Julia: Unexpected Result when Using #parallel for Loop with SharedArray - for-loop

Question

I am trying to learn parallel for loop in Julia, while I was confused by the result of the example code below:
addprocs(4)
#everywhere begin
N = 10
V = SharedArray{Int64,1}(N)
#sync #parallel for i = 1:N
V[i] = i
println(V[i])
end
end
By using println, I tried to identify which worker went through which step. Surprisingly, what I got from the above code was that each worker kept going through the entire iteration until the last worker (worker 3 in my case) finished the for loop:
From worker 4: 1
From worker 4: 2
From worker 4: 3
From worker 5: 1
From worker 5: 2
From worker 5: 3
From worker 5: 4
From worker 5: 5
From worker 5: 6
From worker 5: 4
From worker 5: 5
From worker 5: 6
From worker 2: 7
From worker 2: 8
From worker 2: 4
From worker 2: 5
From worker 2: 6
From worker 2: 9
From worker 2: 10
From worker 4: 1
From worker 4: 2
From worker 2: 7
From worker 2: 8
From worker 3: 9
From worker 5: 4
From worker 5: 5
From worker 5: 6
From worker 3: 10
From worker 3: 7
From worker 3: 8
From worker 3: 7
From worker 3: 8
From worker 3: 9
From worker 3: 10
From worker 2: 4
From worker 2: 5
From worker 2: 6
From worker 3: 7
From worker 3: 8
From worker 4: 3
From worker 4: 9
From worker 4: 10
From worker 4: 1
From worker 4: 2
From worker 4: 3
From worker 4: 1
From worker 4: 2
From worker 4: 3
From worker 5: 9
From worker 5: 10
This is not what I would expect from parallel for loop, as I thought the work should be distributed among the workers and then combined together, rather than being done separately for each of the worker. I mean, should the concept of parallel for loop be that each worker only goes through 2-3 steps?
Is something wrong with my code or I misunderstand the concept of parallel for loop?
Thanks!
Edit: I just realized that there is something to do with #everywhere. After I eliminated #everywhere, things work as expected.

Answer 1

You told julia (through #everywhere) to run the parallel loop on every worker and on the host process, hence you count from 1 to 10 in a parallel fashion 5 times. (Check that in the output you posted every number from 1 to 10 occurs precisely 5 times)
Slightly more detailed: First we note that the total number of processes is nprocs() == 5 (one "host" and 4 workers, check workers()). #everywhere tells Julia to run the content of the begin end block on every process, hence 5 times in our example. The content of the begin end block is "do a parallel loop, counting from 1 to 10". That is exactly what happened.
When you remove the #everywhere you do just a single parallel loop. Thus you will get exactly what you want:
julia> N = 10
V = SharedArray{Int64,1}(N)
#sync #parallel for i = 1:N
V[i] = i
println(V[i])
end
From worker 5: 9
From worker 5: 10
From worker 3: 4
From worker 3: 5
From worker 3: 6
From worker 2: 1
From worker 2: 2
From worker 2: 3
Suggested reading: https://docs.julialang.org/en/stable/manual/parallel-computing