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Erlang / Erlang Advanced Interview questions

What is the two-version code loading rule and what happens when a third version is loaded?

The BEAM keeps at most two versions of a module in memory at once: the current version, which new calls resolve to, and the old version, still executing for any process that was mid-loop when the reload happened. This bounded window is what allows hot code upgrades without an unbounded pile-up of stale code.

code:load_file(my_module).      %% current becomes old, new module becomes current
%% ... later ...
code:load_file(my_module).      %% loading a THIRD version

If a third version is loaded while some process is still running the old (now second-oldest) version, the BEAM has nowhere to put it — so it forcibly kills any process still executing that stale version via code:purge/1, then promotes the new version to current. This is why long-running loops are written to periodically make a fully-qualified call back into themselves (e.g. ?MODULE:loop(State)): it gives them a chance to pick up the newer code before a third reload force-terminates them.

What happens if a third version of a module is loaded while a process still runs the oldest version?
Why do long-running loops typically call back into themselves with a fully-qualified call?

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