Reputation: 11
I am attempting to write a C++ template struct for a splay tree, but when I try to test the code, I am getting very strange results.
This is my code for the template:
template <class T>
struct splaytree {
struct node {
splaytree<T> *tree;
node *p, *c[2];
T v;
int w;
node(T t, splaytree<T> *st) {
v = t;
p = 0;
c[0] = 0;
c[1] = 0;
w = 1;
tree = st;
}
int side() {
return (p->c[1] == this) ? 1:0;
}
void r() {
node *x = this;
int b = x->side();
node *p = x->p;
x->w = p->w;
p->w = x->c[1^b]->w + 1 + p->c[1^b]->w;
x->p = p->p;
p->p = x;
p->c[0^b] = x->c[1^b];
x->c[1^b] = p;
}
void splay() {
node *x = this;
while (x->p) {
node *p = x->p;
if (p == tree->root) x->r();
else if (((x->side())^(p->side()))) {
x->r();
x->r();
}
else {
p->r();
x->r();
}
}
tree->root = this;
}
int index() {
this->splay();
return this->c[0]->w;
}
};
node *root;
splaytree() {
root = 0;
}
void add(T k) {
node x0(k,this);
node *x = &x0;
if (root == 0) {
root = x;
return;
}
node *i = root;
while (i != x) {
int b = (k < i->v) ? 0:1;
if (i->c[b] == 0) {
i->c[b] = x;
i->w++;
x->p = i;
}
i = i->c[b];
}
x->splay();
}
};
I am using this to test it:
int main() {
splaytree<int> st;
st.add(2);
cout << st.root->v << endl;
cout << st.root->v << endl;
st.add(3);
cout << st.root->c[0] << endl;
}
I inserted the element 2 and then printed the value of the root node twice. Somehow the two prints gave me two different values (2 and 10 in Ideone at http://ideone.com/RxZMyA). When I ran the program on my computer, it gave me 2 and 1875691072 instead. In both cases, when inserting a 3 after the 2, the root node's left child was a null pointer when it should be a node object containing 2.
Can someone please tell me why I am getting two different values when printing the same thing twice, and what I can do to make my splaytree.add() function work as intended? Thanks!
Upvotes: 1
Views: 117
Reputation: 7923
After
node x0(k,this);
node *x = &x0;
if (root == 0) {
root = x;
return;
}
root
is an address of a local variable. By the time you print root->v
, x0
is gone out of scope. All bets as to what the root
really points to are off.
Upvotes: 1