获得点的坐标从升压几何多边形 | 获得点的坐标从升压几何多边形

本文介绍了获得点的坐标从升压几何多边形的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个简单的DLL做一些计算与升压几何多边形。（主要路口和差异。）由于DLL将最有可能从C＃code调用，德尔福，谁在那里别人知道，我应该将结果转换成一切都可以处理数组。

I have a simple DLL doing some calculations with Boost Geometry polygons. (Mostly intersections and differences.) Since the DLL will be most likely called from C# code, and from Delphi and who knows from where else, I should convert the result into arrays that everything can handle.

更新：
我已经简化了，有点纠正我的code。新的code看起来完全不同，采用了完全不同的方法（ for_each_point ），并以某种方式仍然不能编译。

UPDATE:I had simplified and somewhat corrected my code. The new code looks completely different, uses a completely different approach (for_each_point), and somehow still doesn't compile.

我的新code：

#include <vector>
#include <boost/range.hpp>
#include <boost/geometry.hpp>
#include <boost/geometry/geometries/polygon.hpp>

using namespace boost::geometry;

typedef boost::geometry::model::point
    <
        double, 2, boost::geometry::cs::spherical_equatorial<boost::geometry::degree>
    > spherical_point;
class PointAggregator {
private :
    double *x, *y;
    int count;

public :
    PointAggregator(int size) {
        x = (double*) malloc(sizeof(double) * size);
        y = (double*) malloc(sizeof(double) * size);
        count = 0;
    }

    ~PointAggregator() {
        free(x);
        free(y);
    }

    inline void operator()(spherical_point& p) {
        x[count] = get<0>(p);
        y[count] = get<1>(p);
        count++;
    }

    void GetResult(double *resultX, double *resultY) {
        resultX = x;
        resultY = y;
    }
};

void VectorToArray(std::vector<model::polygon<spherical_point>> resultVector, double x[], double y[], int *count) {
    int i = 0;
    for (std::vector<model::polygon<spherical_point>>::iterator it = resultVector.begin(); it != resultVector.end(); ++it) {
        if (boost::size(*it) >= 2) {
            *count = boost::size(*it);
            PointAggregator* pa = new PointAggregator(*count);
            boost::geometry::for_each_point(*it, *pa);
            pa->GetResult(x, y);
            delete(pa);
            break;
        }
    }
}

目前的编译错误是：

The current compilation errors are:

错误C2039：类型：不是的boost :: MPL :: eval_if_c成员iterator.hpp 63

错误C3203：类型：非特类模板不能作为模板参数迭代模板参数，预计一个真正的类型difference_type.hpp 25

错误C2955：'的boost ::类型：使用类模板需要模板参数列表difference_type.hpp 25

错误C2955：'的boost :: iterator_difference：使用类模板需要模板参数列表difference_type.hpp 26

哪几个不喜欢看他们有什么关系code的这部分（我的文件名是geometry.cpp），但一切使用升压几何被注释掉了，我仍然得到这些错误，因此， ...

Which ones don't look like they have anything to do with this part of code (my filename is geometry.cpp), but everything else that uses Boost Geometry is commented out and I still get these errors, so...

（我是新的C ++和Boost所以我可能会错过一些基本的概念，而把code从互联网在一起。）
我以为我可以不通过多边形进行迭代很容易，我错过了不平凡的一部分，或者说，一个多边形不能用作环或迭代只是没有我以为的方式本身，或者我没有想法还有什么可以是错误的。我做了什么错了？

(I'm new to C++ and Boost so I might have missed some basic concept while putting code from the internet together.)I assume that I can just not iterate through a polygon that easily and I missed the non-trivial part, or that a polygon can not be used as a ring, or iterations are just not the way I thought they are, or I have no idea what else can be wrong. What did I do wrong?

推荐答案

我发现，需要修复的几件事情：

I found a few things that needed to be fixed:

一个问题我看到的是在你的模板。一定要放空间！

升压范围适用于容器或持有开头的范围，最终对

重新迭代器presents像一个指向对象。获取迭代器的大小将不会做你想做的。您需要为使用整个容器的boost ::大小，或std ::距离（begin_iterator，end_iterator）。

下面是一个编译版本：

#include <vector>
#include <boost/range.hpp>
#include <boost/geometry.hpp>
#include <boost/geometry/geometries/polygon.hpp>

using namespace boost::geometry;

typedef boost::geometry::model::point
    <
        double, 2, boost::geometry::cs::spherical_equatorial<boost::geometry::degree>
    > spherical_point;
class PointAggregator {
private :
    double *x, *y;
    int count;

public :
    PointAggregator(int size) {
        x = (double*) malloc(sizeof(double) * size);
        y = (double*) malloc(sizeof(double) * size);
        count = 0;
    }

    ~PointAggregator() {
        free(x);
        free(y);
    }

    inline void operator()(spherical_point& p) {
        x[count] = get<0>(p);
        y[count] = get<1>(p);
        count++;
    }

    void GetResult(double *resultX, double *resultY) {
        resultX = x;
        resultY = y;
    }
};

// added spaces to the close brackets >> becomes > >
void VectorToArray(std::vector<model::polygon<spherical_point> > resultVector, double x[], double y[], int *count) {
    for (std::vector<model::polygon<spherical_point> >::iterator it = resultVector.begin(); it != resultVector.end(); ++it) {
        if (boost::size(resultVector) >= 2) {
            // getting the size of the whole container
            *count = boost::size(resultVector);
            PointAggregator* pa = new PointAggregator(*count);
            boost::geometry::for_each_point(*it, *pa);
            pa->GetResult(x, y);
            delete(pa);
            break;
        }
    }
}

这篇关于获得点的坐标从升压几何多边形的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！