我是C的新手,我正在为一个项目开发一个XOR链表。我已完成大部分代码,但我似乎无法使列表的删除功能正常工作。它似乎能够删除一些数字,但不能删除传递给函数的任何数字。任何有C经验的人都可以看看,并可能指出我哪里出错了?我一直在研究这个问题并且没有太多运气而且我已经开始了3次以上:(非常感谢任何帮助。谢谢。你可以看到我第一次尝试代码here。我可以只发布一个链接,所以如果您想看到我的第二次尝试,请告诉我,我可以通过电子邮件发送给您或其他。感谢您的时间。
#include <stdio.h>
#include <stdlib.h>
#include "rndm.h"
struct node {
int data;
unsigned long link;
};
struct node *head, *tail, *currN, *prevN, *nextN, *tmp;
void insert(struct node **headN, struct node **tailN, int n);
void delete(struct node **headN, struct node **tailN, int n);
void list(struct node *head, int i);
void nextNode();
void previNode();
//============================================================
void insert(struct node **headN, struct node **tailN, int numN) {
struct node *newnode = malloc(sizeof(struct node));
newnode->link =(unsigned long)(*headN);
newnode->data = numN;
//if empty list
if (*headN == NULL){
*headN = newnode;
currN = *headN;
(*headN)->link = 0;
} else if ((*headN)->link == (unsigned long)NULL){
if (numN <= (*headN)->data){
newnode->link = (unsigned long) *headN;
(*headN)->link = (unsigned long) newnode;
tail = *headN;
*headN = newnode;
nextN = tail;
prevN = NULL;
} else {
newnode->link = (unsigned long) *headN;
(*headN)->link = (unsigned long) newnode;
tail = newnode;
nextN = NULL;
currN = tail;
prevN = *headN;
}
} else {
currN = *headN;
prevN = NULL;
nextN = (struct node *)(currN->link ^ (unsigned long) prevN);
if (numN > tail->data){
while (currN!=tail){
nextNode();
}
newnode->link = (unsigned long) currN;
currN->link = (unsigned long) newnode ^ (unsigned long) prevN;
tail = newnode;
} else if (numN < head->data){
currN->link = (unsigned long) newnode ^ (unsigned long) nextN;
newnode->link = (unsigned long) currN;
*headN = newnode;
nextN = currN;
currN = *headN;
} else {
while (numN > currN->data){
nextNode();
}
newnode->link = (unsigned long) prevN ^ (unsigned long) currN;
prevN->link ^= (unsigned long) currN ^ (unsigned long) newnode;
currN->link ^= (unsigned long) prevN ^ (unsigned long) newnode;
}
}
}
void delete(struct node **headN, struct node **tailN, int numD){
struct node *prevN = NULL;
struct node *currN = *headN;
while ( currN != NULL )
{
struct node *nextN = (struct node *) (currN->link ^ (unsigned long)prevN);
//if the number is found, then delete it
if (currN->data == numD)
{
if(prevN)
{
prevN->link ^= (unsigned long)currN ^ (unsigned long)nextN;
}
else
*headN = nextN;
if(nextN)
{
nextN->link ^= (unsigned long)currN ^ (unsigned long)prevN;
}
else
*tailN = prevN;
free(currN);
break;
}
prevN = currN;
currN = nextN;
}
}
void list(struct node *head, int i){
if(i == 0){
currN = head;
prevN = NULL;
int count = 1;
nextN = (struct node *) (currN->link ^ (unsigned long) prevN);
printf("Linked List in ascending order\n");
while(currN!=NULL){
if(count <= 10){
printf("%-5d", currN->data);
nextNode();
count++;
}
else{
printf("\n");
count = 1;
}
}
}
printf("\n\n");
if(i == 1){
printf("Linked List in descending order\n");
currN = tail;
int count2 = 1;
prevN = (struct node *) currN->link;
nextN = NULL;
while (currN!=NULL){
if(count2 <= 10){
printf("%-5d", currN->data);
previNode();
count2++;
}else{
printf("\n");
count2 = 1;
}
}
}
printf("\n");
}
void nextNode(){
nextN = (struct node *) (currN->link ^ (unsigned long) prevN);
prevN = currN;
currN = nextN;
}
void previNode(){
prevN = (struct node *) (currN->link ^ (unsigned long) nextN);
nextN = currN;
currN = prevN;
}
int main(){
int i, num;
float seed;
head = NULL; tail = NULL; currN = NULL; prevN = NULL; nextN = NULL;
init_seed(1234567);
set_range(1,9999);
//inserting data into the linked list
for ( i=0; i<100; ++i){
num = rndm();
insert( &head, &tail, num );
}
list((struct node*)head, 0);
//delete((struct node**)head, (struct node**)tail, 45);
//delete((struct node**)head, (struct node**)tail, 4040);
//delete((struct node**)head, (struct node**)tail, 9769);
list((struct node*)head, 1);
return 0;
}
答案 0 :(得分:5)
看起来你在网上拿了一些代码并尝试使用它。
代码工作得很好,你只是不知道指针是什么。
你在做:
delete((struct node**)head, (struct node**)tail, 45);
以下是变量head
和tail
的定义:
struct node {
int data;
unsigned long link;
};
struct node *head, *tail, *currN, *prevN, *nextN, *tmp;
delete()
函数的原型是void delete(struct node **headN, struct node **tailN, int numD);
“哦,编译器要求struct node **
,让我们投吧”。这不是它的工作原理。
试试这个:
delete(&head, &tail, 45);
答案 1 :(得分:0)
关于您的代码的说明:
node.link
不应该是unsigned long
,因为这假定了编译器/平台的特性。
编辑2 :(由@Matthew Flaschen建议)使用intptr_t使您的代码更具可移植性。
答案 2 :(得分:-1)
查看delete
函数让我对这个指针操作感到疑惑,顺便说一下,你使用的是参数地址,所以它应该是delete(&head, &tail, 45);
继续......
void delete(struct node **headN, struct node **tailN, int numD) { struct node *prevN = NULL; struct node *currN = *headN; struct node *tempNode = NULL; while ( currN != NULL ) { struct node *nextN = (struct node *) (currN->link ^ (unsigned long)prevN); //if the number is found, then delete it if (currN->data == numD) { if(prevN) prevN->link ^= (unsigned long)currN ^ (unsigned long)nextN; else *headN = nextN; if(nextN) nextN->link ^= (unsigned long)currN ^ (unsigned long)prevN; else *tailN = prevN; /* Sanity check here...you could be screwing up the list ** by calling free(currN) */ tempNode = currN; free(tempNode); /* free(currN); <-- That could be misleading... */ break; } prevN = currN; currN = nextN; } }
对代码的修改是为了确保currN
是一致的,用眼睛看指针操作,这可能是有问题的,因为列表最终可能会被free
打开currN
......只是为了安全......