C "pointer being freed was not allocated" error - c

Question

I'm experimenting with a quadtree implementation found online, but am receiving "pointer being freed was not allocated" errors intermittently as part of the following (stripped down) _node_release function:
static void _node_release(node* node)
{
if(node->subnodes[0])
{
for(size_t i=0;i<4;i++)
{
_node_release(node->subnodes[i]);
free(node->subnodes[i]); // this causes "pointer being freed was not allocated" errors
}
}
// ...
free(node); // the node passed to _node_release is free'd here
}
Would I be correct to assume that node is being incorrectly free'd twice here? The _node_release function free's the node passed to it, but the code is also attempting to free each subsequent subnode after making a recursive call to _node_release.

Answer 1

Yes you are correct with this assumption.
The problem is in your recursive call. Suppose I am a subnode in the recursive call. When I am done calling free on my subnodes, i then execute my last line, free(node) to free myself.
Now i'm the parent of this node. I do the same routine. One of my subnodes, however, was the guy who just freed himself. So when I call free(node->subnodes[i]) I get the error that you see

Answer 2

Yes you are freeing your nodes twice.
1: static void _node_release(node* node)
2: {
3: if(node->subnodes[0])
4: {
5: for(size_t i=0;i<4;i++)
6: {
7: _node_release(node->subnodes[i]);
8: free(node->subnodes[i]); // this causes "pointer being freed was not allocated" errors
9: }
10: }
11: // ...
12:
13: free(node); // the node passed to _node_release is free'd here
14: }
Lets say you have NodeB as a child of NodeA
_free_release(NodeA)
_free_release(NodeB) – line 7
free(NodeB) - line 14
free(NodeB) - line 8
To fix this just remove the free on line 8
static void _node_release(node* node)
{
if(node->subnodes[0])
{
for(size_t i=0;i<4;i++)
{
_node_release(node->subnodes[i]);
}
}
// ...
free(node); // the node passed to _node_release is free'd here
}