我正在尝试实现一个简单的单链接整数列表,这些整数将在Visual Studio c ++ 2010 express中插入时进行排序。
问题在于,当我创建一个新节点并在其上调用.getValue()函数时,会返回正确的数字,但是当我尝试在列表中已经存在的节点上调用getValue()时,某种方式会丢失。该节点可能没有正确插入列表,但我无法找到原因。显示一些看起来像参考值或其他值的其他值而不是正确的值。
我在调试时向监视窗口添加了当前值,但除了要插入的给定值之外,我仍然无法看到任何变量。我是视觉工作室的新手,所以我不确定我是否在那里遗漏了一些东西。这是我的代码:
#include "Node.h";
#include <iostream>
//namespace Linked{
//The first two constructors would be the first in the linked list.
Node::Node(void){
value = 0;
next = 0;
}
Node::Node(int setValue){
value = setValue;
next = 0;
}
Node::Node(int setValue,Node *nextNode){
value = setValue;
next = nextNode;
}
Node * Node::getNext(){
return next;
}
void Node::setNext(Node newNext){
next = &newNext;
}
int Node::getValue(){
return value;
}
bool Node::isEqual(Node check){
return value==check.getValue()&&next == check.getNext();
}
/*
int main(){
int firstInt, secondInt;
std::cin>>firstInt;
Node first = Node(firstInt);
std::cout<<"Enter second int: ";
std::cin>>secondInt;
Node second = Node(secondInt, &first);
std::cout<<"Second: "<<second.getValue()<<"\nFirst: "<<(*second.getNext()).getValue();
system("pause");
}*/
以下是链接列表:
//LinkedList.cpp
LinkedList::LinkedList(void)
{
head = 0;
size = 0;
}
LinkedList::LinkedList(int value)
{
head = &Node(value);
size = 1;
}
void LinkedList::insert(int value){
if(head == 0){
Node newNode = Node(value);
head = &newNode;
std::cout<<"Adding "<<(*head).getValue()<<" as head.\n";
}else{
std::cout<<"Adding ";
Node current = *head;
int numChecked = 0;
while(size<=numChecked && (((*current.getNext()).getValue())<value)){
current = (*(current.getNext()));
numChecked++;
}
if(current.isEqual(*head)&¤t.getValue()<value){
Node newNode = Node(value, ¤t);
std::cout<<newNode.getValue()<<" before the head: "<<current.getValue()<<"\n";
}else{
Node newNode = Node(value,current.getNext());
current.setNext(newNode);
std::cout<<newNode.getValue()<<" after "<<current.getValue()<<"\n";
}
}
size++;
}
void LinkedList::remove(int){
}
void LinkedList::print(){
Node current = *head;
std::cout<<current.getValue()<<" is the head";
int numPrinted = 0;
while(numPrinted<(size-1)){
std::cout<<(current.getValue())<<", ";
current = (*(current.getNext()));
numPrinted++;
}
}
int main(){
int a[5] = {30,20,25,13,2};
LinkedList myList = LinkedList();
int i;
for(i = 0 ; i<5 ; i++){
myList.insert(a[i]);
}
myList.print();
system("pause");
}
非常感谢任何指导!
答案 0 :(得分:2)
当你在insert中创建节点时,你将它们从堆栈中分配出去,这意味着它们会在函数返回后丢失。
让他们离开堆:
Node * newNode=new Node(value);
使用时:
Node newNode=Node(value);
您正在堆栈上分配该对象,这意味着指针:
&newNode
只有在该函数返回之前才有效。如果你使用堆内存这不再是一个问题,但它确实意味着你必须为列表实现一个析构函数,它会遍历并删除每个节点。