[Cython] prange CEP updated

[Dag Sverre Seljebotn]

On 05/04/2011 02:17 PM, mark florisson wrote:
> On 4 May 2011 14:10, Dag Sverre Seljebotn<d.s.seljebotn at astro.uio.no>  wrote:
>> On 05/04/2011 01:59 PM, mark florisson wrote:
>>>
>>> On 4 May 2011 13:54, Dag Sverre Seljebotn<d.s.seljebotn at astro.uio.no>
>>>   wrote:
>>>>
>>>> On 05/04/2011 01:48 PM, mark florisson wrote:
>>>>>
>>>>> On 4 May 2011 13:47, mark florisson<markflorisson88 at gmail.com>      wrote:
>>>>>>
>>>>>> On 4 May 2011 13:45, Dag Sverre Seljebotn<d.s.seljebotn at astro.uio.no>
>>>>>>   wrote:
>>>>
>>>>>>> Look.
>>>>>>>
>>>>>>> i = 42
>>>>>>> for i in prange(n):
>>>>>>>     f(i)
>>>>>>> print i # want 42 whenever n == 0
>>>>>>>
>>>>>>> Now, translate this to:
>>>>>>>
>>>>>>> i = 42;
>>>>>>> #pragma omp parallel for firstprivate(i) lastprivate(i)
>>>>>>> for (temp = 0; ...; ...) {
>>>>>>>     i = ...
>>>>>>> }
>>>>>>> #pragma omp parallel end
>>>>>>> /* At this point, i == 42 if n == 0 */
>>>>>>>
>>>>>>> Am I missing something?
>>>>>>
>>>>>> Yes, 'i' may be uninitialized with nsteps>      0 (this should be valid
>>>>>> code). So if nsteps>      0, we need to initialize 'i' to something to
>>>>>> get
>>>>>> correct behaviour with firstprivate.
>>>>
>>>> This I don't see. I think I need to be spoon-fed on this one.
>>>
>>> So assume this code
>>>
>>> cdef int i
>>>
>>> for i in prange(10): ...
>>>
>>> Now if we transform this without the guard we get
>>>
>>> int i;
>>>
>>> #pragma omp parallel for firstprivate(i) lastprivate(i)
>>> for (...) { ...}
>>>
>>> This is invalid C code, but valid Cython code. So we need to
>>> initialize 'i', but then we get our "leave it unaffected for 0
>>> iterations" paradox. So we need a guard.
>>
>> You mean C code won't compile if i is firstprivate and not initialized?
>> (Sorry, I'm not aware of such things.)
>
> It will compile and warn, but it is technically invalid, as you're
> reading an uninitialized variable, which has undefined behavior. If
> e.g. the variable contains a trap representation on a certain
> architecture, it might halt the program (I'm not sure which
> architecture that would be, but I believe they exist).
>
>> My first instinct is to initialize i to 0xbadabada. After all, its value is
>> not specified -- we're not violating any Cython specs by initializing it to
>> garbage ourselves.
>
> The problem is that we don't know whether the user has initialized the
> variable. So if we want firstprivate to suppress warnings, we should
> assume that the user hasn't and do it ourselves.

I meant that if we don't care about Valgrindability, we can initialize i 
at the top of our function (i.e. where it says "int __pyx_v_i").

>> OTOH, I see that your approach with an if-test is more Valgrind-friendly, so
>> I'm OK with that.
>>
>> Would it work to do
>>
>> if (nsteps>  0) {
>>     #pragma omp parallel
>>     i = 0;
>>     #pragma omp for lastprivate(i)
>>     for (temp = 0; ...) ...
>>     ...
>> }
>
> I'm assuming you mean #pragma omp parallel private(i), otherwise you
> have a race (I'm not sure how much that matters for assignment). In
> any case, with the private() clause 'i' would be uninitialized
> afterwards. In either case it won't do anything useful.

Sorry, I meant that lastprivate(i) should go on the parallel line.

if (nsteps>  0) {
     #pragma omp parallel lastprivate(i)
     i = 0;
     #pragma omp for
     for (temp = 0; ...) ...
     ...
}

won't this silence the warning? At any rate, it's obvious you have a 
better handle on this than me, so I'll shut up now and leave you to it :-)

Dag Sverre