本文整理汇总了C++中channel::empty方法的典型用法代码示例。如果您正苦于以下问题：C++ channel::empty方法的具体用法？C++ channel::empty怎么用？C++ channel::empty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channel的用法示例。
在下文中一共展示了channel::empty方法的2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: getCornerness
/*
* compute cornerness
*/
bool harrisCorners::getCornerness(const channel& fxx,
const channel& fxy,
const channel& fyy,
const float scale,
channel& cornerness,
float& maxCornerness) const {
// we can assume that all channels are connected, but try it out if not
if ((fxx.getMode() != channel::Connected) ||
(fxy.getMode() != channel::Connected) ||
(fyy.getMode() != channel::Connected)) {
setStatusString("Channels not contigous in getCornerness");
return false;
}
if (fxx.empty() || fxy.empty() || fyy.empty()) {
cornerness.clear();
maxCornerness = 0.0f;
return false;
}
int i;
const int end = fxx.rows()*fxx.columns();
const float *const pfxx = &fxx.at(0);
const float *const pfxy = &fxy.at(0);
const float *const pfyy = &fyy.at(0);
cornerness.resize(fxx.size(),0,false,false);
float* pcor = &cornerness.at(0);
float det,trace,txx,txy,tyy,c;
float maxc = 0.0f;
for (i=0;i<end;++i) {
txx=pfxx[i];
txy=pfxy[i];
tyy=pfyy[i];
det=txx*tyy - txy*txy;
trace=txx+tyy;
c = det-scale*trace*trace;
pcor[i]=c;
if (c>maxc) {
maxc=c;
}
}
maxCornerness = maxc;
return true;
}示例2: findCornerMaxima
/*
* find corners with maximal cornerness
*/
bool harrisCorners::findCornerMaxima(const channel& cornerness,
channel& cornersOnly,
pointList& cornerMax) const {
if (cornerness.empty()) {
cornersOnly.clear();
cornerMax.clear();
return true;
}
const parameters& par = getParameters();
const float corner = par.cornerValue/255.0f;
const float noCorner = par.noCornerValue/255.0f;
localMaxima<float> lmax;
localMaxima<float>::parameters lmaxPar(par.localMaximaParameters);
lmaxPar.noMaxValue = noCorner;
lmaxPar.maxNumber = par.maximumCorners;
lmax.setParameters(lmaxPar);
if (lmax.apply(cornerness,cornersOnly,cornerMax)) {
pointList::iterator it;
int i;
for (it=cornerMax.begin(),i=0;
(it!=cornerMax.end());
++it) {
cornersOnly.at(*it) = corner;
}
for (;it!=cornerMax.end();++it) {
cornersOnly.at(*it) = noCorner;
}
return true;
}
return false;
}