By: anon2 (anon.delete@this.anon.com), September 16, 2022 5:49 pm
Room: Moderated Discussions
--- (---.delete@this.redheron.com) on September 16, 2022 11:28 am wrote:
> anon2 (anon.delete@this.anon.com) on September 15, 2022 7:04 pm wrote:
> > Everybody knows the data integrity problems with parity protected write-back arrays. ECC has also
> > seemed to be a difficult problem for L1 data cache that seems like nobody has solved very well.
> >
>
> > What is expensive about L1 ECC which is less costly in
> > L2? Keep in mind you need write-through, so L2 has to
> > receive all the stores.
>
> ECC saves bits (relative to simpler options like replication) by performing math across
> the ENTIRE line. This means that each time you modify an element of L1, something (whether
> in the core itself or in the L1) will have to read the entire line (whether before or
> after the write) to calculate the new ECC value. That's a lot of extra power.
>
> L2 is different because the entire line is written at once
> from the L1 to the L2, so there's a one-time calculation.
Uh, obviously that's not the case in a write-through L1 design.
> anon2 (anon.delete@this.anon.com) on September 15, 2022 7:04 pm wrote:
> > Everybody knows the data integrity problems with parity protected write-back arrays. ECC has also
> > seemed to be a difficult problem for L1 data cache that seems like nobody has solved very well.
> >
>
> > What is expensive about L1 ECC which is less costly in
> > L2? Keep in mind you need write-through, so L2 has to
> > receive all the stores.
>
> ECC saves bits (relative to simpler options like replication) by performing math across
> the ENTIRE line. This means that each time you modify an element of L1, something (whether
> in the core itself or in the L1) will have to read the entire line (whether before or
> after the write) to calculate the new ECC value. That's a lot of extra power.
>
> L2 is different because the entire line is written at once
> from the L1 to the L2, so there's a one-time calculation.
Uh, obviously that's not the case in a write-through L1 design.