C/C++ Bit Twiddling

Question

in the spirit of graphics.stanford.edu/~seander/bithacks.html I need to solve the following problem:

int x; 
int pow2; // always a positive power of 2
int sgn;  // always either 0 or 1
// ...
// ...
if(sgn == 0)
    x -= pow2;
else
    x += pow2;

Of course I need to avoid the conditional. So far the best I came up with is

x -= (1|(~sgn+1))*pow2

but that involves a multiplication which I also would like to avoid. Thanks in advance.

EDIT: Thanks all,

x -= (pow2^-sgn) + sgn

seems to do the trick!

Answer 1

I would try

x -= (pow2 ^ (~sgn+1)) + sgn

or, as suggested by lijie in the comments

x -= (pow2 ^ -sgn) + sgn

If sgn is 0, ~sgn+1 is also 0, so pow2 ^ (~sgn+1) == pow2. If sgn is 1, (~sgn+1) is 0xFFFFFFFF, and (pow2 ^ (~sgn+1)) + sgn == -pow2.

Answer 2

mask = sgn - 1; // generate mask: sgn == 0 => mask = -1, sgn == 1 => mask = 0

x = x + (mask & (-pow2)) + (~mask & (pow2)); // use mask to select +/- pow2 for addition