why is mex parfor slower them mex for?

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I am starting to work with the Parallel Computing Toolbox, and just constructed an FIR filter example to compare for and parfor
coefs = [-0.00393617608745112 -5.95945405003999e-05...] length 1x10498
values = [30.3750000000000 30.3760000000000...] length 1x131000
tic;
outVal = FIRMP(coefs,values);
%outVal = FIRMP_mex(coefs,values);
time = toc;
with function FIRMP
function [result] = FIRMP(coefs, values)
coefLen = length(coefs);
valLen = length(values);
result = zeros(size(values));
(par)for I = 1 : valLen - coefLen;
suma = 0;
for J = 1 : coefLen
suma = suma + coefs(J)*values(I + J);
end
result(I) = suma;
end
end
I used 4 threads and got this results
for : time= 13.5s
parfor: time = 5.5s
It is OK, but if I create C++ mex (matlab CODER) and run again, the result has changed
for : time = 3.1s
parfor: time = 4.3s
why is the 'parfor' in C++ mex slower than 'for'?

采纳的回答

Ryan Livingston
Ryan Livingston 2018-8-29
编辑:Ryan Livingston 2018-8-29
When I try your example on Linux (Debian 9) using GCC I see a good speedup with parfor in generated MEX:
for : time = 1.3s
parfor : time = 0.4s
On Windows 10 using Microsoft Visual Studio 2017, I see a much more modest speedup:
for : time = 1.3s
parfor : time = 1.0s
What compiler and OS are you using?
One thing that may be happening for certain compilers is that each of the parfor loop iterations are very fast. When this is the case, the overhead of managing threads can dominate the loop execution time. This can ruin any possible parallelism gains.
The Coder documentation covers this in some detail:
as does the MATLAB parfor documentation:
  6 个评论
Josef Shrbeny
Josef Shrbeny 2018-9-2
Ryan, Using variables 1 x :n instead of 1 x :inf (both input and local) solved the problem
Now, the 'parfor mex C++' is about 3x faster than 'for mex C++'. (4 threads) Thank you for all your help. You gave me very useful tips and links.
Ryan Livingston
Ryan Livingston 2018-9-3
You're welcome Josef. Glad to hear you found a solution.

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