本文整理汇总了C++中TracePoint::get_location方法的典型用法代码示例。如果您正苦于以下问题：C++ TracePoint::get_location方法的具体用法？C++ TracePoint::get_location怎么用？C++ TracePoint::get_location使用的例子？那么, 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TracePoint的用法示例。
在下文中一共展示了TracePoint::get_location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:
static void
PrintTracePoint(const TracePoint &point, std::ofstream& fs)
{
fs << point.GetTime()
<< " " << point.get_location().longitude
<< " " << point.get_location().latitude
<< " " << point.GetAltitude()
<< " " << point.GetVario()
<< "\n";
}示例2: Result
TriangleSecondLeg::Result
TriangleSecondLeg::Calculate(const TracePoint &c, unsigned best) const
{
// this is a heuristic to remove invalid triangles
// we do as much of this in flat projection for speed
const unsigned df_2 = b.flat_distance(c);
const unsigned df_3 = c.flat_distance(a);
const unsigned df_total = df_1+df_2+df_3;
// require some distance!
if (df_total<20) {
return Result(0, 0);
}
// no point scanning if worst than best
if (df_total<= best) {
return Result(0, 0);
}
const unsigned shortest = min(df_1, min(df_2, df_3));
// require all legs to have distance
if (!shortest) {
return Result(0, 0);
}
if (is_fai && (shortest*4<df_total)) { // fails min < 25% worst-case rule!
return Result(0, 0);
}
const unsigned d = df_3+df_2;
// without FAI rules, allow any triangle
if (!is_fai) {
return Result(d, df_total);
}
if (shortest*25>=df_total*7) {
// passes min > 28% rule,
// this automatically means we pass max > 45% worst-case
return Result(d, df_total);
}
const unsigned longest = max(df_1, max(df_2, df_3));
if (longest*20>df_total*9) { // fails max > 45% worst-case rule!
return Result(0, 0);
}
// passed basic tests, now detailed ones
// find accurate min leg distance
fixed leg(0);
if (df_1 == shortest) {
leg = a.get_location().distance(b.get_location());
} else if (df_2 == shortest) {
leg = b.get_location().distance(c.get_location());
} else if (df_3 == shortest) {
leg = c.get_location().distance(a.get_location());
}
// estimate total distance by scaling.
// this is a slight approximation, but saves having to do
// three accurate distance calculations.
const fixed d_total((df_total*leg)/shortest);
if (d_total>=fixed(500000)) {
// long distance, ok that it failed 28% rule
return Result(d, df_total);
}
return Result(0, 0);
}