小编Ale*_*kis的帖子

假设以下简单代码:

private const int A = 2;
private const int B = 3;


public int Result
    {
        get
        {
            return A * B;
        }
    }

我很多次使用Result属性.
每次都会重新计算产品A*B吗？

我有一个不可变对象,例如笛卡尔空间中的一个节点.该类是不可变的,所以我缓存hashCode非常快速的散列.

private final int hashCode;

private final double x, y, z;

public Node(final double x, final double y, final double z)
{
    this.x = x;
    this.y = y;
    this.z = z;
    this.hashCode = Objects.hashCode(this.x, this.y, this.z);
}

@Override
public boolean equals(final Object obj)
{
    if (this == obj) { return true; }
    if (obj == null) { return false; }
    if (!(obj instanceof Node)) { return false; }
    final Node other = (Node) obj;
    return Objects.equal(this.x, other.x) && …

我试图将Howard stack_allocHinnant 's 与boost rtrees 一起使用，如以下示例所示：

#include "stack_alloc.h"
#include <boost/geometry/index/rtree.hpp>

using NodePoint = boost::geometry::model::point<double, 2, boost::geometry::cs::cartesian>;
using Linear = boost::geometry::index::linear<8, 2>;
using RTree =
    boost::geometry::index::rtree<NodePoint, Linear, boost::geometry::index::indexable<NodePoint>,
                                  boost::geometry::index::equal_to<NodePoint>,
                                  stack_alloc<NodePoint, 100>>;

int main()
{
    RTree my_tree{};

    return 0;
}

无法使用相当大的模板错误堆栈进行编译。我认为问题的核心是：

/usr/local/include/boost/geometry/index/detail/rtree/node/variant_static.hpp:26:7：错误：无效使用了不完整的类型'class boost :: geometry :: index :: detail :: rtree： ：allocators，100>，boost :: geometry :: model :: point，boost :: geometry :: index :: linear <8，2>，boost :: geometry :: model :: box>，boost :: geometry： ：index :: detail :: rtree :: node_variant_static_tag>'

这是带有完整错误的 …