本文整理汇总了C++中simtk::Vector::nrow方法的典型用法代码示例。如果您正苦于以下问题：C++ Vector::nrow方法的具体用法？C++ Vector::nrow怎么用？C++ Vector::nrow使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类simtk::Vector的用法示例。
在下文中一共展示了Vector::nrow方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: log
/*Detailed Computational Costs
Comparisons Div Mult Additions Assignments
splineGuess log(n,2) 2 3 1
(n currently set to 100)
Newton Iter
f 21 20 13
df 20 19 11
update 4 1 3 6
total 4 1 41 42 30
\endverbatim
To evaluate u to SimTK::Eps*100 this typically involves 2 Newton
iterations, yielding a total cost of
\verbatim
Comparisons Div Mult Additions Assignments
eval U 7+8=15 2 82 42 60
*/
double SegmentedQuinticBezierToolkit::calcU(double ax, const SimTK::Vector& bezierPtsX,
const SimTK::Spline& splineUX, double tol,
int maxIter)
{
//Check to make sure that ax is in the curve domain
double minX = 1e100;
double maxX = -1e100;
for(int i=0; i<bezierPtsX.nrow(); i++){
if(bezierPtsX(i) > maxX)
maxX = bezierPtsX(i);
if(bezierPtsX(i) < minX)
minX = bezierPtsX(i);
}
SimTK_ERRCHK_ALWAYS( ax >= minX && ax <= maxX,
"SegmentedQuinticBezierToolkit::calcU",
"Error: input ax was not in the domain of the Bezier curve specified \n"
"by the control points in bezierPtsX.");
double u = splineUX.calcValue(ax);
u = clampU(u);
double f = 0;
f = calcQuinticBezierCurveVal(u,bezierPtsX)-ax;
double df= 0;
double du= 0;
int iter = 0;
bool pathologic = false;
//Newton iterate to the desired tolerance
while(abs(f) > tol && iter < maxIter && pathologic == false){
//Take a Newton step
df = calcQuinticBezierCurveDerivU(u,bezierPtsX,1);
if(abs(df) > 0){
du = -f/df;
u = u + du;
u = clampU(u);
f = calcQuinticBezierCurveVal(u,bezierPtsX)-ax;
}else{
pathologic = true;
}
iter++;
}
//Check for convergence
SimTK_ERRCHK2_ALWAYS( (f <= tol),
"SegmentedQuinticBezierToolkit::calcU",
"Error: desired tolerance of %f on U not met by the Newton iteration."
" A tolerance of %f was reached.",tol, f);
SimTK_ERRCHK_ALWAYS( (pathologic==false),
"SegmentedQuinticBezierToolkit::calcU",
"Error: Newton iteration went pathologic: df = 0 to machine precision.");
//Return the value
return u;
}