https://gcc.gnu.org/bugzilla/show_bug.cgi?id=120284
--- Comment #4 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
(In reply to Huiba Li from comment #3)
> > Note the 0 there rather than r. r in the input means any register while 0
> > means it needs to match the same register as the 0th operand which in this
> > case is the output operand.
>
> Thanks for your quick response. Using "0" does resolve the issue, as what
> "+r" does.
>
> And actually, I don't need the input register to be the same as the output
> register. All I need here is making x both input and output to the assembly.
> So in theory, should it also be OK to write ```asm volatile("" : "=r"(x) :
> "r"(x))```?
>
> BTW, the "+r" form and "0" form both produce code like this:
>
> ...
> <+64>: movq %r15, %rax
> <+67>: movq %rax, %r15
> vpbroadcastd (%rax), %zmm17
> ...
>
> This also looks sub-optimal.
Marking x as an output without tieing it to another register will have garbage
in the variable after the inline-asm. That is explicitly mentioned.
