Parallel Pool; variable does not exist after running

Hello,

When I run the following simple script there is no problem and I'll get i and e at the end of the loop

for i=1:30
e=i/2
end

However, when I want to use parallel pool for the same loop, but with parfor:

parfor i=1:30
e=i/2
end

I don't get any thing when running the loop finishes. Am I missing something here?

Answer 1

[Raymond Norris]

Hi Parsa,

A for loop runs in a serial fashion, one iteration after the other, with the same reproducable results each time (more or less). In your example, we know e will always equal 15. A parfor loop runs its iterations in any order. Therefore, the last value of i could be 14 (and not 30) and therefore, e would equal 7 (and not 15). For this reason, after a parfor, i and e are not know (more about this in a second).

MATLAB must clasify each of the variables within the body of a parfor. You can read more about classification here

https://www.mathworks.com/help/coder/ug/classification-of-variables-in-parfor-loops.html

Based on this classification, we deem i to be a loop variable and e to be a temporary variables, nether of which persist after the completion of a parfor. Sliced output and reduction variables can be assigned and persist after a parfor.

Raymond

[Walter Roberson]

in order for a change inside parfor to be visible outside the parfor, then one of two things must be true :

1. the variable must indexed by the parfor loop control variable; or
2. the variable must be acting as a reduction variable

An example of a reduction is

   t = 0
   parfor k = 1 : 50
     t = t + x.^k./factorial(k) ;
   end

parfor is able to figure out that the result for t would be the same no matter what order the loop ran, except for round-off error, and permits that case.

[Raymond Norris]

I make brief mention at the end about sliced output and reduction variables. In the OP's example, e is neither of them, though I suspect it's a toy example and not the actual code.

[Parsa Ghadermazi]

Thanks a lot. Problem solved!