Interesting. I only know a little assembler, so I can't tell what -03 is producing there, but I know enough to see that it's doing some sort of loop bouncing between even/odd label choice. Is that now a constant-size binary (if still linear time)?
What GCC is doing at -O3 is encoding even (resp. odd) numbers in a bitmap.
43690 = 0xAAAA = 0b1010101010101010
21845 = 0x5555 = 0b0101010101010101
Then
sal rax, cl
test eax, 43690
jne .L3
// .L3
mov edi, OFFSET FLAT:.LC1 // This is a pointer to the string "odd"
call puts
jmp .L2
is equivalent to C:
if ((1 << number) & 0xAAAA) {
puts("odd"); // x64 is little endian
}
So execution is constant time (it only needs to check against two constants), there's no loop.
It should be advantageous compared to a jump table because it doesn't have to do two indirect jumps (though you now have branches that might mispredict).
