Question about parfor nested loop

Hi,

I have a question about the nested parfor loop. Please see the following codes:

clc
clear
parfor i=1:3    
    for j=1:3
        if j<2
            B(i,j)=j;
        end
    end
end

The result of this code should be B=[1;1;1], but the Matlab gives B=[1,1,1;1,1,1;1,1,1]. I have been confused by this problem for a long time and thank you in advance if you have some time to look into the problem.

Haonan

[Ive J]

Well, you may wanna retry, because this is just impossible!

[Walter Roberson]

When I test, I get [1;1;1] as you expected.

[haonan He]

Thanks for your answer. I have tried many times and via different versions of Matlab, including 2016b and 2019a, but the results are the same [1,1,1;1,1,1;1,1,1], which is not expected. I think it is the issue with parallel programming. Please see the following picture for the Matlab result. Thanks very much.

[Rik]

clc
clear
parfor i=1:3    
    for j=1:3
        if j<2
            B(i,j)=j;
        end
    end
end
B

B = 3×1

1 1 1

[Ive J]

Can you also check this?

B = (0);
parfor ii=1:3    
    for jj=1:3
        if jj<2
            B(ii,jj)=jj;
        end
    end
end
disp(size(B))

[haonan He]

Thanks Rik. I reran the code on Matlab 2020b and the result was correct, but it was not as expected when I used old versions, e.g, 2016a 2019a. So maybe it is the bug of the old versions.

[haonan He]

Thanks Ive. I think the problem may be caused by different versions. When I ran your code on old versions, there was a system error "Index beyond matrix dimension", but it is correct on the lateset version 2020b. So maybe it is the bug of the old versions.

Answer 1

[Matt J]

So maybe it is the bug of the old versions.

Yes, it appears to be version-related. I get the same thing in R2018a. Regardless, it is not good practice to use parfor to loop over sliced variables whose slices, in this case the B(i,:), are not allocated in advance. This cleaner version, for example, does not have the issue:

clc
clear
B=nan(3); %pre-allocate
parfor i=1:3    
    for j=1:3
        if j<2
            B(i,j)=j;
        end
    end
end

[haonan He]

Thanks Matt. Yes, the code does not generate the error.