新手提问，非常简单，关于链表的反向输出

[mm1543]

写了一个验证的代码，不知道问题在那里，请教各位，多谢！
我想实现的是输入一些数字把它们链接起来，然后把这个单向链表的指针翻转，然后再输出

#include <iostream>
struct node
{
int data;
node* next;
};

void reverse(node*& head);

void main()
{
using namespace std;
node anode;
cin>>anode.data;
node* h=&anode;
node *pre;
for( int i=1; i<5; i++)
{
node anode2;
cin>>anode2.data;
node *n=&anode2;
if (i==1)
{
h->next=n;
}
else
{
pre->next =n;
}
pre=n;

}

reverse(h);
for (int j=0; j<5; j++)
{
cout<<(h+j)->data;
}
return ;
}

void reverse(node*& head)
{
node * pre, *bak, *now;
if (head==NULL)
return;
pre=head;
now=head;
bak=head->next;
while (bak!=NULL)
{
now=bak;
bak=now->next;
now->next=pre;
pre=now;
}
}

[bargio_susie]

#include  <iostream>
using namespace std;

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

void reverse(node*& head);

void main()
{
	node anode;
    cin>>anode.data;

	node *h = &anode;
    node *pre = h;

	for( int i=1; i <5; i++)
	{
		node *anode2 = new node;
		cin>>anode2->data;
		anode2->next = NULL;
		node *n = anode2;

		if (i == 1)
		{
			h->next = n;
		}
		else
		{
			pre->next = n;
		}
		pre = n;
		
	}
	
	reverse(h);
//	for (int j=0; j <5; j++)
//	{
//		cout << (h+j)->data;
//}
	while (h != NULL)
	{
		cout << h->data << ' ';
		h = h->next;
	}

	return;
}

void reverse(node*& head)
{
	node *bak, *now;

	if (head==NULL)
		return;

	now = head;
	bak = head->next;
	head->next = NULL;
	
	while (bak != NULL)
	{
        now = bak->next;
		bak->next = head;
		head = bak;
		bak = now;
	}
}

[ryfdizuo]

//更改较大；
#include  <iostream> 
struct node  
{ 
	int data; 
	node* next; 
	node():next(NULL)
	{}
}; 

void reverse(node*& head); 
void DeleteList(node* head);

void main() 
{ 
	using namespace std; 

	node* h=new node; 
	cin>>h->data; 

	node *pre; 
	for( int i=1; i <5; i++) 
	{ 
		node*n = new node; 
		cin>>n->data; 
		
		if (i==1) 
		{ 
			h->next=n; 
			pre=h->next;
		} 
		else 
		{ 
			pre->next=n;
			pre=pre->next;
		} 
	} 
	reverse(h); 
	node *p=h;
	while (p)
	{
		cout<<p->data<<endl;
		p = p->next;
	}
	 
	DeleteList(h);
	return ; 
} 

void reverse(node*& head) 
{ 
	node * pre, *bak, *now; 
	if (head==NULL) 
		return; 
	pre=head; 
	now=head; 
	bak=head->next; 
	while (bak!=NULL) 
	{ 
		now=bak; 
		bak=now->next; 
		now->next=pre; 
		pre=now; 
	} 
	head->next=NULL;		//打断原来head节点和第二个节点之间的联系；
	head = now;				//更新头节点；
} 
void DeleteList(node* head)
{
	node *p;
	while (head)
	{
		p=head;
		head=head->next;
		delete p;
	}
}

[mm1543]

非常感谢你们，马上给你们分，呵呵