error C3203: 'set' : unspecialized class template can't be used as a template

[lucky6969b]

I have tried both const QuadNode* and const QuadNode& to no avail.
Thanks
Jack

std::map<std::string, std::set<QuadNode*, QuadNodeSortedComparator>> activePool;

struct QuadNodeSortedComparator 
{
	bool operator()(const QuadNode& node1, const QuadNode& node2) const {	  
		return std::tie(node1.x, node1.z) < std::tie(node2.x, node2.z);

	}

};

[tiliavirga]

Since the set is holding pointers I think you should access the members by ->, like node1->x rather than node1.x.

[monarch_dodra]

I have tried both const QuadNode* and const QuadNode& to no avail.
Thanks
Jack

std::map<std::string, std::set<QuadNode*, QuadNodeSortedComparator>> activePool;

struct QuadNodeSortedComparator 
{
	bool operator()(const QuadNode& node1, const QuadNode& node2) const {	  
		return std::tie(node1.x, node1.z) < std::tie(node2.x, node2.z);

	}

};

Please provide the actual code that you used to produce the error. You just provided a snipet, and all it would produce is "error: 'QuadNodeSortedComparator' was not declared in this scope".

Appart from the fact that you are using a function that takes "const QuadNode&" to compare "QuadNode*" objects, I see nothing wrong with the code, provided you use the correct forward declarations etc.

But the devil is in the details:
- What is your "QuadNode" definition?
- What are your includes?
- What and where are your forward declarations?

Please provide a short but *complete* example that reproduces your issue.

[lucky6969b]

Hello, I change it back to QuadNode instead of QuadNode*, everything works great now.
Thanks for helping
Jack

[tiliavirga]

I change it back to QuadNode instead of QuadNode*, everything works great now.

And the reason for that is that the set now holds QuadNode objects rather than QuadNode pointers. In the first case it's okay to access QuadNode members using . whereas in the second case you must use -> instead. Also in the second case it's kind of overkill to pass in the pointers by const reference to the () operator.