中值表达式求值（C语言解法）

#include 
#include 
using namespace std;
#define MAXSIZE 100
typedef struct LNode
{
    int OPNT[MAXSIZE];
    int top1;

} LNode; //操作数栈
typedef struct SLNode
{
    char OPRT[MAXSIZE];
    int top2;

} SLNode; //运算符栈；
void Init_Stack1(LNode &Q)
{
    Q.top1 = 0;//栈顶指针永远指向栈顶元素的上一个
}
void Init_Stack2(SLNode &S)
{
    S.top2 = 0;
}
bool Push_stack1(LNode &Q, int &x)
{
    if (Q.top1 < MAXSIZE)
    {
        Q.OPNT[Q.top1] = x;
        Q.top1++;
        return true;
    }
    else
    {
        return false;
    }
}
bool Push_stack2(SLNode &S, char &x)
{
    if (S.top2 < MAXSIZE)
    {
        S.OPRT[S.top2] = x;
        S.top2++;
        return true;
    }
    else
    {
        return false;
    }
}
bool Get_head1(LNode &Q, int &x)
{
    if (0 < Q.top1 && Q.top1 < MAXSIZE)
    {
        x = Q.OPNT[Q.top1 - 1];
        return true;
    }
    else
    {
        return false;
    }
}
bool Get_head2(SLNode &S, char &x)
{
    if (0 < S.top2 && S.top2 < MAXSIZE)
    {
        x = S.OPRT[S.top2 - 1];
        return true;
    }
    else
    {
        return false;
    }
}
bool Pop_stack1(LNode &Q, int &x)
{
    if (Q.top1 != 0)
    {
        Q.top1--;
        x = Q.OPNT[Q.top1];
        return true;
    }
    else
    {
        return false;
    }
}
bool Pop_stack2(SLNode &S, char &x)
{
    if (S.top2 != 0)
    {
        S.top2--;
        x = S.OPRT[S.top2];
        return true;
    }
    else
    {
        return false;
    }
}
bool Priority(char a, char b)
{
    //用于判断运算符优先级的高低，优先级高于或等于就返回true，否则为false
    if ((a == '+' || a == '-') && (b == '+' || b == '-' || b == '*' || b == '/'))
    {
        return true; //尽管后一个与前一个一个相同但还是后一个优先级高
    }
    else if ((a == '*' || a == '/') && (b == '*' || b == '/'))
    {
        return true;
    }
    else
    {
        return false;
    }
}
int CalculateTwo(int n1, int n2, char x)
{
    if (x == '+')
    {
        return n1 + n2;
    }
    if (x == '-')
    {
        return n2 - n1;
    }
    if (x == '*')
    {
        return n1 * n2;
    }
    if (x == '/')
    {
        return n2 / n1;
    }
}
void Calculate(char str[], int &e)
{
    LNode Q;
    SLNode S;
    Init_Stack1(Q);
    Init_Stack2(S);
    int i, n1, n2, n = 0, sum; //用来算存进OPNT栈中的数;
    char t;                    //存一个操作符；
    for (i = 0; str[i] != '#'; i++)
    {
        if ('0' <= str[i] && str[i] <= '9')
        {
            //n = 0;
            while (1)
            {
                n = n * 10;
                n += (str[i] - '0');
                if ('0' <= str[i + 1] && str[i + 1] <= '9')
                {
                    i++;
                }
                else
                    break;
            }
            Push_stack1(Q, n);
            n = 0;
        }
        else if (str[i] != ')')
        {
            while (Get_head2(S, t) && t != '(' && Priority(str[i], t))
            {
                Pop_stack2(S, t); //后来的比原来的优先级低就弹出后再放回；
                Pop_stack1(Q, n1);
                Pop_stack1(Q, n2);
                sum = CalculateTwo(n1, n2, t);
                Push_stack1(Q, sum);
            }
            Push_stack2(S, str[i]);
        }
        else
        {
            while (Pop_stack2(S, t))
            {
                if (t == '(')
                {
                    break;
                }
                else
                {
                    Pop_stack1(Q, n1);
                    Pop_stack1(Q, n2);
                    sum = CalculateTwo(n1, n2, t);
                    Push_stack1(Q, sum);
                }
            }
        }
    }
    while (Pop_stack2(S, t) && Pop_stack1(Q, n1) && Pop_stack1(Q, n2))
    {
        sum = CalculateTwo(n1, n2, t);//栈中还有元素的把栈中元素用完
        Push_stack1(Q, sum);
    }
    Pop_stack1(Q, e);
}
int main()
{
    char str[MAXSIZE];
    int e;
    cin >> str;
    // push
    //cout< 
等以后学会了STL在来补STL的算法