Linus Torvalds (torvalds.delete@this.linux-foundation.org) on September 12, 2021 11:10 am wrote:

> The tags don't need to be very big - you just want to be able to find the physically indexed
> L2 entry from a virtually indexed L1 entry. And assuming your L2 is inclusive of the L1
> (which seems a pretty reasonable assumption), that means that you only need enough bits
> to find that existing L2 entry so that you can do writeback to the right L2 entry.

Yeah that tagging is one way to do it - but it also wastes L1 size chunk of L2-cache.

Traditional inclusive cache is system where lower level cache is a subset of higher level. So 256KB L2 with 32KB L1 with inclusive layout means that there 256KB worth of tags for L2 but only 224KB of data cache. That missing 32KB is that L1 cache. There's no problem mapping virtually indexed L1 into physically indexed L2 but L2 has to have at least same or higher associativity with L1. So L1 cache is always mapped to same known L2 tags.

And how is mapping virtually indexed cache to physically indexed possible? By interleaving data so it fits. 8-way set cache can be seen as 8-direct mapped caches interleaved 8-way, so with 64 byte cache lines every way has addresses mapped 512 bytes apart. With that known 512 bit spread its possible to map that cache into L2 because it fits into bits that do not change in address translation, and if cache configurations are compatible with addressing - like they are if they share associativity.