I think there is a restriction of the smallest value of the -m option. --- 11年8月25日，周四, Oktaka Com <[email protected]> 写道：
发件人: Oktaka Com <[email protected]> 主题: Slab Allocator question 收件人: "memcached" <[email protected]> 日期: 2011年8月25日,周四,上午2:12 Hi all, 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? Thanks,
