Question

这是我的xor链表实现的代码

#include<iostream>

using namespace std;

struct node
{
    int v;
    node *next;
};

node *start=NULL;
node *end=NULL;

node *newnode(int v)
{
    node *np=new node;
    np->v=v;
    np->next=NULL;
    return np;
}

//returns the xor of two nodes
node *xor(node *a,node *b)
{
    return (node*)((long long)(a)^(long long)(b));
}

void insert(node *current,node *prev,node *np)
{
    if(start==NULL)
    {
        np->next=xor(prev,NULL);
        start=np;
        end=np;
    }
    else
    {
        insert(xor(prev,current->next),current,np);
    }
}

void displayforward(node *start,node *prev)
{
    if(start==NULL) return ;
    cout<<start->v<<"-> ";
    displayforward(xor(start->next,prev),start);
}

void displayBackward(node *end, node *prev){
    if(end == NULL) return;

    cout<< end->v<<" -> ";
    displayBackward( xor(end->next, prev), end);
}

int main()
{
    int a[] = {1,2,3,4,5,6,7,8,9,10}, n = 10;

    for(int i=0; i < n; i++){
        node *prev = NULL;
        insert(start, prev, newnode(a[i]));
    }

    cout<<"Forward: \n";
    node *prev=NULL;
    displayforward(start, prev);

    cout<<"\nBackward: \n";
    displayBackward(end, prev);

    return 0;
}

但是当我编译它时，它会给我这样的错误

1>------ Build started: Project: xor_list, Configuration: Debug Win32 ------
1>  xor_list.cpp
1>c:\users\daviti\documents\visual studio 2010\projects\xor_list\xor_list\xor_list.cpp(30): error C2872: 'end' : ambiguous symbol
1>          could be 'c:\users\daviti\documents\visual studio 2010\projects\xor_list\xor_list\xor_list.cpp(9) : node *end'
1>          or       'end'
1>c:\users\daviti\documents\visual studio 2010\projects\xor_list\xor_list\xor_list.cpp(69): error C2872: 'end' : ambiguous symbol
1>          could be 'c:\users\daviti\documents\visual studio 2010\projects\xor_list\xor_list\xor_list.cpp(9) : node *end'
1>          or       'end'
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

为什么它是一个暧昧的符号结束？这个错误意味着什么？

Answer 1

编译器不确定是使用end全局变量还是std::end符号，这恰好是将迭代器返回到容器末尾的函数。将任何东西分配给该符号没有多大意义是另一回事。删除using namespace std;指令以解决此特定问题。我建议用using std::cout;替换它以避免污染命名空间。

Answer 2

项目中似乎还有另一个名为end的符号。您必须重命名该符号。

或许您有一个将end定义为关键字的扩展程序或插件。

Answer 3

当您使用

时，您必须更改变量名称以消除歧义

using namespace std;

或您必须在不包含std::end的情况下将结尾声明为using namespace std;。

Answer 4

这是我使用单个XOR指针实现的Doubly LL。请提出你的意见。谢谢！

#include <stdio.h>
#include <stdlib.h>

static struct node *nodeAlloc();
static struct node *xor(struct node *a, struct node *b);
static void insertFirst(int);
static void insertLast(int);
void dumpList(struct node *, struct node *);

struct node {
    int value;
    struct node *np;
};

struct node *head;
struct node *tail;

static struct node *nodeAlloc() {
    return (struct node *)malloc(sizeof(struct node));
}

static struct node *xor(struct node *prev, struct node *next) {
    return (struct node*)((unsigned long)prev ^ (unsigned long)next);
}

static void insertFirst(int value) {
    struct node *x;
    x = nodeAlloc();
    x->value = value;

    if(head == NULL && tail == NULL) {
        x->np = NULL;
        head = x;
        tail = x;
    } else {
        x->np = xor(NULL, head);
        head->np = xor(x, head->np);

        head = x;                
    }
}

static void insertLast(int value) {
    struct node *x;
    x = nodeAlloc();
    x->value = value;

    if(head == NULL && tail == NULL) {
        x->np = NULL;
        head = x;
        tail = x;
    } else {
        x->np = xor(tail, NULL);
        tail->np = xor(tail->np, x);

        tail = x;
    }
}

void dumpList(struct node *node, struct node *prev) {
    if(node != NULL) {
        printf("\n%d", node->value);
        dumpList(xor(node->np, prev), node);
    }
}


//------------------

int main() {
    insertFirst(7);
    insertFirst(5);
    insertFirst(4);
    insertFirst(2);
    insertLast(8);
    insertLast(9);
    insertLast(10);


    printf("\n------------------\nForward List:");

    dumpList(head, NULL);

    printf("\n------------------\nBackward List:");

    dumpList(tail, NULL);   

    return 0;
}

xor链表实现

4 个答案: