如何在计算器 C 中检查表达式是否被 0 除？ 浮点异常

Question

I tried to make simple calculator supporting brackets and classic operators +,-,*,/ and I also need to eliminate division by zero in it. I may not be able to cope with it, I need to check if there is division by zero in the expression and if so function return 0. Now it returns Floating point exception . Can someone help?

My code:

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

#define MAX_SIZE 1024

int insert_operand(int *operand, int * top_num, int num) /* data is pushed into the data stack*/
{
     (*top_num) ++;
     operand[*top_num] = num; /*save data*/

     return 0; /*Exit normally*/
}

int insert_oper (char * oper , int *top_oper , char ch)
{
     (*top_oper)++;
     oper[*top_oper] = ch; /*save operator*/

     return 0; /*Exit normally*/
}

int compare(char *oper , int *top_oper , char ch) /* compare the priority of the operating server*/
{

     if((oper[*top_oper] == '-' || oper[*top_oper] == '+') /*Determine whether the current priority is higher than the priority of the operator at the top of the stack*/
             && (ch == '*' || ch == '/'))
     {
         return 0;
     }

     else if(*top_oper == -1 || ch == '('
             || (oper[*top_oper] == '(' && ch != ')')) /*Determine whether the operator stack is empty; whether the top operator is '('*/
     {
         return 0;
     }

     else if (oper[*top_oper] =='(' && ch == ')')
     {
         (*top_oper)--;
         return 1;
     }

     else
     {
         return -1; /*Operate the operator*/
     }

}

int deal_date(int *operand ,char *oper ,int *top_num, int *top_oper) /*perform data operation*/
{
     int num_1 = operand[*top_num]; /*Take out two data from the data stack*/
     int num_2 = operand[*top_num - 1];

     int value = 0;

     if(oper[*top_oper] == '+')
     {
         value = num_1 + num_2;
     }

     else if(oper[*top_oper] == '-')
     {
         value = num_2 - num_1;
     }

     else if(oper[*top_oper] == '*')
     {
         value = num_2 * num_1;
     }

     else if(oper[*top_oper] == '/')
     {
         value = num_2 / num_1;
     }

     (*top_num) --; /*Move the top of the data stack down one bit*/
     operand[*top_num] = value; /*Push the obtained value into the data stack*/
     (*top_oper) --; /*Move the top of the operator stack down one bit*/

    return value;
}

void compress(char *stx)    /* The additional function */
{
    char work[101];
    int i = strlen(stx);

    strcpy(work, stx);

    for (int j = 0; j < i; j++)
    {
        stx[j] = 0;
    }

    i = 0;

    for (int j = 0; j < (int)strlen(work); j++)
    {
        if (work[j] != ' ')
        {
            stx[i] = work[j];
            i++;
        }
    }
}

int main()
{
     int operand[MAX_SIZE] = {0}; /*data stack, initialize*/
     int top_num = -1;

     char oper[MAX_SIZE] = {0}; /*operator stack, initialize*/
     int top_oper = -1;

     char *str = (char *) malloc (sizeof(char) * 100); /*get expression (without ＝)*/
     //scanf("%s", str);
     scanf("%[^\n]", str);  /* Refined the scanf call to receive all characters prior to the newline/return character */

     compress(str);         /* Added this function to remove spaces */

     char* temp;
     char dest[MAX_SIZE];
     int num = 0;

     int i = 0;
     while(*str != '\0')
     {
         temp = dest;

         while(*str >= '0' && *str <= '9') /*judging whether it is data*/
         {
             *temp = *str;
             str++;
             temp++;
         } /*Encounter a symbol to exit*/

         if(*str != '(' && *(temp - 1) != '\0') /*Determine whether the symbol is '('*/
         {
             *temp = '\0';

             num = atoi(dest); /*convert string to number*/
             insert_operand(operand, &top_num,num); /*Push data into the data stack*/
         }

          while(1)
          {
              i = compare(oper,&top_oper,*str); /*judgment operator priority*/

             if(i == 0)
             {
                 insert_oper(oper,&top_oper,*str); /*press operator*/
                 break;
             }

             else if(i == 1) /*judging whether the expression in brackets ends*/
             {
                 str++;
             }

             else if(i == -1) /* data processing */
             {
                 deal_date(operand,oper,&top_num,&top_oper);
             }

          }

         str ++; /* point to the next character of the expression */
     }

     printf("%d\n",operand[0]); /*output result*/

     return 0;

}

Thanks for all answers

Answer 1

I would detect if the divisor is zero along these lines:

    else if(oper[*top_oper] == '/')
    {
        value = num_1 ? num_2 / num_1 : 0;
    }

or perhaps more verbose:

    else if(oper[*top_oper] == '/')
    {
        if(num_1)
            value = num_2 / num_1;
        else
            value = 0;
    }

and example session:

1 / 2
0