我试图在C ++中实现一个简单的二叉树,但是指针给我带来了麻烦。尽管我从未为指针分配结构,但它不会被评估为NULL,因此,实际上永远不会分配并且一切都会中断。
代码:
#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;
struct BTNode {
int data;
BTNode *left;
BTNode *right;
};
int main() {
BTNode *tree;
srand(time(NULL));
int input[11];
for(int i=0;i<11;i++) {
input[i] = 1 + rand()%50;
}
for(int i=0;i<11;i++) { //populate binary tree
cout << "Populating tree " << i << "\n";
if(!tree) {
tree = new BTNode();
cout << "Initialising tree\n";
tree->data = input[i];
} else {
cout << "Setting tree pointer\n";
BTNode *curnode = tree;
while(true) {
if(curnode->data >= input[i]) {
if(!curnode->left) {
curnode->left = new BTNode();
curnode->left->data = input[i];
cout << "Put " << input[i] << " in left node of " << curnode->data << ".\n";
break;
}
else {
cout << "Moving on to left node...\n";
curnode = curnode->left;
}
}
else {
if(!curnode->right) {
curnode->right = new BTNode();
curnode->right->data = input[i];
cout << "Put " << input[i] << " in right node of " << curnode->data << ".\n";
break;
}
else {
cout << "Moving on to right node...\n";
curnode = curnode->right;
}
}
}
}
}
}
给出输出
Populating tree 0
Setting tree pointer
Moving on to right node...
Segmentation fault (core dumped)
答案 0 :(得分:9)
因为它没有初始化。
局部变量未初始化为nullptr
或其他任何内容。它们可以具有任何值*,这几乎意味着 - 始终初始化您的值。
在gcc中使用-Wall
等警告选项有助于避免这些错误,因为您会收到有关此类问题的通知。
*根据规范不完全正确,但是有用的简化。