https://gcc.gnu.org/bugzilla/show_bug.cgi?id=85090
Jakub Jelinek <jakub at gcc dot gnu.org> changed:
What |Removed |Added
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CC| |jakub at gcc dot gnu.org,
| |vmakarov at gcc dot gnu.org
--- Comment #5 from Jakub Jelinek <jakub at gcc dot gnu.org> ---
I guess it depends on what exactly a normal subreg on lhs means.
The documentation says:
When used as an lvalue, 'subreg' is a word-based accessor.
Storing to a 'subreg' modifies all the words of REG that
overlap the 'subreg', but it leaves the other words of REG
alone.
but in this case, we really don't have a GPR, but rather a single much larger
(512-bit) register. Does it still imply that for -m32 (subreg:SI (reg:V32HI)
0)
sets just the low 32 bits of the large register and doesn't modify anything
else,
and for -m64 the same means set low 32 bits, have the 32 bits above it
undefined and the rest of bits unmodified?
Seems store_bit_field treats it that way, but perhaps IRA does not?
