[erlang-questions] refactoring a very large record

[Jesper Louis Andersen]

2011/10/20 Björn-Egil Dahlberg <wallentin.dahlberg@REDACTED>:

> But rather then "mark-and-sweep" that is hinted at here, erlang runtime
> system uses a copying collector.
> Without going in to gritty details, anything that can be reached from a
> rootset (stack, registers and a few other things) are live data and copied
> to a new heap. The old heap is then deallocated.

Phew. Copying collectors are much faster :)

In addition, you only pay for live data with them. Dead data is
reclaimed for free.

-- 
J.