Related to this answer: https://stackoverflow.com/a/11227902/4714970
In the above answer, it's mentioned how you can avoid branch prediction fails by avoiding branches.
The user demonstrates this by replacing:
if (data[c] >= 128)
{
sum += data[c];
}
With:
int t = (data[c] - 128) >> 31;
sum += ~t & data[c];
How are these two equivalent (for the specific data set, not strictly equivalent)?
What are some general ways I can do similar things in similar situations? Would it always be by using >>
and ~
?
int t = (data[c] - 128) >> 31;
The trick here is that if data[c] >= 128
, then data[c] - 128
is nonnegative, otherwise it is negative. The highest bit in an int
, the sign bit, is 1 if and only if that number is negative. >>
is a shift that extends the sign bit, so shifting right by 31 makes the whole result 0 if it used to be nonnegative, and all 1 bits (which represents -1) if it used to be negative. So t
is 0
if data[c] >= 128
, and -1
otherwise. ~t
switches these possibilities, so ~t
is -1
if data[c] >= 128
, and 0
otherwise.
x & (-1)
is always equal to x
, and x & 0
is always equal to 0
. So sum += ~t & data[c]
increases sum
by 0
if data[c] < 128
, and by data[c]
otherwise.
Many of these tricks can be applied elsewhere. This trick can certainly be generally applied to have a number be 0
if and only if one value is greater than or equal to another value, and -1
otherwise, and you can mess with it some more to get <=
, <
, and so on. Bit twiddling like this is a common approach to making mathematical operations branch-free, though it's certainly not always going to be built out of the same operations; ^
(xor) and |
(or) also come into play sometimes.