C++ Geom::rot90方法代码示例

/**
 *
 * \todo
 * FIXME: This can be done using linear operations, more stably and
 *  faster.  method: transform A and C into B's space, A should be
 *  negative and B should be positive in the orthogonal component.  I
 *  think this is equivalent to
 *  dot(A, rot90(B))*dot(C, rot90(B)) == -1.
 *    -- njh
 */
static bool
vectors_are_clockwise (Geom::Point A, Geom::Point B, Geom::Point C)
{
    using Geom::rot90;
    double ab_s = dot(A, rot90(B));
    double ab_c = dot(A, B);
    double bc_s = dot(B, rot90(C));
    double bc_c = dot(B, C);
    double ca_s = dot(C, rot90(A));
    double ca_c = dot(C, A);

    double ab_a = acos (ab_c);

    if (ab_c <= -1.0) {
        ab_a = M_PI;
    }

    if (ab_c >= 1.0) {
        ab_a = 0;
    }

    if (ab_s < 0) {
        ab_a = 2 * M_PI - ab_a;
    }

    double bc_a = acos (bc_c);

    if (bc_c <= -1.0) {
        bc_a = M_PI;
    }

    if (bc_c >= 1.0) {
        bc_a = 0;
    }

    if (bc_s < 0) {
        bc_a = 2 * M_PI - bc_a;
    }

    double ca_a = acos (ca_c);

    if (ca_c <= -1.0) {
        ca_a = M_PI;
    }

    if (ca_c >= 1.0) {
        ca_a = 0;
    }

    if (ca_s < 0) {
        ca_a = 2 * M_PI - ca_a;
    }

    double lim = 2 * M_PI - ca_a;

    if (ab_a < lim) {
        return true;
    }

    return false;
}