Your questions are tough for a couple of reasons:
- The answer depends a lot on the microarchitecture of the processor
which can vary significantly from generation to generation. - These are fine-grained details which Intel doesn’t generally release to the public.
Nevertheless, I’ll try to answer…
When multiple uops are ready in the reservation station, in what order are they scheduled to ports?
It should be the oldest [see below], but your mileage may vary. The P6 microarchitecture (used in the Pentium Pro, 2 & 3) used a reservation station with five schedulers (one per execution port); the schedulers used a priority pointer as a place to start scanning for ready uops to dispatch. It was only pseudo FIFO so it’s entirely possible that the oldest ready instruction was not always scheduled. In the NetBurst microarchitecture (used in Pentium 4), they ditched the unified reservation station and used two uop queues instead. These were proper collapsing priority queues so the schedulers were guaranteed to get the oldest ready instruction. The Core architecture returned to a reservation station and I would hazard an educated guess that they used the collapsing priority queue, but I can’t find a source to confirm this. If somebody has a definitive answer, I’m all ears.
When a uop can go to multiple ports (like the add and lea in the example above), how is it decided which port is chosen?
That’s tricky to know. The best I could find is a patent from Intel describing such a mechanism. Essentially, they keep a counter for each port that has redundant functional units. When the uops leave the front end to the reservation station, they are assigned a dispatch port. If it has to decide between multiple redundant execution units, the counters are used to distribute the work evenly. Counters are incremented and decremented as uops enter and leave the reservation station respectively.
Naturally this is just a heuristic and does not guarantee a perfect conflict-free schedule, however, I could still see it working with your toy example. The instructions which can only go to one port would ultimately influence the scheduler to dispatch the “less restricted” uops to other ports.
In any case, the presence of a patent doesn’t necessarily imply that the idea was adopted (although that said, one of the authors was also a tech lead of the Pentium 4, so who knows?)
If any of the answers involve a concept like oldest to choose among uops, how is it defined? Age since it was delivered to the RS? Age since it became ready? How are ties broken? Does program order ever come into it?
Since uops are inserted into the reservation station in order, oldest here does indeed refer to time it entered the reservation station, i.e. oldest in program order.
By the way, I would take those IACA results with a grain of salt as they may not reflect the nuances of the real hardware. On Haswell, there is a hardware counter called uops_executed_port that can tell you how many cycles in your thread were uops issues to ports 0-7. Maybe you could leverage these to get a better understanding of your program?