptr *VAptr)
>> +{
>> +#ifdef MR_THREAD_SAFE
>> +    MercuryLock *lock = VA0->lock;
>> +#endif
>> +    int         ret;
>> +
>> +    ML_maybe_lock(lock);
>> +
>> +    ret = ML_va_set(VA0, I, X, VAptr);
>> +
>> +    ML_maybe_unlock(lock);
>> +
>> +    return ret;
>> +}
>> +
>> +static int
>>  ML_va_set(ML_va_ptr VA0, MR_Integer I, MR_Word X, ML_va_ptr *VAptr)
>>  {
>>      ML_va_ptr VA1;
>> @@ -691,6 +761,9 @@ ML_va_set(ML_va_ptr VA0, MR_Integer I, MR_Word X,
>> ML_va_ptr *VAptr)
>>          VA1->index      = -1;
>>          VA1->value      = (MR_Word) NULL;
>>          VA1->rest.array = VA0->rest.array;
>> +#ifdef MR_THREAD_SAFE
>> +        VA1->lock       = VA0->lock;
>> +#endif
>>
>>          VA0->index     = I;
>>          VA0->value     = VA0->rest.array->elements[I];
>
> I'm concerned about the copying of this lock.  Is it a copy or do VA0->lock and
> VA1->lock both refer to the same lock.  VA0->lock is currently locked and will
> be unlocked by ML_va_set_dolock.  Will this unlock VA1->lock?  (which it
> should).

Here we create VA1 as the latest version of the array, VA0 being the
latest version before that.  Both lead to the same underlying array,
so they need to share the same lock.  So ML_va_set_dolock will indeed
unlock it.

Thanks for looking.

Peter

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