In the last week or so, I've been attempting to implement pathfinding for my first time. I've been reading a tutorial on the general concept and have been trying to implement it on my own. For the Closed List and Open List, I use std::vector. The guide says once I find the optimal node to delete it from the Open List and add it to the Closed List. I'm doing that with this code:

closedList.push_back(openList[lowF]);
openList.erase(openList.begin()+lowF);

The issue is that the wrong node is being deleted from the Open List, causing the pathfinding calculations to flow in an infinite loop. For Instance:

Closed List = (9,4)

Open List = (9,3) (10,4)

After the code I posted above executes, the vectors look like this:

Closed List = (9,4) (9,3)

Open List = (9,3)

While they should look like this

Closed List = (9,4) (9,3)

Open List = (10,4)

I checked the lowF value, and it is 0. So openList.erase(openList.begin()+lowF) should be the same as openList.erase(openList.begin()). For some reason, that line of code is deleting the second element in the vector instead of the first. I'm attaching a error log file I printed out. I'd appreciate any help. ty

int lowF = 0;
Node tmp;
Node currentNode(Position(currentPos.x, currentPos.y), 0);
vector<Node> openList;
vector<Node> closedList;
closedList.push_back(currentNode);

while(closedList[closedList.size()-1].pos.x != pathDest.x && closedList[closedList.size()-1].pos.y != pathDest.y)
{
getChildNodes(closedList[closedList.size()-1], closedList, openList, wallCords);

for(int i=0;i<openList.size();i++)
{
if(openList[lowF].getF(pathDest) > openList.getF(pathDest))
lowF = i;
}
closedList.push_back(openList[lowF]);
openList.erase(openList.begin()+(lowF));
}

path.push_front(closedList[closedList.size()-1]);
tmp = closedList[closedList.size()-1];

while(!(tmp == currentNode))
{
tmp = *tmp.parent;
path.push_front(tmp);
}
path.push_front(tmp);

Here's the entire function. My node class contains a Node* parent, int g, and a Position class which has x and y integers

Yes, it does. openList is passed as vector<Node>& openList. Here's the general algorithm for it. The functions in the if statement's body don't affect either list.

Node* tmp;
if(!hasWall(parentNode.pos.x, parentNode.pos.y-1, wallCords) && !onClosedList(Position(parentNode.pos.x, parentNode.pos.y-1), closedList))
{
tmp = new Node(Position(parentNode.pos.x, parentNode.pos.y-1), &parentNode);
openList.push_back(*tmp);
}

Is there a copy constructor on the Nodes? Not sure what would cause this but that's the only thing I can think of, everything else looks pretty typical.

Edit:

Check assignment constructor also. Failing that you could just break on it in the debugger and just keep clicking "step into" through all the compiler mumbo jumbo to see if anything comes up.

Edit #2 (for reals this time) :

It's an assignment operator, not a constructor. I think that's what's broken, there's a typo in there that's keeping stuff from getting moved over. vector.erase(iter) moves everything after iter over, and iter is basically just a pointer. If this is no good, you can use a vector of pointers to Nodes, or a std::list of nodes (I'm not sure that's guaranteed not to copy on erase, but I have never seen it do it).

This is the first time I've actually ever used operator overloading before. I didn't really kow what I was doing. Here's the function:

Node Node::operator=(Node copy)
{
return copy;
}

It seems like that operator= isn't working, so I'll fix it and try to get it to work.

yeah that's a no op assignment. So node1 = node2 doesn't do anything. Not sure a memcpy(this, &copy, sizeof(Node)) is legal, probably better to just assign all the fields. You don't need a deep copy though, at least not for this purpose, you could just copy the pointers, since the old copy is going to be dead in a nanosecond.

Edit:

whoops, I realized that was bad advice. If the destructor of frees an allocated resource, you are going to be dereferencing an invalid pointer (probably, unless that memory gets reused) when you later try to use the copy. Sorry about that, hope it didn't cause any problems.