我正在尝试为单个链接列表创建insert_at_position函数,但是在列表末尾插入新节点时遇到内存错误。
这是我的代码:
#include <iostream>
#include <memory>
#include <utility>
struct Node {
int data;
std::unique_ptr<Node> next = nullptr;
Node(const int& x, std::unique_ptr<Node>&& p = nullptr)
: data(x)
, next(std::move(p)) {}
};
std::unique_ptr<Node> head = nullptr;
Node* tail = nullptr;
void print() {
auto temp = head.get();
while (temp) {
std::cout << temp->data << " ";
temp = temp->next.get();
}
std::cout << "\n";
}
void push_back(const int& theData) {
std::unique_ptr<Node> newNode = std::make_unique<Node>(theData);
if (!head) {
head = std::move(newNode);
tail = head.get();
}
else {
tail->next = std::move(newNode);
tail = tail->next.get();
}
}
void push_front(const int& theData) {
std::unique_ptr<Node> newNode = std::make_unique<Node>(theData);
newNode->next = std::move(head);
head = std::move(newNode);
if (!tail) {
tail = head.get();
}
}
void insert_at_position(int pos, const int& theData) {
if (pos == 1) {
push_front(theData);
return;
}
auto newNode = std::make_unique<Node>(theData);
auto current = head.get();
for (int i = 1; i < pos; i++) {
current = current->next.get();
}
if (current != nullptr) {
newNode->next = std::move(current->next);
current->next = std::move(newNode);
}
else {
push_back(theData);
}
}
void pop_front() {
head = std::move(head->next);
}
void pop_back() {
if (!head) return;
auto current = head.get();
Node* previous = nullptr;
if (current->next != nullptr) {
previous = current;
current = current->next.get();
}
if (previous != nullptr) {
previous->next = nullptr;
}
else {
head = nullptr;
}
tail = previous;
previous->next = nullptr;
}
void erase(int pos, const int& theData) {
}
int main() {
push_back(2);
push_back(4);
push_back(6);
print();
push_front(1);
print();
pop_front();
print();
/*pop_back();
print();*/
insert_at_position(1, 1);
print();
insert_at_position(8, 4);
print();
//insert_at_position(3, 2);
//print();
//
std::cin.get();
}
错误在这一点上发生:insert_at_position(8, 4);
我的逻辑是在最后一个if语句之后,以处理我在列表中间插入某个随机位置的情况,我将在结尾处进行调用push_back函数。
答案 0 :(得分:1)
insert_at_position(8, 4)
尝试将位置8插入到一个短得多的列表中。内部的循环对此没有任何保护,并且在到达列表末尾时会很高兴地取消引用空指针。