[erlang-questions] refactoring a very large record
Jesper Louis Andersen
Thu Oct 20 22:18:09 CEST 2011
2011/10/20 Björn-Egil Dahlberg <>:
> 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.
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