2016-10-11 79 views
1

尝试从r-tree中删除值时出现编译错误。我还将一个原始指针与框一起存储,这似乎是导致问题 - 如果我存储int,字符串或shared_ptr,则不会收到错误。在boost R树中存储原始指针的问题

我没有选择切换到shared_ptr,因为所有这些来自遗留库。是否有另一种解决方法?

我有树的定义如下:

namespace bg = boost::geometry; 
namespace bgi = boost::geometry::index; 
namespace bgm = boost::geometry::model; 

typedef boost::geometry::model::point<float, 2, bg::cs::cartesian> point_t; 
typedef boost::geometry::model::box<point_t> box_t; 
typedef std::pair<box_t, Data*> value_t; 

boost::geometry::index::rtree<value_t, boost::geometry::index::quadratic<16>> rtree; 

而失败的代码如下:

while(!rtree.empty()) { 
    auto it = rtree.begin(); 
    auto value = *it; 
    rtree.remove(value); // <-- this is where the error appears. 
} 

而且错误如下:

...../boost/geometry/index/equal_to.hpp:127:60: error: ambiguous class template instantiation for 'struct boost::geometry::index::detail::equals<NdsInstance*, void>' 
     && detail::equals<T2>::apply(l.second, r.second); 
                 ^
...../boost/geometry/index/equal_to.hpp:28:8: error: candidates are: struct boost::geometry::index::detail::equals<Geometry*, Tag> 
struct equals<Geometry *, Tag> 
    ^
...../boost/geometry/index/equal_to.hpp:37:8: error:     struct boost::geometry::index::detail::equals<T, void> 
struct equals<T, void> 
    ^
...../boost/geometry/index/equal_to.hpp:127:60: error: incomplete type 'boost::geometry::index::detail::equals<NdsInstance*, void>' used in nested name specifier 
     && detail::equals<T2>::apply(l.second, r.second); 
                 ^

全部代码示例可在Colliru上找到。我正在使用gcc 4.9.3和boost 1.62(与boost 1.61相同的错误)。

回答

1

我结束了创建的原始指针的包装:

struct wrapData { 
    public: 
    wrapData(Data *data) { _data = data; } 
    operator Data*() const { return _data; } 
    private: 
    Data *_data; 
}; 

typedef std::pair<box_t, wrapData> value_t; 
1

我得到了同样的问题,我找到另一种方式来转转:

重新定义equal_to函子(RTREE的第四个模板参数)

示例代码:

#include <boost/geometry.hpp> 

namespace bg = boost::geometry; 
namespace bgi = boost::geometry::index; 
namespace bgm = boost::geometry::model; 

using point = bgm::point<double, 2, bg::cs::spherical_equatorial<bg::degree>>; 
using value_type = std::pair<point, int*>; 

struct my_equal { 
    using result_type = bool; 
    bool operator() (value_type const& v1, value_type const& v2) const { 
     return bg::equals(v1.first, v2.first) && v1.second == v2.second;} 
}; 

using rtree = bgi::rtree<value_type, bgi::quadratic<16>, bgi::indexable<value_type>, my_equal>; 

int main() { 
    int a,b; 
    rtree rtree; 
    rtree.insert(std::make_pair(point(45,45), &a)); 
    rtree.insert(std::make_pair(point(45,45), &b)); 
    rtree.remove(std::make_pair(point(45,45), &b)); 

    return 0; 
} 

工作的gcc 6.2.1提高1.61.0(也GCC 4.9.3,推动1.58.0)

this ticket

+0

启发我喜欢你的解决方案比我的好,它使大量的清洁剂来访问后台的数据。 –