问题：循环不能结束，一直输出‘#’。为什么？求高人指点迷津！


代码如下
#include<stdio.h>
#include<malloc.h>
typedef struct linknode
{
    char data;
    float data1;
    struct linknode *next;
}LiStack;

/*构造一个栈S*/
void InitStack(LiStack *&s)
{
    s=(LiStack*)malloc(sizeof(LiStack));
    s->next=NULL;
}


/*入栈*/
void Push (LiStack *&s,char e)
{
    LiStack *p;
    p=(LiStack *)malloc(sizeof(LiStack));
    p->data=e;
    p->next=s->next;
    s->next=p;
}


/*出栈*/
int Pop(LiStack *&s,char &e)
{
    LiStack *p;
    if(s->next==NULL)
        return 0;
    p=s->next;
    e=p->data;  //注意不能更换顺序
    s->next=p->next;
    free(p);
    return 1;
}


/*取栈顶元素*/
int GetTop(LiStack *s,char &e)
{   
   
    if(s->next==NULL)
        return 0;
    e=s->next->data;
    printf("%c",e);
        return 1;
}




/*入栈*/
void Push1 (LiStack *&s,float e)
{
    LiStack *p;
    p=(LiStack *)malloc(sizeof(LiStack));
    p->data=e;
    p->next=s->next;
    s->next=p;
}


/*出栈*/
float Pop1(LiStack *&s,float &e)
{
    LiStack *p;
    if(s->next==NULL)
        return 0;
    p=s->next;
    e=p->data;  //注意不能更换顺序
    s->next=p->next;
    free(p);
    return 1;
}


/*取栈顶元素*/
float GetTop1(LiStack *s,float &e)
{   
   
    if(s->next==NULL)
        return 0;
    e=s->next->data;
    printf("%c",e);
        return 1;
}


/*构造数组存储运算符号*/
char OP[7]={'+','-','*','/','(',')','#'};
/*开数组存储对运算符进行比较*/
char d[7][7]={      {'>','>','<','<','<','>','>'},   {'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},   {'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=',' '},   {'>','>','>','>',' ','>','>'},
{'<','<','<','<','<',' ','='}
};
/*计算运算数*/
float Operate(float a, char theta, float b)
{   
    switch( theta )
    {    case '+': return(a+b);
    case '-': return(a-b);
    case '*': return(a*b);
    case '/': return(a/b);
    default : return 0;
    }
}   
/*查找运算符是否相同*/
int In (char Test, char* Op)
{
    for(int i=0; i<7; i++)
        if(Test == Op[i])
            return 1;
        return 0;
}
/*比较运算符*/
char Precede(char a, char b)
{
    int i, j;
    for(i=0; i<7; i++)
        if(a==OP[i])  
            break;
        for(j=0; j<7; j++)
            if(b==OP[j])  
                break;
            return(d[i][j]);
}
/*计算表达式*/
float EvaluateExpression()
{
    LiStack *OPTR,*OPND;
    char c,e,theta;
   
    float a,b,value;
   
    InitStack(OPTR); Push(OPTR,'#');
    InitStack(OPND); c=getchar();
    while(c!='#'||GetTop(OPTR,e)!='#')
    {
        if(!In(c,OP))
        {Push1(OPND,c); c=getchar();}
        else
            switch(Precede(GetTop(OPTR,e),c))
        {     case '<': Push(OPTR,c); c=getchar(); break;
              case '=': Pop(OPTR,e); c=getchar(); break;
              case '>': Pop(OPTR,theta);
                  Pop1(OPND,b);
                  Pop1(OPND,a);
                  Push1( OPND,Operate(a,theta,b) );
                  break;
        }
    }
    return GetTop1(OPND,value);
}

int main()
{
    float value;
    printf("请依次输入表达式:");
    value=EvaluateExpression();
    printf("计算表达式的结果：%.2f\n",value);
    return 0;
}

批评下
给0分  这态度就不端正  没体现出问题的紧急程度

你好。请问怎么修改好呢