kFpRegPair requires a bit of special case logic in allocation and spill
logic, since it requires adjacent fp registers, with the low register
being an even-numbered one.

We make use of a shift+and trick with the reg list to check this. Given
a bit mask of registers available, AND this with a right-shift-by-one
version of itself, and then AND it again with with a bit mask of even
numbered fp registers, we can get a bit mask where all available even
registers are set.

For example, given 00011011, where 1 means register is available:

  00011011
&  0001101 (right shift by 1)
----------
  00001001
& 01010101 (mask of even registers)
----------
  00000001

We see that only the first fp register is available, based on the
adjacency requirement.

The mask of even registers is defined to be uint64_t, to workarond shift
overflow issues when non-unsigned values are used, and also to make this
work on all platforms and compilers without implicit conversion warnings.
It is a bit wasteful, since we really only need storage_t, but since it
is a constexpr it should be okay.

The spilling case tries to be a bit smart, since there are three possible
cases (described in the code), two of which we can spill one register,
and in the last case we have to spill two registers. We try to
distinguish between these cases in order to minimize spills. The dumb
thing here is in the last case, we arbitrarily pick the first two
registers (d0, d1) to spill.

Bug: v8:9909
Change-Id: I408f9f79b1f3c505f2fd73bad15923188ae7aaf4
Reviewed-on: https://chromium-review.googlesource.com/c/v8/v8/+/1975033
Commit-Queue: Zhi An Ng <zhin@chromium.org>
Reviewed-by: Clemens Backes <clemensb@chromium.org>
Cr-Commit-Position: refs/heads/master@{#65546}