> On Jul 14, 2025, at 6:11 PM, Bradley J Lucier <lucier@xxxxxxxxxx> wrote:
>
> I also found
>
> static unsigned int
> execute_hardreg_pre (void)
> {
> #ifdef HARDREG_PRE_REGNOS
> doing_hardreg_pre_p = true;
> unsigned int regnos[] = HARDREG_PRE_REGNOS;
> /* It's possible to avoid this loop, but it isn't worth doing so until
> hardreg PRE is used for multiple hardregs. */
> for (int i = 0; regnos[i] != 0; i++)
> {
> int changed;
> current_hardreg_regno = regnos[i];
> if (dump_file)
> fprintf(dump_file, "Entering hardreg PRE for regno %d\n",
> current_hardreg_regno);
> delete_unreachable_blocks ();
> df_analyze ();
> changed = one_pre_gcse_pass ();
> if (changed)
> cleanup_cfg (0);
> }
> doing_hardreg_pre_p = false;
> #endif
> return 0;
> }
> ...
> bool
> pass_hardreg_pre::gate (function * ARG_UNUSED (fun))
> {
> #ifdef HARDREG_PRE_REGNOS
> return optimize > 0
> && !fun->calls_setjmp;
> #else
> return false;
> #endif
> }
>
> and
>
> heine:~/programs/gcc/gcc-mainline/gcc> grep -R HARDREG_PRE_REGNOS config
> config/aarch64/aarch64.h:#define HARDREG_PRE_REGNOS { FPM_REGNUM, 0 }
>
> so aarch64 seems to be the only architecture that defines HARDREG_PRE_REGNOS.
This is all very interesting, and leads to more questions.
I looked at a sample dump file and what seems to be the summary statistics for the largest routine were
====================
;; Function ___H___num (___H___num, funcdef_no=66, decl_uid=11088, cgraph_uid=524, symbol_order=523)
Entering hardreg PRE for regno 84
starting the processing of deferred insns
ending the processing of deferred insns
df_analyze called
df_worklist_dataflow_doublequeue: n_basic_blocks 31567 n_edges 51599 count 31791 ( 1)
Expression hash table (54357 buckets, 0 entries)
PRE GCSE of ___H___num, 31567 basic blocks, 2524104 bytes needed, 0 substs, 0 insns created
starting the processing of deferred insns
ending the processing of deferred insns
====================
Does this mean that the hardreg_pre pass didn’t actually change anything (“0 substs, 0 insns created”)?
It appears from Arm’s documentation that FPM_REGNUM controls certain properties of FP8 operations. Does this pass need to be run if (as in my case) the routine uses no FP8 instructions?
Thanks.
Brad
