How to overcome the windows stopped working error message? - c

Question

I'm writing a program for a simple calculator that accepts a string (without spaces) with two operators and one operand and outputs the result. My program compiles just fine but during runtime, I'm getting the windows stopped working error message. What could be the reason for it? I think it's good something to do with my program's logic. Here's my code:
#include <stdio.h>
#include <math.h>
int main()
{
int operand_1 = 0, operand_2 = 0, i = 0, result = 0;
char string[10], *ptr, operation;
ptr = string;
printf("Enter the expression:\n");
gets(string);
//puts(string);
while(*ptr==0||*ptr==1||*ptr==2||*ptr==3||*ptr==4||*ptr==5||*ptr==6||*ptr==7||*ptr==8||*ptr==9)
{
operand_1 = operand_1 + ((int)*ptr)*pow(10,i);
i++;
ptr++;
}
operation = *ptr;
ptr++;
i=0;
while(ptr!=NULL)
{
operand_2 = operand_2 + ((int)*ptr)*pow(10,i);
i++;
ptr++;
}
switch(operation)
{
case '+':
result = operand_1 + operand_2;
case '-':
result = operand_1 + operand_2;
case '*':
result = operand_1 * operand_2;
case '/':
result = operand_1 / operand_2;
case '%':
result = operand_1 % operand_2;
}
printf("%d",result);
}

Answer 1

I've resolved the problem and also improved the logic of the program so that it works fine.
#include <stdio.h>
#include <math.h>
int main()
{
while(1)
{
int operand_1 = 0, operand_2 = 0, i = 0, result = 0;
char string[10], *ptr, operation;
ptr = string;
printf("\nEnter the expression:");
scanf("%s",string);
if (*ptr=='q' || *ptr=='Q' && *(ptr+1)=='u' || *(ptr+1)=='U' && *(ptr+2)=='i' || *(ptr+2)=='I' && *(ptr+3)=='t' || *(ptr+3)=='T')
{
printf("\nBye");
break;
}
while(*ptr>='0' && *ptr<='9')
{
operand_1 = operand_1*10 + (*ptr - 48);
i++;
ptr++;
}
operation = *ptr;
ptr++;
i=0;
while(*ptr>='0' && *ptr<='9')
{
operand_2 = operand_2*10 + (*ptr-48);
i++;
ptr++;
}
if(operand_2 == 0)
{
printf("\nDivision by zero");
continue;
}
switch(operation)
{
case '+':
result = operand_1 + operand_2;
break;
case '-':
result = operand_1 - operand_2;
break;
case '*':
result = operand_1 * operand_2;
break;
case '/':
result = operand_1 / operand_2;
break;
case '%':
result = operand_1 % operand_2;
break;
}
printf("\nResult = %d",result);
}
}
Hope it helps someone in need :)