parfor variable cannot be classified

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Jeremy Rutman
Jeremy Rutman 2015-11-11
编辑： Jeremy Rutman 2015-11-17
as far as i can tell, the following code should be parallelizable, however I get a 'cannot be classified' error no matter how I twist it. This is paraphrased code . It is 'theoretically parallelizable' i blv., as the result is independent of execution order.
a{1}=''
a{2}='hi'
a{3}=''
a{4}='bye'
parfor k = 1:10
j=mod(k,5)
b=a{j}
if isempty(b)
a{j} = j+3
end
end
along these lines, what can this possibly mean:
"Notably, the assignments to the variables i, t, and u do not affect variables with the same name in the context of the parfor statement. "
http://www.mathworks.com/help/distcomp/parfor.html?searchHighlight=parfor%20using%20objects%20{}
a more concise statement of the problem is that the following, which is in principle parallelizable for arbitrary 'index' functions f,g (ie positive integer functions in range of the array in question), cannot in fact be compiled due to matlab compiler limitations.
parfor k=1:10
index=f(k)
array(index)=g(index)
end
however as walter roberson points out below this limitation can be somewhat avoided in this case by the following:
parfor k=1:10
index=f(k)
temparray(k)=g(index)
indices(k)=index
end
for k=1:10
array(indices(k)) = temparray(k)
end
  2 个评论
Walter Roberson
Walter Roberson 2015-11-16
You do realize that you original code tries to index a{0} ?
Jeremy Rutman
Jeremy Rutman 2015-11-17
yes, i took out a bunch of extranneous code in an attempt to make the question as clear as poss. and took out a '+1'

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Walter Roberson
Walter Roberson 2015-11-12
When k = 1 or k = 6 both times j = 1 so you have multiple iterations that would be accessing the same elements of a{}. That is not permitted
The indexing of the loop variable needs to be in-line in the access, not calculated in a previous statement and then used, so that parfor can verify that there is no possibility of overlap. mod() is not one of the permitted operations in the indexing for parfor.
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Walter Roberson
Walter Roberson 2015-11-17
We do not inherently know that g(f(x)) will be the same for the non-unique values of f. g could be a function that returns different values at different times. You would need to know that g() is either an array that is invariant during the operation, or a function that has no side effects.
None of the parallelizing compilers I have ever seen have ever been happy with code of this form that might write the same location multiple times but with the same result each time.
Jeremy Rutman
Jeremy Rutman 2015-11-17
编辑：Jeremy Rutman 2015-11-17
ok if the function g changes with time then this would be trouble. In the meantime I adopted your solution and have run into another issue , maybe you can take a look

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