https://gcc.gnu.org/bugzilla/show_bug.cgi?id=87871
--- Comment #21 from Wilco <wilco at gcc dot gnu.org> --- (In reply to Vladimir Makarov from comment #20) > (In reply to Wilco from comment #19) > > (In reply to Peter Bergner from comment #18) > > > (In reply to Segher Boessenkool from comment #15) > > > > Popping a5(r116,l0) -- assign reg 3 > > > > Popping a3(r112,l0) -- assign reg 4 > > > > Popping a2(r114,l0) -- assign reg 3 > > > > Popping a0(r111,l0) -- assign reg 0 > > > > Popping a4(r117,l0) -- assign reg 0 > > > > Popping a1(r113,l0) -- assign reg 2 > > > > Assigning 4 to a5r116 > > > > Disposition: > > > > 0:r111 l0 0 3:r112 l0 4 1:r113 l0 2 2:r114 l0 > > > > 3 > > > > 5:r116 l0 4 4:r117 l0 0 > > > > > > > > > > > > r116 does not conflict with *any* other pseudo. It is alive in the > > > > first > > > > two insns of the function, which are > > > > > > So we initially assign r3 to r116 presumably because it has the same cost > > > as > > > the other gprs and it occurs first in REG_ALLOC_ORDER. Then > > > improve_allocation() decides that r4 is a better hard reg and switches the > > > assignment to that. I'm not sure why it wouldn't choose r0 there instead. > > > > I would expect that r116 has a strong preference for r0 given the r116 = mov > > r0 and thus allocating r116 to r0 should have the lowest cost by a large > > margin. > > p116 conflicts with hr0. Therefore it can not get hr0. p112 is connected > with p116. p112 got hr4 and p116 got 3. Assigning 4 to 116 is profitable. > Therefore assignment of p116 is changed to 4. > > The question is why p116 conflicts with hr0. Before RA we have That's a bug since register copies should not create a conflict. It's one of the most basic optimization of register allocator. And there is also the question why we do move r0 into a virtual register but not assign the virtual register to an argument register.
