> I am currently fiddling with the slab allocator and one thing messes > my understanding of the concept. Think about the following scenario: > > 1) start memcached with ./memcached -m 1 (with 1 Mb of memory limit) > 2) set mykey 0 0 11 --> allocates a 1 Mb slab and a chunk is returned > for about the smallest possible chunk > 3) set mykey2 0 0 200 -- ? > > Now I cannot understand what happens in the 3rd step? There is no > memory left for another slab error returned or something else? I have > tried to test this on my machine but somehow cannot interpret the > results correctly. 2nd step does allocate 1 MB of slab with 96 bytes > of chunk size? I am right?
http://code.google.com/p/memcached/issues/detail?id=95 will be fixed at some point...
