Are load ops deallocated from the RS when they dispatch, complete or some other time?

怎甘沉沦 提交于 2020-02-24 00:37:41

问题


On modern Intel1 x86, are load uops freed from the RS (Reservation Station) at the point they dispatch2, or when they complete3, or somewhere in-between4?


1 I am also interested in AMD Zen and sequels, so feel free to include that too, but for the purposes of making the question manageable I limit it to Intel. Also, AMD seems to have a somewhat different load pipeline from Intel which may make investigating this on AMD a separate task.

2 Dispatch here means leave the RS for execution.

3 Complete here means when the load data returns and is ready to satisfy dependent uops.

4 Or even somewhere outside of the range of time defined by these two events, which seems unlikely but possible.


回答1:


The following experiments suggest that the uops are deallocated at some point before the load completes. While this is not a complete answer to your question, it might provide some interesting insights.

On Skylake, there is a 33-entry reservation station for loads (see https://stackoverflow.com/a/58575898/10461973). This should also be the case for the Coffee Lake i7-8700K, which is used for the following experiments.

We assume that R14 contains a valid memory address.

clflush [R14]
clflush [R14+512]
mfence

# start measuring cycles

mov RAX, [R14]
mov RAX, [R14]
...
mov RAX, [R14]

mov RBX, [R14+512]

# stop measuring cycles

mov RAX, [R14] is unrolled 35 times. A load from memory takes at least about 280 cycles on this system. If the load uops stayed in the 33-entry reservation station until completion, the last load could only start after more than 280 cycles and would need another ~280cycles. However, the total measured time for this experiment is only about 340 cycles. This indicates that the load uops leave the RS at some time before completion.

In contrast, the following experiments shows a case where most uops are forced to stay in the reservation until the first load completes:

mov RAX, R14
mov [RAX], RAX
clflush [R14]
clflush [R14+512]
mfence

# start measuring cycles

mov RAX, [RAX]
mov RAX, [RAX]
...
mov RAX, [RAX]

mov RBX, [R14+512]

# stop measuring cycles

The first 35 loads now have dependencies on each other. The measured time for this experiment is about 600 cycles.

The experiments were performed with all but one core disabled, and with the CPU governor set to performance (cpupower frequency-set --governor performance).

Here are the nanoBench commands I used:

./nanoBench.sh -unroll 1 -basic -asm_init "clflush [R14]; clflush [R14+512]; mfence" -asm "mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RAX, [R14]; mov RBX, [R14+512]"

./nanoBench.sh -unroll 1 -basic -asm_init "mov RAX, R14; mov [RAX], RAX; clflush [R14]; clflush [R14+512]; mfence" -asm "mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RAX, [RAX]; mov RBX, [R14+512]"



来源:https://stackoverflow.com/questions/59905395/are-load-ops-deallocated-from-the-rs-when-they-dispatch-complete-or-some-other

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