fetchNext idx always returning 1

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Kazi Siddiqui
Kazi Siddiqui 2020-7-6
回答： Edric Ellis 2020-7-6
The documentation says that fetchNext "returns the linear index of that future in array F as idx". Here is my code:
Script experiment.m:
parpool('threads');
f(1:3)=parallel.FevalFuture;
for i=1:3
f(i)=parfeval(@worker, 1, i);
end
for i=1:3
[idx, x]=fetchNext(f(i));
disp(""+idx+" "+x);
end
Function worker.m:
function x = worker(x)
x=x+1;
end
Output:
1 2
1 3
1 4
Why is idx always 1? What am I doing wrong?

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采纳的回答

Edric Ellis
Edric Ellis 2020-7-6
The first output of fetchNext tells you which of the futures you passed in as an input has just completed. In your case, you are calling fetchNext(f(i)) - so you are always calling fetchNext with a single future, therefore the index is always 1.
You should pass all the futures into fetchNext, like this:
for i=1:3
[idx, x]=fetchNext(f);
disp(""+idx+" "+x);
end

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