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#include <iostream>
struct TreeNode{
TreeNode(int aVal){
value = aVal;
pLeft = NULL;
pRight = NULL;
}
int value;
TreeNode* pLeft;
TreeNode* pRight;
};
int main(){
TreeNode* node1 = new TreeNode(1);
TreeNode* node2 = new TreeNode(2);
printf("node1:%lld,node2:%lld\n ", node1,node2);
TreeNode* node3 = NULL;
node3 = node1;
node1 = node2;
printf("node1:%lld,node2:%lld, node3:%lld\n ", node1,node2,node3);
}
int a = 0;
int b = 1;
int* p2 = &b;
int* p1 = &a;
int* p3 = p1;
int* p1 = p2;
p1指向的是a的地址,p3这时也指向a的地址,最后一句使p1指向p2所指的地址,也就是b的地址。
那这么一来,p3指向不指向p2是猫关系都没有。又没有传递性。
懂了吧。