I'm trying to implement an N-ary Tree in c++ using the glib, but as I'm not a c++ expert, I'm having some problems finding out how to use it right. Does anybody have a simple example written in C++ to help me understand how to use the basic functions? I'm having special problems with g_node_traverse, I just can't get the GNodeTraverseFunc right.
You can find the description of the N-ary Tree here: http://developer.gnome.org/glib/stable/glib-N-ary-Trees.html
I found some examples in c, but I couldn't manage to translate them correctly into c++ here:
http://www.ibm.com/developerworks/linux/tutorials/l-glib/section7.html
Tried with the last piece of code for n-ary trees.
I appreciate your help.
Well, I have managed to run some code. The problem was basically the casts that were needed because Gnome uses gpointers and my data is to be stored in a struct. So my code is:
gboolean iter(GNode* n, gpointer data) {
node s=*(node *)(n->data);
int ID=g_node_depth(n);
if (G_NODE_IS_ROOT(n)==true)
{
std::cout<<"Node "<<ID<<" is a Root"<<std::endl;
}
else if (G_NODE_IS_LEAF(n)==true)
{
std::cout<<"Node "<<ID<<" is a Leaf"<<std::endl;
}
std::cout<<"Speed of Node "<<ID<<" is: "<<s.v<<std::endl;
return FALSE;
}
int main(){
node prueba,prueba1;
prueba.phi=0;
prueba.v=1;
prueba.x=50;
prueba.y=100; //Position in y
prueba1.phi=90;
prueba1.v=6;
prueba1.x=30;
prueba1.y=90;
GNode * root = g_node_new((gpointer) &prueba);
g_node_append(root, g_node_new((gpointer) &prueba1));
g_node_traverse(root, G_PRE_ORDER, G_TRAVERSE_ALL, -1, iter, NULL);
return 0;
}
Where my struct is:
struct state {
double x; //Position in x of a car
double y; //Position in y "
double phi; //Yaw angle of a car
double v; //Speed of a car
};
struct node {
double x;
double y;
double phi;
double v;
std::vector <state > trajectory;
};
The idea is to have the whole previous "trajectory" stored at each node so if I choose a random node/leaf I don't have to reconstruct the trajectory but just take it.
This code works now. It might be improved and I'm open to any comment.
I hope it's useful for someone.