tcp_init函数的一点疑问

jjww

[code:1]
if (num_physpages >= (128 * 1024))
                goal = num_physpages >> (21 - PAGE_SHIFT);
        else
                goal = num_physpages >> (23 - PAGE_SHIFT);

        for (order = 0; (1UL << order) < goal; order++)
                ;
        do {
                tcp_ehash_size = (1UL << order) * PAGE_SIZE /
                        sizeof(struct tcp_ehash_bucket);
                tcp_ehash_size >>= 1;
                while (tcp_ehash_size & (tcp_ehash_size - 1))
                        tcp_ehash_size--;
                tcp_ehash = (struct tcp_ehash_bucket *)
                        __get_free_pages(GFP_ATOMIC, order);
        } while (!tcp_ehash && --order > 0);
[/code:1]
这段代码是确定order值，最终通过order值，确定可以存储的tcp_ehash_bucket结构数目，通过__get_free_pages函数分配该结构所需的页面空间。
但是我不知道order为什麽需要这样确定？依据是什么？哪里有相关的参考信息？

zhiwood

order is number of pages. it has to be 2(x) which is required by binary buddy allocator the kernel uses.

jjww

从上面这段代码看，并不能保证是order是2的倍数呀。
另外，为什麽内存大于512M时是21-12呢？小于时是23－12呢？

zhiwood

sorry, order is the exponent of 2, 2^order is the number of the pages requested.

zhiwood

guess: greater than 512MB, ehash size can be 1/512 of total memory
          less than 512, ehash size can be 1/2048 of the total.

jjww

thanks!