急急急！！栈链插入与删除运算
如何补充？

#include<iostream>
#define NULL 0
using namespace std;
typedef int datatype;
typedef struct node
{
datatype node *next;
struct node *next;
}linkstack;
linkstack *top;
linkstack push_lstack(linkstack *top,datatype x) //进栈
{
linkstack *p;
p=(node*)malloc(sizeof(linkstack));
p->data=x;
p->next =top;
top=p;
return(top);
}
linkstack pop_lstack(linkstack *top,datatype *datap) //出栈
{
linkstack *p;
if(top==null)
{
cout<<"空栈!";
return (NULL);
}
else
{
*datap=top->data;
p=top;
top=top->next;
free(p);
return(top);
}
}
int main()
{

}
[解决办法]

#include "stdafx.h"
Status InitStack(stack* s)
{
 //构造一个空栈
 s->base = (SElemType*) malloc(STACK_INIT_SIZE*sizeof(SElemType));
 if(s->base==NULL) return ERROR;
 s->top = s->base ; //表示空栈
 s->stacksize = STACK_INIT_SIZE ;
 return OK;
}

Status DestroyStack(stack* s)
{
  //销毁栈s，s不再存在
    free(s->base);
    s->top =s->base =NULL;
    free(s); 
    return 0;
}

Status ClearStack(stack* s)
{
  //把栈s清空
    s->top = s->base ;
    return OK;
}

int StackEmpty(stack* s)
{
 //若栈s为空，则返回true, 否则返回false
  if(s->top == s->base) return 1;
  return 0;
}
 
int StackLength(stack* s)
{
  //返回s的元素的个数，即栈的长度
    return s->top - s->base;
}

Status GetTop(stack* s,SElemType* e)
{
 // 若栈不空，则用e返回s的栈顶元素，操作成功返回0 ,失败返回-1
    if(s->top==s->base) return ERROR;
    *e= *(s->top-1);
    return OK;
}

Status Push(stack* s,SElemType e)
{//插入元素e为新栈顶的元素
    if(s->top-s->base >= s->stacksize)
    { //栈，追加存储空间
        s->base  = (SElemType* ) realloc(s->base,(s->stacksize+STACKINCRAMENT)*sizeof(SElemType));
        if(s->base==NULL) return ERROR;
        s->top = s->base +s->stacksize;
        s->stacksize += STACKINCRAMENT;
    }
    *(s->top)++ = e;
    return OK;
}

Status Pop(stack* s,SElemType* e)
{ //若栈不空，则删除s的栈顶元素，用e返回其值 
 //操作成功返回0 ，失败返回-1
    if(s->top==s->base) return ERROR;
    *e = *(--s->top);
    return OK;
}


Status StackTraverse(stack* s)
{
 //从栈顶依次对栈中的每一个元素进行遍历
    SElemType* temp = s->top;
    for(temp--;temp!=s->base-1;temp--)
    {
       printf("%d ",*temp);
    }
    printf("\n");
    return OK;
}