27 Sep
2011
27 Sep
'11
5:02 a.m.
On 26 Sep 2011, at 23:14, Arseniy Alekseyev wrote:
Garbage collection takes amortized O(1) per allocation, doesn't it?
No. For Mark-Sweep GC, the cost is proportional to (H+R) / (H-R) where H is the total heap size R is the reachable (i.e. live) heap This formula amortises the cost of a collection over the amount of free space recovered. For two-space copying collection, the cost is proportional to R / ((H/2)-R) In both cases, as R approaches H (or H/2), the cost of GC becomes rather large. So in essence, the more live data you have, the more GC will cost. Regards, Malcolm