Is z an object here? How to interpret the code?
이전 댓글 표시
data = rand(4,4);
means = zeros(1,4);
parfor I = 1:4
% The below usage of structures
% is flagged by Code Analyzer
z.mean = mean(data(:,I));
means(I) = z.mean;
end
disp(means)
Hi all, I found the above code was used to demonstrate incorrect use of parfor construct with MATLAB which can flag compiler errors.
On executing as expected with the code I encountered the an error stating The variable z in a parfor cannot be classified. See Parallel for Loops in MATLAB, "Overview".
what I fail to understand is 1) what is the datatype of z? is it a class an object ??? 2) but the code can be easily modified to handle the problem by replacing z.mean by z and the program runs perfectly. That makes me wonder why using z.mean is required??
답변 (1개)
James Tursa
2017년 9월 13일
편집: James Tursa
2017년 9월 13일
0 개 추천
z is a struct. It has a field named "mean".
댓글 수: 2
Aketh Thimmasandra-Maregowda
2017년 9월 13일
편집: Aketh Thimmasandra-Maregowda
2017년 9월 13일
James Tursa
2017년 9월 13일
편집: James Tursa
2017년 9월 13일
I don't know why a struct would be needed for that either. Just use z.
Where did the z.mean code come from in the first place? Did you inherit this code from someone or download it from somewhere? Does it complain if you do a direct assignment:
means(I) = mean(data(:,I));
Is this just an exercise in finding out why parfor doesn't like the struct? (Seems counterproductive to use it for mean calculations)
카테고리
도움말 센터 및 File Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!