details of MESI protocol for multicore processors

情到浓时终转凉″ 提交于 2019-12-06 07:56:55

The reason you are having trouble finding these answers is because the traditional protocols were not defined for hierarchical cache architectures so the MESI protocol by itself doesn't define what will happen when you have an L1 and an L2 cache. It depends on three other system attributes.

If the L2 is designed to be exclusive of the L1 (i.e., it is guaranteed that L2 and L1 can never have common cache lines), then any line in the L1 will be invalid state (basically not present) in the L2.

If the L2 is inclusive of the L1, i.e., every line in the L1 must have an entry in the L2 as well, the entry in the L2 will contain a descriptor stating which L1 cache has the line in E state.

Whether or not the value is written out to L2 or memory on a read from E or W stage depends on whether your system supports cache-to-cache transfers or not. In old day, when each chip was a single core, and core-to-core communication was as expensive as read/write to memory, systems would write the data to memory and make the other processor read it (this allowed them to not support cache-to-cache transfers). In multi-core, talking via memory is insanely expensive compared to talking to other cores on-chip, so almost all multi-core chips today support cache-to-cache transfer. Thus, a read from E or W stage is not serviced by writing to memory.

I hope this helps.

atoMerz

I found this. It might help.
Third comment here might also be useful.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!