inline TYPE VECTOR_CMOV( TYPE a, TYPE b, TYPE c, TYPE d){
TYPE z = VECTOR_EQUAL ( a, b );
// return _mm_blendv_epi8( d, c, z );
TYPE g = VECTOR_AND ( c, z );
TYPE h = VECTOR_ANDNOT( z, d );
return VECTOR_OR ( g, h );
}
/**
* Returns the intersections of two vectors of circles.
* @ret: the number of intersection [0,1,2]
*/
static inline VECTOR
__attribute__((__always_inline__,__gnu_inline__,__nonnull__,__artificial__))
circle_get_intersection_ps(const VECTOR p1x, const VECTOR p1y, const VECTOR p2x,
const VECTOR p2y, const VECTOR r1, const VECTOR r2,
VECTOR *restrict retx, VECTOR *restrict rety)
{
VECTOR d = distance(p1x, p1y, p2x, p2y);
// no solutions, the circles are separate || the circles are coincident || no solutions because one circle is contained within the other
// => infinite number of solutions possible
VECTOR one_sol = VECTOR_GE(r1 + r2, d);
one_sol = VECTOR_AND(one_sol, VECTOR_LE(VECTOR_ABS(r1 - r2), d));
one_sol = VECTOR_AND(one_sol, VECTOR_NE(VECTOR_ZERO(), d));
VECTOR a = (r1*r1 - r2*r2 + d*d) / (d + d);
VECTOR v = r1*r1 - a*a;
VECTOR h = VECTOR_SQRT(v);
VECTOR dx = (p2x - p1x) / d;
VECTOR dy = (p2y - p1y) / d;
VECTOR p3x = p1x + a * dx;
VECTOR p3y = p1y + a * dy;
dx *= h;
dy *= h;
VECTOR p4x = p3x + dy;
VECTOR p4y = p3y - dx;
inline TYPE VECTOR_GET_SIGN_BIT(TYPE a, TYPE m){
TYPE b = VECTOR_AND(a, m);
return b;
}
//
// ON INVERSE LE SIGNE DE LA DONNEE EN FONCTION DU SIGNE
//
inline TYPE VECTOR_invSIGN2( TYPE a, TYPE z){
TYPE g = VECTOR_AND ( a, z );
TYPE h = VECTOR_ANDNOT( VECTOR_NEG(a), z );
return VECTOR_OR ( g, h );
}
