Thanks. I didn't know about macroexpand(). To me macros often feel like black magic.
On 17 June 2015 at 14:22, David Gold <[email protected]> wrote: > 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. >> > -- When an engineer says that something can't be done, it's a code phrase that means it's not fun to do.
