求多位数的运算表达式结果—栈

因为单片机实训的项目是关于计算器，所以需要多位数来进行连续混合运算，所以需要用到栈。但是和上学期做的题还不太一样，所以

---

整数、负数、小数的连续混合运算

#include<iostream>
#include<string>
#include<stdlib.h>
#include<stdio.h>
using namespace std;

const char oper[7] = { '+', '-', '*', '/', '(', ')', '=' };
#define OK 1
#define ERROR 0
#define OVERFLOW -2
typedef char SElemType;
typedef int Status;
typedef struct SNode
{
    double data;
    struct SNode *next;
} SNode, *LinkStack;

double h;

Status InitStack(LinkStack &S)
{
    S = NULL;
    return OK;
}
bool StackEmpty(LinkStack S)
{
    if (!S)
        return true;
    return false;
}
Status Push(LinkStack &S, SElemType e)
{
    SNode *p = new SNode;
    if (!p)
    {
        return OVERFLOW;
    }
    p->data = e;
    p->next = S;
    S = p;
    return OK;
}


Status Pushint(LinkStack &S,double e)
{
    SNode *p = new SNode;
    if (!p)
    {
        return OVERFLOW;
    }
    p->data = e;
    p->next = S;
    S = p;
    return OK;
}


Status Pop(LinkStack &S, SElemType &e)
{
    SNode *p;
    if (!S)
        return ERROR;
    e = S->data;
    p = S;
    S = S->next;
    delete p;
    return OK;
}
Status Popint(LinkStack &S,double &e)
{
    SNode *p;
    if (!S)
        return ERROR;
    e = S->data;
    p = S;
    S = S->next;
    delete p;
    return OK;
}


double GetTop(LinkStack &S)
{
    if (!S)
        return ERROR;


    return S->data;
}
bool In(char ch)  //判断ch是否为运算符
{
    for (int i = 0; i < 7; i++)
    {
        if (ch == oper[i])
        {
            return true;
        }
    }
    return false;
}
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 '<';
    }
    else
        return '>';
}

double Operate(double first, char theta, double second)  //计算两数运算结果
{
    switch (theta)
    {
    case '+':
        return (first) + (second);
    case '-':
        return (first) - (second);
    case '*':
        return (first) * (second);
    case '/':
        double m = (first)*1.0 / (second);
        return m;
    }
    return 0;
}


//算法3.22　表达式求值
void 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;
    char ch, theta,x, top;
    double a,b;
    string c;
    InitStack(OPND); //初始化OPND栈
    InitStack(OPTR); //初始化OPTR栈
    Push(OPTR, '='); //将表达式起始符“=”压入OPTR栈
    cin >> 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);
            c+=ch;
            cin >> ch;
        } //ch不是运算符则进OPND栈
        else
        {
            if(!c.empty())
            {
                double y = atof(c.c_str());
                Pushint(OPND,y);
                c.erase();
            }


            switch (Precede(GetTop(OPTR), ch)) //比较OPTR的栈顶元素和ch的优先级
            {
            case '<':
                Push(OPTR, ch);
                cin >> 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
                Popint(OPND, b);
                Popint(OPND, a); //弹出OPND栈顶的两个运算数
                Pushint(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);
                cin >> ch; //弹出OPTR栈顶的“(”，读入下一字符ch
                break;
            }//switch
        }
    } //while
   h = GetTop(OPND); //OPND栈顶元素即为表达式求值结果

}

int main()
{
    while (1)
    {
            cout << "请输入要计算的表达式（操作数和结果都在整数的范围内，以=结束）：" << endl;
            EvaluateExpression();//算法3.22　表达式求值
            //cout << "计算结果为 " << res<<endl;
            printf("%.2lf\n", h);
            //break;
    }
}