On 04/19/2016 04:07 PM, Emilio G. Cota wrote: > + ht_avg_len = qht_avg_bucket_chain_length(&tcg_ctx.tb_ctx.htable, > &ht_heads); > + cpu_fprintf(f, "TB hash avg chain %0.5f buckets\n", ht_avg_len); > + cpu_fprintf(f, "TB hash size %zu head buckets\n", ht_heads);
I think the accounting is questionable here. Consider the following data: TB count 230467/671088 TB invalidate count 25915 TB hash avg chain 1.03073 buckets TB hash size 131072 head buckets This means that we've got 230467 - 25915 = 204552 active TB's, installed into a hash table with 131072 heads. For a perfectly uniform distribution of TBs, that would be an average chain length of 204552 / 131072 = 1.56. In order to get the average down to 1.03, there need to be a substantial number of heads with zero entries. I think perhaps it might be more enlightening to separately account for empty and non-empty heads. E.g. TB hash buckets used xxxx/131072 TB hash avg chain yyyy where xxxx is the number of non-empty heads, and yyyy = |TBs| / xxxx. I also wonder if it wouldn't be better to size the hash table as appropriate for the maximum number of allowable TBs. r~
