C语言栈—多位数的运算式求和

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

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

double h;//结果

double e1;//数栈
char e2;//运算符栈

LinkStack InitStack(LinkStack S)
{
    S = NULL;
    return S;
}

int StackEmpty(LinkStack S)
{
    if (!S)
        return 1;
    return 0;
}

LinkStack Push(LinkStack S)
{
    SNode *p = (LinkStack)malloc(sizeof(SNode));
    if (!p)
    {
        return OVERFLOW;
    }
    p->data = e2;
    p->next = S;
    S = p;
    return S;
}


LinkStack Pushint(LinkStack S)
{
    SNode *p = (LinkStack)malloc(sizeof(SNode));
    if (!p)
    {
        return OVERFLOW;
    }
    p->data = e1;
    p->next = S;
    S = p;
    return S;
}


LinkStack Pop(LinkStack S)
{
    SNode *p;
    if (!S)
        return ERROR;
    e2 = S->data;
    p = S;
    S = S->next;
    free(p);
    return S;
}

LinkStack Popint(LinkStack S)
{
    SNode *p;
    if (!S)
        return ERROR;
    e1 = S->data;
    p = S;
    S = S->next;
    free(p);
    return S;
}


double GetTop(LinkStack S)
{
    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 1;
        }
    }
    return 0;
}

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 '/':
        return (first)*1.0 / (second);
    }
    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;
    double every_num[10];//存数

    int num = 0;//每个数的位数

    InitStack(OPND); //初始化OPND栈
    InitStack(OPTR); //初始化OPTR栈
    e2 = '=';
    OPTR = Push(OPTR); //将表达式起始符“=”压入OPTR栈
    scanf("%c", &ch);
    if(ch == 'n')
        scanf("%c", &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))
        {
            every_num[num++] = ch - '0';
            scanf("%c", &ch);
        } //ch不是运算符则进OPND栈
        else
        {
            if(num > 0)
            {
                int i;
                double y = 0;

                for(i = num-1; i >= 0; i--){
                     y += every_num[i] * pow(10, num-i-1);
                }

                num = 0;

                e1 = y;
                OPND = Pushint(OPND);
            }


            switch (Precede(GetTop(OPTR), ch)) //比较OPTR的栈顶元素和ch的优先级
            {
            case '<':
                e2 = ch;
                OPTR = Push(OPTR);
                scanf("%c", &ch); //当前字符ch压入OPTR栈，读入下一字符ch
                break;
            case '>':
                OPTR = Pop(OPTR); //弹出OPTR栈顶的运算符
                theta = e2;
                OPND = Popint(OPND);
                b = e1;
                OPND = Popint(OPND); //弹出OPND栈顶的两个运算数
                a = e1;
                e1 = Operate(a, theta, b);
                OPND = Pushint(OPND); //将运算结果压入OPND栈
                break;
            case '=': //OPTR的栈顶元素是“(”且ch是“)”
                OPTR = Pop(OPTR);
                x = e2;
                scanf("%c", &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)
    {
     //       printf("%dn", 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）：);
            EvaluateExpression();//算法3.22　表达式求值
            //cout << "计算结果为 " << res<<endl;
            printf("%.2lfn", h);
            //break;
    }
}

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