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页面: https://www.xiaolincoding.com/os/3_memory/linux_mem2.html#_5-4-min-free-kbytes-的计算逻辑 里面的
比如:当前 NUMA 节点中有两个 zone :ZONE_DMA 和 ZONE_NORMAL,内存容量大小分别是:100 M 和 800 M。那么 ZONE_DMA 与 ZONE_NORMAL 之间的比例就是 1 :8。
根据这个比例,ZONE_DMA 区域里的 WMARK_MIN 水位线就是:min_free_kbytes * 1 / 8 。ZONE_NORMAL 区域里的 WMARK_MIN 水位线就是:min_free_kbytes * 7 / 8。
这里的1/8应该是 1/9 , 而7/8应该是8/9
根据函数 __setup_per_zone_wmarks,lowmem_pages 算的是ZONE_DMA 和 ZONE_NORMAL 的总page,对应上面是900M内存,而 tmp = (u64)pages_min * zone_managed_pages(zone); do_div(tmp, lowmem_pages);
zone_managed_pages(zone) 算的是当前的zone的page,对应的是100M或800M
所以对应的比例应该是 1/9 或者 8/9
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页面:
https://www.xiaolincoding.com/os/3_memory/linux_mem2.html#_5-4-min-free-kbytes-的计算逻辑
里面的
这里的1/8应该是 1/9 , 而7/8应该是8/9
根据函数 __setup_per_zone_wmarks,lowmem_pages 算的是ZONE_DMA 和 ZONE_NORMAL 的总page,对应上面是900M内存,而
tmp = (u64)pages_min * zone_managed_pages(zone);
do_div(tmp, lowmem_pages);
zone_managed_pages(zone) 算的是当前的zone的page,对应的是100M或800M
所以对应的比例应该是 1/9 或者 8/9
The text was updated successfully, but these errors were encountered: