Have you tried macroexpanding the expression? Doing so yields
julia> macroexpand(:( for i = 1:N
@sync @parallel for j = (i + 1):N
tmp[j] = i * j
end
end ))
:(for i = 1:N # line 2:
begin # task.jl, line 342:
Base.sync_begin() # line 343:
#6#v = begin # multi.jl, line 1487:
Base.pfor($(Expr(:localize, :(()->begin # expr.jl,
line 113:
begin # multi.jl, line 1460:
function (#7#lo::Base.Int,#8#hi::Base.Int) # multi.jl, line
1461:
for j = (i + 1:N)[#7#lo:#8#hi] # line 1462:
begin # line 3:
tmp[j] = i * j
end
end
end
end
end))),Base.length(i + 1:N))
end # line 344:
Base.sync_end() # line 345:
#6#v
end
end)
It looks like @parallel does the work of setting up a properly formatted
call to Base.pfor. In particular, it builds an Expr object with head
:localize and argument a zero-arg anonymous function, and then passes the
interpolation of that expression along with `Base.length(i + 1:N)` to
Base.pfor. The body of the anonymous function declares another function
with arguments `#7#lo`, `#8#hi`. The latter variables somehow annotate the
delimiters of your inner loop, which gets reproduced inside the body of the
declared function. I'm *guessing* that the anonymous function is used as a
vehicle to pass the code of the annotated inner loop to Base.pfor without
executing it beforehand. But I could be wrong.
Then @sync just wraps all the above between calls to `Base.sync_begin` and
`Base.sync_end`.
I also should note I have zero experience with Julia's parallel machinery
and am entirely unfamiliar with the internals of Base.pfor. I just enjoy
trying to figure out macros.
On Wednesday, June 17, 2015 at 5:49:58 AM UTC-4, Daniel Carrera wrote:
>
>
> On Wednesday, 17 June 2015 10:28:37 UTC+2, Nils Gudat wrote:
>>
>> I haven't used @everywhere in combination with begin..end blocks, I
>> usually pair @sync with @parallel - see an example here
>> <https://github.com/nilshg/LearningModels/blob/master/NHL/NHL_6_Bellman.jl>,
>> where I've parallelized the entire nested loop ranging from lines 25 to 47.
>>
>
>
> Aha! Thanks. Copying your example I was able to produce this:
>
> N = 5
> tmp = SharedArray(Int, (N))
>
> for i = 1:N
> # Compute tmp in parallel #
> @sync @parallel for j = (i + 1):N
> tmp[j] = i * j
> end
>
> # Consume tmp in serial #
> for j = (i + 1):N
> println(tmp[j])
> end
> end
>
>
> This seems to work correctly and gives the same answer as the serial code.
> Can you help me understand how it works? What does "@sync @parallel" do? I
> feel like I half-understand it, but the concept is not clear in my head.
>
> Thanks.
>
> Daniel.
>
