Doing a full 64-bit decomposition is really stupid especially for
simple values like 0 and -1.

But if we are going to optimize this, go all the way and try for all 2
and 3 instruction sequences not requiring a temporary register as
well.

First we do the easy cases where it's a zero or sign extended 32-bit
number (sethi+or, sethi+xor, respectively).

Then we try to find a range of set bits we can load simply then shift
up into place, in various ways.

Then we try negating the constant and see if we can do a simple
sequence using that with a xor at the end.  (f.e. the range of set
bits can't be loaded simply, but for the negated value it can)

The final optimized strategy involves 4 instructions sequences not
needing a temporary register.

Otherwise we sadly fully decompose using a temp..

Example, from ALU64_XOR_K: 0x0000ffffffff0000 ^ 0x0 = 0x0000ffffffff0000:

0000000000000000 <foo>:
   0:   9d e3 bf 50     save  %sp, -176, %sp
   4:   01 00 00 00     nop
   8:   90 10 00 18     mov  %i0, %o0
   c:   13 3f ff ff     sethi  %hi(0xfffffc00), %o1
  10:   92 12 63 ff     or  %o1, 0x3ff, %o1     ! ffffffff <foo+0xffffffff>
  14:   93 2a 70 10     sllx  %o1, 0x10, %o1
  18:   15 3f ff ff     sethi  %hi(0xfffffc00), %o2
  1c:   94 12 a3 ff     or  %o2, 0x3ff, %o2     ! ffffffff <foo+0xffffffff>
  20:   95 2a b0 10     sllx  %o2, 0x10, %o2
  24:   92 1a 60 00     xor  %o1, 0, %o1
  28:   12 e2 40 8a     cxbe  %o1, %o2, 38 <foo+0x38>
  2c:   9a 10 20 02     mov  2, %o5
  30:   10 60 00 03     b,pn   %xcc, 3c <foo+0x3c>
  34:   01 00 00 00     nop
  38:   9a 10 20 01     mov  1, %o5     ! 1 <foo+0x1>
  3c:   81 c7 e0 08     ret
  40:   91 eb 40 00     restore  %o5, %g0, %o0
Signed-off-by: David S. Miller <davem@davemloft.net>