Question

我是新手，以提升和“沉重”的模板。我已经玩了几天，并试图将数组传递给惊人的boost::geometry::convex_hull函数。没有运气。

我准备了以下示例：

#include <boost/geometry.hpp>
#include <boost/geometry/geometries/polygon.hpp>
#include <boost/geometry/geometries/adapted/boost_tuple.hpp>
#include <boost/geometry/geometries/register/point.hpp>
#include <iostream>

using namespace std;
namespace bg = boost::geometry;

BOOST_GEOMETRY_REGISTER_BOOST_TUPLE_CS(cs::cartesian)

int main()
{
    typedef boost::tuple<float, float> point;
    typedef bg::model::polygon<point> polygon;

    polygon poly, hull;

    // define rectangle, this will become also our convex hull
    bg::append( poly, point( 0.0, 0.0 ) );
    bg::append( poly, point( 1.0, 0.0 ) );
    bg::append( poly, point( 1.0, 1.0 ) );
//    bg::append( poly, point( 2.0, 2.0 ) );
    bg::append( poly, point( 1.0, 0.0 ) );

    // mid point, which should not be part of the hull
    bg::append( poly, point( 0.5, 0.5 ) );

    // The above poly would ideally constructed with: 
    // float myInputData[] = { 0.0, 0.0,   1.0, 0.0,   1.0, 1.0,   2.0, 2.0,   1.0, 0.0  };
    // and then used like: bg::convex_hull( myInputData, hull );

    bg::convex_hull( poly, hull );

    cout << "convex hull is:\n";
    vector<boost::tuples::tuple<float, float> >::iterator it;

    for( it = hull.outer().begin(); it != hull.outer().end(); ++it )
      cout << "(" << bg::get<0>(*it) << "/" << bg::get<1>(*it) << ")\n";
}

附加bg :: append的点应该是一个简单的c数组，如：

float mydataArray[ 20 ];

其中数据布局的格式为[x1，y1，x2，y2 ...] 所以理想的函数是：bg :: append_points（arrayOfAlternating_X_Y_coordinates）

我还想将结果（hull）作为数组指针传递给OpenGL，而不是迭代并通过float读出浮点数（因此，目标是：摆脱bg :: get＆lt; 0＆gt;（* it ））。

结果也应该来自[x1，y1，x2，y2 ...]

我有种感觉，这种提升并不能提供我想要的功能。那么，我摆脱循环的想法有什么问题吗？我如何编写一个可以被bg :: convex_hull函数使用的智能访问类？

Answer 1

除the other answer中显示的polygon<>模型的机制外，您可以将polygon<>替换为ring<>模型，因为似乎没有内环涉及？

这样，你可以直接初始化到你的戒指中：

typedef bg::model::ring<point> ring;

ring poly {
    { 0.0, 0.0 },
    //{ 0.5, 0.5 }, // mid point, which should not be part of the hull
    { 1.0, 0.0 },
    { 1.0, 1.0 },
    { 2.0, 2.0 },
    { 1.0, 0.0 },
};

API调用可以是：

point const* hull_array_ptr = &hull.front();

call_API(hull_array_ptr, hull.size());

样本

<强> Live On Coliru

#include <boost/geometry.hpp>
#include <boost/geometry/geometries/ring.hpp>
#include <boost/geometry/io/io.hpp>
#include <boost/geometry/geometries/adapted/boost_tuple.hpp>
#include <boost/geometry/geometries/register/point.hpp>
#include <iostream>

using namespace std;
namespace bg = boost::geometry;

BOOST_GEOMETRY_REGISTER_BOOST_TUPLE_CS(cs::cartesian)
typedef boost::tuple<float, float> point;

void dump(point const& p) { cout << get<0>(p) << " " << get<1>(p) << ","; }

void call_API(point const* arr, size_t n) {
    cout << "hull (API):";
    for_each(arr, arr+n, dump);
    cout << "\n";
}

int main()
{
    typedef bg::model::ring<point> ring;

    ring poly {
        { 0.0, 0.0 },
        //{ 0.5, 0.5 }, // mid point, which should not be part of the hull
        { 1.0, 0.0 },
        { 1.0, 1.0 },
        { 2.0, 2.0 },
        { 1.0, 0.0 },
    };

    cout << "raw:       " << bg::wkt(poly) << "\n";
    bg::correct(poly);
    cout << "corrected: " << bg::wkt(poly) << "\n";

    ring hull;
    bg::convex_hull(poly, hull);

    cout << "hull:      " << bg::wkt(hull) << "\n";

    point const* hull_array_ptr = &hull.front();

    call_API(hull_array_ptr, hull.size());
}

再次打印：

raw:       POLYGON((0 0,1 0,1 1,2 2,1 0))
corrected: POLYGON((0 0,1 0,1 1,2 2,1 0,0 0))
hull:      POLYGON((0 0,2 2,1 0,0 0))
hull (API):0 0,2 2,1 0,0 0,

Answer 2

好的，第三次是魅力，对吧

OP：感谢您的努力，但这不是您想要的解决方案。我澄清了这个问题。目标是为输入和输出提供一个扁平的c型浮子阵列。打印出来的数据应该是这样的：
float[]
我： 花车意味着什么？

OP：交替的x和y坐标，[...]

如果你做了必要的事情

尺寸/填充检查
对齐覆盖

您可以在原始template<typename T> using compact_point = boost::tuple<T, T>; template<typename T> using compact_ring = boost::iterator_range<T*>; static_assert(sizeof(compact_point<float>) == 2*sizeof(float), ""); static_assert(alignof(compact_point<float>) >= alignof(float), ""); BOOST_GEOMETRY_REGISTER_BOOST_TUPLE_CS(cs::cartesian) BOOST_GEOMETRY_REGISTER_RING_TEMPLATED(compact_ring) using point = compact_point<float>; using ring = compact_ring<point>;上创建一个环形作为一系列类型惩罚点：

template <typename T, size_t N>
compact_ring<compact_point<T> > as_compact_ring(T (&arr)[N]) {
    auto f = reinterpret_cast<point*>(+arr);
    return { f, f + N/2 };
}

template <typename T>
boost::iterator_range<T const*> as_compact_points(std::vector<compact_point<T> > const& r) {
    auto f = reinterpret_cast<T const*>(&r[0]);
    return { f, f + r.size()*2 };
}

注意请注意，这仅适用于只读环

转换例程：

int main() {
    alignas(compact_point<float>) float ringdata[] {
        0.0, 0.0, // clockwise rect
        0.0, 2.0,
        //
        1.0, 1.0, // dent...
        //
        2.0, 2.0,
        2.0, 0.0,
        0.0, 0.0,
    };

    ring poly = as_compact_ring(ringdata);

    cout << "raw: " << bg::wkt(poly) << "\n";

    std::string reason;
    if (!bg::is_valid(poly, reason)) {
        std::cout << "NOT VALID: " << reason << "\n";
        return 255;
    }

    bg::model::ring<point> hull; // not a range proxy though
    bg::convex_hull(poly, hull);

    cout << "hull:" << bg::wkt(hull) << "\n";

    // force back:
    auto view = as_compact_points(hull);
    float const* rawhull = &*view.begin();

    call_API(rawhull, 2*hull.size());
}

你去了，你现在可以申请：

演示

raw: POLYGON((0 0,0 2,1 1,2 2,2 0,0 0))
hull:POLYGON((0 0,0 2,2 2,2 0,0 0))
hull (API):0 0 0 2 2 2 2 0 0 0

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打印

class ThisFails {
    func foo() -> PA {
        return X().test()   // error: generic parameter 'U' could not be inferred
    }
}

X().test() as PA            // error: generic parameter 'U' could not be inferred

Answer 3

问。附加bg :: append的点应该在一个数组中，这样理想的函数就是：bg::append_points(arrayOf_X_Y_coordinates)

您可以一次指定多边形的外环：

point raw[] = {
    { 0.0, 0.0 },
    //{ 0.5, 0.5 }, // mid point, which should not be part of the hull
    { 1.0, 0.0 },
    { 1.0, 1.0 },
 // { 2.0, 2.0 },
    { 1.0, 0.0 },
};

// define rectangle, this will become also our convex hull
poly.outer().assign(begin(raw), end(raw));

Q。我还希望将结果（hull）作为数组指针传递给OpenGL，而不是迭代并通过float读取坐标float（使用bg :: get＆lt; 0＆gt;（*它））。

您可以传递初始矢量元素的地址：

point const* hull_array_ptr = &hull.outer().front();

call_API(hull_array_ptr, hull.outer().size());

样本

注意确保您的多边形有效，满足convex_hull算法的前提条件！ correct在此处执行必要的修复：

<强> Live On Coliru

#include <boost/geometry.hpp>
#include <boost/geometry/geometries/polygon.hpp>
#include <boost/geometry/io/io.hpp>
#include <boost/geometry/geometries/adapted/boost_tuple.hpp>
#include <boost/geometry/geometries/register/point.hpp>
#include <iostream>

using namespace std;
namespace bg = boost::geometry;

BOOST_GEOMETRY_REGISTER_BOOST_TUPLE_CS(cs::cartesian)
typedef boost::tuple<float, float> point;

void dump(point const& p) { cout << get<0>(p) << " " << get<1>(p) << ","; }

void call_API(point const* arr, size_t n) {
    cout << "hull (API):";
    for_each(arr, arr+n, dump);
    cout << "\n";
}

int main()
{
    typedef bg::model::polygon<point> polygon;

    polygon poly, hull;

    point raw[] = {
        { 0.0, 0.0 },
        //{ 0.5, 0.5 }, // mid point, which should not be part of the hull
        { 1.0, 0.0 },
        { 1.0, 1.0 },
        { 2.0, 2.0 },
        { 1.0, 0.0 },
    };

    poly.outer().assign(begin(raw), end(raw));
    cout << "raw:       " << bg::wkt(poly) << "\n";
    bg::correct(poly);
    cout << "corrected: " << bg::wkt(poly) << "\n";

    bg::convex_hull(poly, hull);

    cout << "hull:      " << bg::wkt(hull) << "\n";

    point const* hull_array_ptr = &hull.outer().front();

    call_API(hull_array_ptr, hull.outer().size());
}

输出：

raw:       POLYGON((0 0,1 0,1 1,2 2,1 0,0 0))
corrected: POLYGON((0 0,1 0,1 1,2 2,1 0,0 0))
hull:      POLYGON((0 0,2 2,1 0,0 0))
hull (API):0 0,2 2,1 0,0 0,

如何使用数组的增强来计算凸包，而不是分别设置每个点？

3 个答案:

样本

转换例程：

演示

样本