Object defined recursively?

[timboyk12]

Hi everyone,

I'm working on an object (a Barnes-Hut tree) that contains a body, an octant, and 8 additional BHTrees inside of it. I originally wrote the code in java, where this worked quite simply. But now, I'm getting an error for each of the fields: error: field 'bTNW' has incomplete type

Here's the code:

#ifndef BHTREE_H
#define BHTREE_H

#include <iostream>
#include <math.h>
#include "Body.h"
#include "Oct.h"

using namespace std;

class BHTree {
Body body;
Oct oct;
BHTree bTNW;
BHTree bTNE;
BHTree bTSW;
BHTree bTSE;
BHTree bBNW;
BHTree bBNE;
BHTree bBSW;
BHTree bBSE;

public:
void set_values (Oct q);
// bool isExternal(BHTree t);
// void insert(Body b);
// void updateForce(Body b, double t);

};

void BHTree::set_values(Oct q){
oct=q;
}


#endif


I imagine it has something to do with C++'s forward declarations that don't really apply in java to functions and classes, but I'm not sure. Any help would be appreciated!

[laserlight]

It is not possible to define the BHTree class with BHTree members. What you can do is to define BHTree* members.

[nuzzle]

I originally wrote the code in java, where this worked quite simply.

It's because in Java you worked with pointers without much thinking about it (because every object is represented by a pointer (reference) to the object). This variable declaration in Java

BHTree x;

corresponds to this,

BHTree* x;

in C++.

In both cases you need to allocate objects dynamically using new. In C++ you must also delete the objects when you're finished with them. In Java you didn't need to because it has automatic garbage collection.

[treuss]

I'm working on an object (a Barnes-Hut tree) that contains a body, an octant, and 8 additional BHTrees inside of it.

If the object contains a body, an octant and 8 objects, each of them containing a body, an octant and 8 objects, each of them containing a body, an octant and 8 objects, each of them ..... I don't think you have enough memory to create such an object. Actually I don't think anyone has.