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14933dc8d9
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> |
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