parfor loop with continue gives incorrect results
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Consider the following code:
N = 1000;
failed = false( 1, N );
values = cell( 1, N );
n_failed = 0;
parfor idx = 1:N
try
if ( rand() > 0.7 )
n_failed = n_failed + 1;
error( '' );
end
catch err
failed(idx) = true;
continue;
end
values{idx} = rand( 1e3, 1 );
end
fprintf( 'N failed 1: %d\n\n', sum(failed) );
fprintf( 'N failed 2: %d\n', n_failed );
If I run this on my machine (macOS 10.12, r2017a), `sum(failed)` is 0, while `n_failed` is, as expected, ~300. What am I missing here? I don't see anything in the documentation about `continue` not being supported in a parfor loop?
댓글 수: 4
답변 (1개)
Matt J
2018년 10월 9일
편집: Matt J
2018년 10월 9일
Don't pass an empty string '' to error() if you want the catch block to be triggered. An empty string apparently does not result in an error being thrown.
In other words, this works fine:
N = 1000;
failed = false( 1, N );
n_failed = 0;
parfor i = 1:N
try
if ( rand() > 0.7 )
n_failed = n_failed + 1;
error('an error');
end
catch
failed(i) = true;
continue
end
end
fprintf( 'N failed 1: %d\n\n', sum(failed) );
fprintf( 'N failed 2: %d\n', n_failed );
댓글 수: 9
Edric Ellis
2018년 10월 11일
There's an existing problem in the parfor machinery that causes the values assignment to fail. The problem relates to the combination of try / catch and the assignment to values. You can trick the parfor machinery into operating correctly by changing how it analyses values. The following should work:
N = 1000;
failed = false( 1, N );
values = cell( 1, N );
n_failed = 0;
parfor idx = 1:N
% Dummy reference to "values(idx)":
if false
values(idx);
end
try
if ( rand() > 0.7 )
n_failed = n_failed + 1;
assert(false);
end
catch err
failed(idx) = true;
continue;
end
values{idx} = rand( 1e3, 1 );
end
fprintf( 'N failed 1: %d\n\n', sum(failed) );
fprintf( 'N failed 2: %d\n', n_failed );
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