fetchNext idx always returning 1

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?

Answer 1

[Edric Ellis]

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