How to do
STORE in an out-of-order processor.
When does memory write happen?
STORE it happens at commit. Has to be done at that point for the same reasons commit happens in program order.
Where does load get data? From a load-store queue. (This seems like ROB for memory insts)
Load-Store Queue (LSQ)
Structure like ROB for
STORE with 4 parameters
|Load or Store||Address||Value (if any)||Complete?|
LOAD address matches a previous
STORE address in the LSQ the value is copied from the
STORE LSQ entry (Store-to-load forwarding)
If an inst doesn’t have an address there are options on how to handle further instructions:
- Do everything in-order
- Wait for all previous store addresses to be known (not completed)
- This is because we can compare with them and know weather or not we need their val or need to go to memory
- Go anyway
- In this case
STOREs will also check if their address matches any further in the list and those instructions then go into recovery because they have the wrong values.
- In this case
Go anyway is usually used
Store to Load Forwarding
LOAD: Which earlier store do we get value from? STORE: Which later load do I give my value to?
When loads commit they only then store in the register. When store commits it only then writes to the data cache/memory.
LSQ, ROB and RS
All instructions get a ROB entry. LSQ acts as a RS for LOAD/STORE. So if there isn’t a ROB entry and a LSQ entry available the IQ stalls.
- Compute address
- Produce value (for STORE 1 and 2 can be done in either order)
- Write Result (LOAD)
- Commit (Free ROB and LSQ entries) a. STORE: write to memory