我之前尝试过问我的问题,但我认为我问问题的方式不合适。所以我在这里再次尝试:(我仍然不知道什么主题合适)
首先我定义
typedef boost::variant<point, Line, Vertex> vec_variant;
typedef std::vector<vec_variant> vec;
我编写函数取决于它返回点、线、顶点甚至它们的组合的情况。
vec my_func::select_T(const mesh::section& s, const char* model) const
{
vec new_vec;
.
.
.
.
//loop over my lines
else if ( strcmp(model , "Line") == 0 )
{
for(section::lineIterator ite = s.beginLine(); ite != s.endLine(); ++ite )
{
Line ed = *ite;
Point p0 = ed.point(0);
Point p1 = ed.point(1);
Point p0_modified ( /* some modification */ );
Point p1_modified ( /* some modification */ );
if( /* some other conditions */ )
{
new_vec.push_back(ed);
new_vec.push_back(p0_modified); //note before pushing back any point
new_vec.push_back(p1_modified); //first I pushed back line
}
else if ( /* some other conditions */ )
{
.
.
.
vertex m = .......;
new_vec.push_back(ed);
new_vec.push_back(m); //note before pushing back any point
//first I pushed back line
}
}
}
}
return new_vec;
}
所以最后我们可能会有这样的东西 {ed, p0_modified, p0_modified, ed, m, ed, m, ed, p0_modified, p0_modified, ed, m, ....} {Line,Point,Point,Line,Vertex ,线,顶点,线,点,点,线,顶点, ...}
现在我在代码的另一部分(不同的文件)调用这个函数
first I defined a visitor:
template<typename T>
struct T_visitor : public boost::static_visitor<>
{
T_visitor(std::vector<T>& v) : zeroVector(v) {}
template<typename U>
void operator () (const U&) {}
void operator () (const T& value)
{
zeroVector.push_back(value);
}
private:
std::vector<T>& zeroVector;
};
我在这里调用了上面的函数:
void func_2( /*......*/ )
{
. //we can use above visitor to store each type (point, line, Vertex) in vactor<point or line or Vertex)
. //we do not know what the new_vec is at the end of the loop. the only thing we know is that after each line there
. //would be either two points or one vertex
.
const char *model = "Edge";
.
.//How to find line ed and corresponded points?
.
create_line( Point& p0_modified, Point& p1_modified, Line& ed); //two modified points and line
.
.//How to find point m and corresponded line?
.
create_point( m(0), m(1), m(2), Line& ed); //point m coordinates and line
.
.
}