请教一下这个C语言的链栈运算的实现为什么不能运行？


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

#define OK 1
#define ERROR 0
#define OVERLOW  -2
const char oper[7] = {'+' , '-' , '*' , '/' , '(' , ')', '#',};
typedef char SElem;
typedef int status;

typedef struct Snode{
    int data ;
    struct Snode* next ;
}Snode,*LinkStack;

status InitStack (LinkStack s){//链栈的初始化 
    (s)=NULL;
    return OK;
}

int stackempty (LinkStack s)//判空 
{
    if(!s)
    return OK;
    return ERROR;
}
status Push (LinkStack *s, SElem e)//入栈 
{
    Snode *p = (LinkStack )malloc(sizeof(LinkStack));
    if( !p )
    {
        return OVERLOW;
    }
    p->data = e;
    p->next = (*s);
    (*s) = p;
    return OK;
}

status Pop (LinkStack *s,SElem *e)//弹出栈 
{
    Snode *p;
    if( !(*s) )
    return ERROR;
    (*e) = (*s)->data;
    p = (*s);
    (*s) = (*s) ->next;
    free(p);
    return OK;
}

status GetTop(LinkStack *s)//取数据 
{
    if( !(*s) )
    return ERROR;
    return (*s)->data;
}

int  In( char ch)//判断ch是否为运算符 
{
    int i;
    for( i = 0 ; i < 7 ; i++ )
    {
        if ( ch == oper [i] )
            return OK;
    }
    return ERROR ;
}

char Precede (char theta1, char theta2)//判断运算符优先级 
{
    if((theta1 == '(' && theta2 == ')' ) || (theta1 == '#' && theta2 =='#' ))
    {
        return '=' ;
    }
    else if (theta1 == '(' || theta1 == '#' || theta2 == '(' || (theta1 == '+' || theta1 == '-' )
                                                            && (theta2 == '*' || theta2 == '/' ))
    {
        return '<' ;//theta1的运算符优先级小于theta2 
      }
    else return '>' ;

}

char Operate (char first , char theta ,char second)//计算两数运算的结果     
{
    switch (theta){
    case '+':
    return (first -'0') + (second -'0') + 48;
    case '-':
    return (first -'0') - (second -'0') + 48;
    case '*':
    return (first - '0') * (second -'0') + 48;
    case '/':
    return (first -'0') / (second -'0') + 48;
    }
    return 0;
}

char EvaluateExpression() {//算术表达式求值的算符优先算法，设OPTR和OPND分别为运算符栈和操作数栈
    LinkStack OPTR, OPND;//POTR寄存运算符，OPND寄存操作数据或运算结果
    char ch, theta, a, b, x, top;
    InitStack(OPND); 
    InitStack(OPTR); 
    Push(&OPTR, '#'); //将表达式起始符“#”压入OPTR栈
    scanf("%s",ch); 
    while (ch != '#' || (GetTop(&OPTR) != '#')) //表达式没有扫描完毕或OPTR的栈顶元素不为“#”
    {
        if (!In(ch)) {
            Push(&OPND, ch);
            scanf("%s",ch);
        } //ch不是运算符则进OPND栈
        else
            switch (Precede(GetTop(&OPTR), ch)) //比较OPTR的栈顶元素和ch的优先级
            {
            case '<':
            Push(&OPTR, ch);
            scanf("%s",ch); //当前字符ch压入OPTR栈，读入下一字符ch
            break;
            case '>':
            Pop(&OPTR, &theta); //弹出OPTR栈顶的运算符
            Pop(&OPND, &b);
            Pop(&OPND, &a); //弹出OPND栈顶的两个运算数
            Push(&OPND, Operate(a, theta, b)); //将运算结果压入OPND栈
            break;
            case '=': //OPTR的栈顶元素是“(”且ch是“)”
            Pop(&OPTR, &x);
            scanf("%s",ch); //弹出OPTR栈顶的“(”，读入下一字符ch
            break;
        } 
}
        return GetTop(&OPND); //OPND栈顶元素即为表达式求值结果
}


int menu() {
    int c;
    printf("0-9以内的多项式计算\n");
    printf("1.计算\n");
    printf("0.退出\n");
    printf("选择：\n");
    scanf("%d",&c);
    return c;
}

int main() {
    while (1) {
        switch (menu()) {
        case 1: {
            printf("请输入要计算的表达式（操作数和结果都在0-9的范围内，以#结束）：\n");
            char res = EvaluateExpression();//算法3.22　表达式求值
            printf("计算结果为：%d\n",(res-48));
        }
            break;
        case 0:
            printf("退出成功\n");
            exit(0);
        default:
            break;
        }
    }
}