c语言实现逆波兰表达式！

C语言实现逆波兰表达式(栈的应用)

 #include<iostream>
 #include<cstdio>
 using namespace std;
 const int MAXSIZE = 110;
 char a[MAXSIZE];
 double operNum[MAXSIZE];
 ​
 double getSum(int* i){//地址传递，可以在边求值时边改变i的原值。如若是值传递，会导致值的重复算 
     double sum = 0.0;
     int k = 0;
     while(a[*i] >= '0' && a[*i] <= '9'){
         sum = sum * 10 + a[*i] - '0';
         (*i)++;
     }
     if(a[*i] == '.'){
         (*i)++;
         while(a[*i] >= '0' && a[*i] <= '9'){
             sum = sum * 10 + a[*i] - '0'; 
             (*i)++;
             k++;
         }
     }
     while(k!=0){
         sum /= 10;
         k--;
     }
     return sum;
 }
 int main(){
     gets(a);//这样可以读入空格，要求以'#'结尾
     
     int i=0;
     int top=0;
     double x1,x2;
     while (a[i] != '#'){
         if(a[i] >= '0' && a[i] <= '9'){
             double f = getSum(&i);
             operNum[top++] = f;
         }else if(a[i] == ' '){
             i++;
         }else if(a[i] == '+'){
             x1 = operNum[--top];
             x2 = operNum[--top];
             operNum[top++] = x1 + x2;
             i++;
         }else if(a[i] == '-'){
             x1 = operNum[--top];
             x2 = operNum[--top];
             operNum[top++] = x2 - x1;
             i++;
         }else if(a[i] == '*'){
             x1 = operNum[--top];
             x2 = operNum[--top];
             operNum[top++] = x1 * x2;
             i++;
         }else if(a[i] == '/'){
             x1 = operNum[--top];
             x2 = operNum[--top];
             operNum[top++] = x2 / x1;
             i++;
         }
     }
     
     cout<<operNum[0]<<endl;
     return 0;
 }