| |
| // inline utilities |
| /* Returns 0 if negative, 1 if zero, 2 if positive |
| */ |
| inline int side(double x) { |
| return (x > 0) + (x >= 0); |
| } |
| |
| /* Returns 1 if negative, 2 if zero, 4 if positive |
| */ |
| inline int sideBit(double x) { |
| return 1 << side(x); |
| } |
| |
| /* Given the set [0, 1, 2, 3], and two of the four members, compute an XOR mask |
| that computes the other two. Note that: |
| |
| one ^ two == 3 for (0, 3), (1, 2) |
| one ^ two < 3 for (0, 1), (0, 2), (1, 3), (2, 3) |
| 3 - (one ^ two) is either 0, 1, or 2 |
| 1 >> 3 - (one ^ two) is either 0 or 1 |
| thus: |
| returned == 2 for (0, 3), (1, 2) |
| returned == 3 for (0, 1), (0, 2), (1, 3), (2, 3) |
| given that: |
| (0, 3) ^ 2 -> (2, 1) (1, 2) ^ 2 -> (3, 0) |
| (0, 1) ^ 3 -> (3, 2) (0, 2) ^ 3 -> (3, 1) (1, 3) ^ 3 -> (2, 0) (2, 3) ^ 3 -> (1, 0) |
| */ |
| inline int other_two(int one, int two) { |
| return 1 >> 3 - (one ^ two) ^ 3; |
| } |
| |
| /* Returns -1 if negative, 0 if zero, 1 if positive |
| */ |
| inline int sign(double x) { |
| return (x > 0) - (x < 0); |
| } |
| |
| inline double interp(double A, double B, double t) { |
| return A + (B - A) * t; |
| } |