https://gcc.gnu.org/bugzilla/show_bug.cgi?id=104110
Andrew Pinski <pinskia at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
Last reconfirmed| |2022-01-19
Keywords|ra |
Status|UNCONFIRMED |NEW
Ever confirmed|0 |1
--- Comment #3 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
Rs6000 has:
/* On the RS/6000, if it is valid in the insn, it is free. */
case CONST_INT:
if (((outer_code == SET
|| outer_code == PLUS
|| outer_code == MINUS)
&& (satisfies_constraint_I (x)
|| satisfies_constraint_L (x)))
While aarch64 has:
/* If an instruction can incorporate a constant within the
instruction, the instruction's expression avoids calling
rtx_cost() on the constant. If rtx_cost() is called on a
constant, then it is usually because the constant must be
moved into a register by one or more instructions.
The exception is constant 0, which can be expressed
as XZR/WZR and is therefore free. The exception to this is
if we have (set (reg) (const0_rtx)) in which case we must cost
the move. However, we can catch that when we cost the SET, so
we don't need to consider that here. */
if (x == const0_rtx)
*cost = 0;
else
{
/* To an approximation, building any other constant is
proportionally expensive to the number of instructions
required to build that constant. This is true whether we
are compiling for SPEED or otherwise. */
if (!is_a <scalar_int_mode> (mode, &int_mode))
int_mode = word_mode;
*cost = COSTS_N_INSNS (aarch64_internal_mov_immediate
(NULL_RTX, x, false, int_mode));
}
I think rs6000 definition should be used instead otherwise we get the case
where (set (reg) (reg)) is the same cost as (set (reg) (const_int)) so GCC
decides it is better to do the (set (reg) (reg)) instead of staying with the
(set (reg) (const_int)).
