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