下面一个用C编写的简单计算器的程序：
帮我重点分析一下函数push和pop，以及结构体在程序中期的作用？

#include <stdio.h> /*库函数*/
struct s_node { int data; struct s_node *next; };
typedef struct s_node s_list;
typedef s_list *link;
link operator=NULL;
link operand=NULL;
link push(link stack,int value)
{ link newnode;
    newnode=(link*) malloc(sizeof(s_list));
    if(!newnode)
    { printf("\nMemory allocation failure!!!");
      return NULL; }
    newnode->data=value;
    newnode->next=stack;
    stack=newnode;
    return stack; }
link pop(link stack,int *value)
{ link top;
    if(stack !=NULL)
    { top=stack; stack=stack->next; *value=top->data;
     free(top); return stack; }
    else *value=-1; }
int empty(link stack)
{ if(stack==NULL)
    return 1;
    else return 0; }
int is_operator(char operator)
{ switch (operator)
{ case '+': case '-': case '*': case '/': return 1; default:return 0; } }
int priority(char operator)
{ switch(operator)
    { case '+': case '-' : return 1; case '*': case '/' : return 2; default: return 0; } }
int two_result(int operator,int operand1,int operand2)
{ switch(operator)
    { case '+':return(operand2+operand1);
    case '-':return(operand2-operand1);
     case '*':return(operand2*operand1);
     case '/':return(operand2/operand1); }
}
void main()
{ char expression[50]; int position=0; int op=0; int operand1=0; int operand2=0; int evaluate=0;
printf("\nPlease input the inorder expression:");
gets(expression);
while(expression[position]!='\0'&&expression[position]!='\n')
{ if(is_operator(expression[position]))
    { if(!empty(operator))
        while(priority(expression[position])<= priority(operator->data)&& !empty(operator))
        {    operand=pop(operand,&operand1);
            operand=pop(operand,&operand2);
            operator=pop(operator,&op);
            operand=push(operand,two_result(op,operand1,operand2));
        }
        operator=push(operator,expression[position]); }
    else operand=push(operand,expression[position]-48);
    position++; }
while(!empty(operator))
{
    operator=pop(operator,&op);
    operand=pop(operand,&operand1);
    operand=pop(operand,&operand2);
    operand=push(operand,two_result(op,operand1,operand2));
}
operand=pop(operand,&evaluate);
printf("The expression [%s] result is '%d' ",expression,evaluate); getch(); }

这程序太牛了！