C-堆栈中的简单表达式计算器＆＃x27;应用

Question

编写一个求解算术表达式，以字符串并打印其值的程序。使用两个堆栈，一个用于操作数，另一个用于操作员。仅考虑 +和 *操作员。还考虑1位操作数。考虑操作员在解决表达式方面的优先级。

我的代码无法正常工作，但是即使我一直在寻找几天，我也看不到问题。非常感谢您的帮助！

# include<stdio.h>
# include<stdlib.h>
# include<string.h>
# define MAX 40

int *p1, *p2; // pointers rows operands, operators
int *to1, *to2; // tops rows operands, operators
int *bo1, *bo2; // bases rows operands, operators

void push1(int i), push2(char j); //operating, operator - stack
int pop1(void), pop2(void); //operand, operator - unstack

void push1 (int i)
{
    if (p1>bo1)
    {
        printf("Stack of operands full \n");
    }
    *p1=i;
    p1++;
}
void push2 (char j)
{
    if (p2>bo2)
    {
        printf("Operator stack full \n");
    }
    *p2=j;
    p2++;
}

int pop1()
{
    p1--;
    if (p1<to1)
    {
        printf("Empty operand stack \n");
        return 0;
    }
    return *p1;
}

int pop2()
{
    p2--;
    if (p2<to2)
    {
        printf("Empty operator stack \n");
        return 0;
    }
    return *p2;
}

void main()
{
    int a,b, c, result, i, st;
    char o, s[MAX];
    p1= malloc(MAX*sizeof(int)); //allocate memory stack 1
    p2= malloc(MAX*sizeof(char)); //allocate memory stack 2

    to1=p1;
    to2=p2;
    bo1=p1+MAX-1;
    bo2=p2+MAX-1;

    printf("Enter the expression\n");
    gets(s);
    st=strlen(s);

    push1(s[0]); //put the first digit
    for (i=0; i<(st-1)/2+1; i++)
    {push1(s[2*(i+1)]); //stack the next digit
     push2(s[2*i+1]); //stack the next operator
         if (s[2*i+1]=='*') //if it's multiplication...
         {
             a=pop1(); // get the last two digits
             b=pop1();
             o=pop2(); //delete the operator
             push1(a*b); // replace with product
         }
    }
        
    while(p1>to1)
         {
             a=pop1();
             b=pop1();
             c=a+b;
             push1(c);
         }

    result=pop1();
    printf("Result=%d", result);
}

Answer 1

走下输入字符串应该很简单，

  for(int i = 0; i < st; i++){
      char next = s[i];
       ....
  }

为什么所有复杂的2*，-1， /2+1 ....？