Just reread that section. So the 'set' is essentially a block of four entries(that contain four addresses), selected by the low 3-bits of the tag's address. Within those four entries, it uses one of them to allocate/read/write/use/free a (new) entry.

So it's the same as the 8 entry four ways with a 80386?

I've just confused the way and entry in my TLB implementation?

So:

There are 32 entries in total. 4 entries for each set(set selected by the low 3-bits of the LA address). Then lookups/writes, allocation and deallocation are only on those four entries(ways), determined by the 3-bits of the LA?

So that 'way' are the four entries that are selected by the set(which is the least significant 3-bits of the LA). Then the lookup results in a tag, way(which of the four had a result) and translated address for that way and set combination?

Is that correct?

Edit: So this is the opposite of a 80386, which performs lookups on a set(2-bits) of 8-entries within a set?

https://pdos.csail.mit.edu/6.828/2014/reading … i386/s10_06.htm

So, essentially the 80486 doubled the set size(doubled the set lookup size to select what to hit possibly) and halved the entries that result for said set(4 instead of 😎?

So the 80386 has 4 sets of 8 entries to use, while to 80486 has 8 sets of 4 entries to use?