this way u can do it in c .... creation and printing of generic lisk list .
void(List **p,void *data, unsigned int n)
{
List *temp;
int i;
/* Error check is ignored */
temp = malloc(sizeof(List));
temp->data = malloc(n);
for (i = 0; i < n; i++)
*(char *)(temp->data + i) = *(char *)(data + i);
temp->next = *p;
*p = temp;
}
void(List *p,void (*f)(void*))
{
while (p)
{
(*f)(p->data);
p = p->next;
}
}
void printstr(void *str)
{
printf(" \"%s\"", (char *)str);
}
Regads
Nishant Pandey
On Thu, May 31, 2012 at 1:15 PM, Hassan Monfared <[email protected]>wrote:
> Why don't you use templates ?
> ----
> template <class T>
> class LNode
> {
> public:
> LNode(T pData,LNode *pNext):data(pData),next(pNext){}
> T data;
> LNode *next;
> };
> template <class T>
> class LList
> {
> protected:
> LNode<T> *head;
> LNode<T> *tail;
> public:
> LList()
> {
> head=tail=NULL;
> }
>
> void push_back(T pData)
> {
> if(head==NULL)
> {
> head=tail=new LNode<T>(pData,NULL);
> return;
> }
> tail->next=new LNode<T>(pData,NULL);
> tail=tail->next;
> }
> void push_front(T pData)
> {
> if(head==NULL)
> {
> head=tail=new LNode<T>(pData,NULL);
> return;
> }
> LNode<T> *nnode=new LNode<T>(pData,head);
> head=nnode;
> }
> void Print()
> {
> LNode<int> *cur=head;
> while(cur)
> {
> cout<<cur->data<<',';
> cur=cur->next;
> }
> cout << endl;
> }
> void Reverse()
> {
> LNode<T> *cur=head;
> LNode<T> *prev=NULL;
> LNode<T> *tmp;
> while(cur)
> {
> tmp=cur->next;
> cur->next=prev;
> prev=cur;
> cur=tmp;
> }
> tmp=head;
> head=prev;
> tail=tmp;
> }
> };
>
> Regards
>
> On Thu, May 31, 2012 at 8:49 AM, mahendra sengar <[email protected]>wrote:
>
>> how to implement generioc linked list..using void pointer...i havent
>> used void pointer much so, m not able to use it properly in linked
>> list..please help asap !!!
>>
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>>
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Cheers,
Nishant Pandey |Specialist Tools Development |
[email protected]<[email protected]> |
+91-9911258345
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