对不起,我现在编辑了我的问题。请注意粗体字。C++什么是定义递归构造函数的标准方法?
我确实需要一个递归构造函数,同时定义一个kdtree类。但我恐怕我没有按照正确的方式去做。 我该如何更优雅地做到这一点?
这是我的代码,使用这个指针,它编译,并运行良好。 不要做任何事情,只是显示递归构造函数应该看起来像的简短概念。
#include <iostream>
using namespace std;
class foo
{
public:
int a, b;
foo(unsigned int k)//this piece of code just shows the brief idea of what i'm trying to do.
{
if (k)
*this = foo(--k);
else
a = k, b = k;
}
};
int main()
{
foo f(3);
cout << f.a << f.b << endl;
getchar();
}
这是我的kdtree示例代码。 这就是我正在努力实现的,仍然不编译,我会稍后编辑它。
class kdtree
{
public:
int16_t count;//数组里面可以只存mask和key生成的unique_key,因为树结构,和count可以后期生成
int16_t key;
int16_t mask;
inline bool is_full()
{
return mask + count == 0x8000;
};
shared_ptr<kdtree> left, right;
kdtree(){}
kdtree(int x1, int y1, int z1, int x2, int y2, int z2, int _x = 0, int _y = 0, int _z = 0, int len = 0, int ikey = 0x8000)
{
int i = 0x80 >> len/3, j = 0x4000 >> len;
if ((x2 - x1)*(y2 - y1)*(z2 - z1) == j << 10)
{
count = j << 1;
key = ikey;
mask = ~ikey^(ikey - 1);
return;
}
switch (len++ % 3)
{
case 0:
if (x1 < _x&&x2 < _x)
{
*this = kdtree(x1, y1, z1, x2, y2, z2, _x, _y, _z, len, ikey -= j);
return;
}
if (x1 >= _x&&x2 >= _x)
{
*this = kdtree(x1, y1, z1, x2, y2, z2, _x + i, _y, _z, len, ikey += j);
return;
}
left = shared_ptr<kdtree>(new kdtree(x1, y1, z1, _x, y2, z2, _x, _y, _z, len, ikey -= j));
right = shared_ptr<kdtree>(new kdtree(_x, y1, z1, x2, y2, z2, _x + i, _y, _z, len, key += j));
count = j << 1;
key = ikey;
mask = ~ikey^(ikey - 1);
return;
case 1:
if (y1 < _y&&y2 < _y)
{
*this = kdtree(x1, y1, z1, x2, y2, z2, _x, _y, _z, len, ikey -= j);
return;
}
if (y1 >= _y&&y2 >= _y)
{
*this = kdtree(x1, y1, z1, x2, y2, z2, _x, _y + i, _z, len, ikey += j);
return;
}
left = shared_ptr<kdtree>(new kdtree(x1, y1, z1, x2, y2, z2, _x, _y, _z, len, ikey -= j));
right = shared_ptr<kdtree>(new kdtree(x1, y1, z1, x2, y2, z2, _x, _y + i, _z, len, ikey += j));
count = j << 1;
key = ikey;
mask = ~ikey^(ikey - 1);
return;
case 2:
if (x1 < _x&&x2 < _x)
{
*this = kdtree(x1, y1, z1, x2, y2, z2, _x, _y, _z, len, ikey);
return;
}
if (x1 >= _x&&x2 >= _x)
{
*this = kdtree(x1, y1, z1, x2, y2, z2, _x, _y, _z + i, len, ikey + j);
}
left = shared_ptr<kdtree>(new kdtree(x1, y1, z1, x2, y2, _z, _x, _y, _z, len, ikey));
right = shared_ptr<kdtree>(new kdtree(x1, y1, _z, x2, y2, z2, _x, _y, _z + i, len, ikey + j));
count = j << 1;
key = ikey;
mask = ~ikey^(ikey - 1);
return;
}
}
};
对我来说看起来很荒谬,你想实现什么样的目标? –
你想达到什么目的?你可以将a和b设置为0 :-) –
我真的不相信你会需要这样一个奇怪的东西。这应该是一个XY问题。 –