Need an easy way to return the value of a single task in `par` about parseq HOT 4 CLOSED

Out of curiosity, what happens if you do:

Tasks.seq(t1, Tasks.par(t2, t3), t2)

Does it run t2 again? Or do you get an error that t2 already executed? Or does it immediately return the value in t2?

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The current behavior of seq is that it is resolved as soon as the last element in the sequence is resolved. Since t2 will be resolved early in the sequence, the whole sequence with finish early with or without task t3 being run at all. Coming back to parseq a bit fresh, I don't think this is a very intuitive behavior. We do need early finish for use cases with timeouts and such, but it seems that the intuitive behavior for seq would probably be to run the whole sequence to completion and then return the result.

In general, if you try to run a task twice in parseq it will only actually be run once. The second time is a no-op. If you add a listener for the task after it has been resolved the listener is notified immediately.

seq(t1, par(t2, t3), t2) was my first intuitive approach to this problem as well, but I discarded it given the behavior we have in place. It's worth a closer examination.

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Agreed on a closer examination. From a "principle of least surprise", I'd expect seq(t1, par(t2, t3), t2) to only return after t1, t2, and t3 are all done.

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This is solved by withSideEffect in parseq 2.x:
t1.withSideEffect(t2).andThen(t2)

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