表达式计算：运行失败，不知道什么原因

Question

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
const int SIZE = 50;
class Stack_number {
private:
    int num[SIZE];
    int top1;
public:
    Stack_number();
    //~Stack_number();
    void push(int q);
    int pop();
    int get_top();
};
class Stack_operator {
private:
    char op[SIZE];
    int top2;
public:
    Stack_operator();
    //~Stack_operator();
    void push(char ch);
    char pop();
    char get_top();
};
bool In(char ch); //判断输入的符号是否是运算符！
char Precede(char ch1, char ch2); //比较运算符优先级别！
int Operate(int i, char ch, int j);//运算！
int main() {
    Stack_number OPND;
    Stack_operator OPTR;
    int nu, n = 0, p = 0, q = 0;
    char c, o;
    char ch = getchar(); //getchar()是stdio.h中的库函数，它的作用是从stdin流中读入一个字符
                         //也就是说，如果stdin有数据的话不用输入它就可以直接读取了
                         //第一次getchar()时，确实需要人工的输入，但是如果你输了多个字符，以后的getchar()再执行时就会直接从缓冲区中读取了
                         //实际上是 输入设备->内存缓冲区->getchar()
    while (ch != '#' || OPTR.get_top() != '#') {
        if (!In(ch)) {
            //while (!In(ch)) {
            nu = int(ch) - 48;
            //n++;//n用来记录当前读入数字的位数！
            OPND.push(nu);
            ch = getchar();
            //}
        }
        else {
            o = OPTR.get_top();
            switch (Precede(o, ch))
            {
            case'<':
                OPTR.push(ch);
                ch = getchar();
                break;
            case'>':
                p = OPND.pop();
                q = OPND.pop();
                c = OPTR.pop();
                OPND.push(Operate(q, c, p));
                ch = getchar();
                break;
            case'=':
                OPTR.pop();
                ch = getchar();
                break;
            }
        }
    }
    cout << OPND.get_top();
    return 0;
}


void Stack_number::push(int q) {
    if (top1 < SIZE)
        num[top1++] = q;

}
int Stack_number::pop() {
    if (top1 > 0)
        return num[top1--];
}
int Stack_number::get_top() {
    if (top1 > 0)
        return num[top1-1];
}

void Stack_operator::push(char ch) {
    if (top2 < SIZE)
        op[top2++] = ch;
}
char Stack_operator::pop() {
    if (top2 > 0)
        return op[top2--];
}
char Stack_operator::get_top() {
    if (top2 > 0)
        return op[top2-1];
}

//判断输入的符号是否是运算符！
bool In(char ch) {
    if (ch == '-' || ch == '+' || ch == '*' || ch == '/' || ch == '(' || ch == ')' || ch == '#')
        return true;
    else
        return false;
}
//比较运算符优先级别！
char Precede(char ch1, char ch2) {
    int ang[2] = { 0,0 };
    char op[2] = { ch1,ch2 };
    char aa[7][7] = {
        {'>','>','<','<','<','>','>'},
        {'>','>','<','<','<','>','>'},
        {'>','>','>','>','<','>','>'},
        {'>','>','>','>','<','>','>'},
        {'<','<','<','<','<','=',' '},
        {'>','>','>','>',' ','>','>'},
        {'<','<','<','<','<',' ','='}
    };  //不要忘了 ；！
    for (int i = 0; i < 2; i++) {

        switch (op[i]) {//利用op[i]记录θ1和θ2！！！
        case '+':ang[i] = 0;
            break;
        case'-':ang[i] = 1;
            break;
        case '*':ang[i] = 2;
            break;
        case'/':ang[i] = 3;
            break;
        case '(':ang[i] = 4;
            break;
        case')':ang[i] = 5;
            break;
        case'#':ang[i] = 6;
            break;
        }
        //绝妙！i=0时 ang[0]=0~6来确定aa数列的第几行！
         //i=1时ang[1] = 0~6来确定aa数列的第几列！
    }
    return aa[ang[0]][ang[1]];//格外注意该处格式较模糊复杂！
}
//进行二元运算！
int Operate(int i, char ch, int j) {
    switch (ch) {
    case'+':return i + j;
        break;
    case'-':return i - j;
        break;
    case'*':return i * j;
        break;
    case'/':return i / j;
        break;
        //default: 
            //break;
    }
}
Stack_number::Stack_number() {
    num[0] = 0;
    top1 = 0;
}
//Stack_number::~Stack_number() {

//}
Stack_operator::Stack_operator() {
    op[0] = '#';
    top2 = 1;
}
//Stack_operator::~Stack_operator(){

//}

Answer 1

求助求助- -~！