finding a repeating integer in an array - c

Question

Hi managed to make my program checked if i have repeated 8's. I also want to print out how many repeated 8's i have stored in the array.
#include <stdio.h>
#include <stdbool.h>
int main(void)
{
bool digit_seen[10] = {false};
int digit=0;
long n;
printf("Enter number: ");
scanf("%ld", &n);
while (n>0)
{
digit = n % 10;
if (digit_seen[digit])
{
break;
}
digit_seen[digit] = true;
n/=10;
}
if (n>0 && digit ==8)
{
printf("Repeated 8's");
}
else
{
printf("No 8's found");
}
return 0;
}

Answer 1

If I understand your problem, you want to know the number of occurences a 8 is in your number. Like 88 has 2 8s. If that's the case, I don't see why you use a boolean array. First, you need a counter. Second, you need to know if digit is 8 for every digit and increment this counter if that's an 8. Here's an example :
#include <stdio.h>
#include <stdbool.h>
int main(void)
{
int digit=0;
long n;
int counter = 0;
printf("Enter number: ");
scanf("%ld", &n);
while (n>0)
{
digit = n % 10;
if(digit == 8)
{
counter++;
}
n/=10;
}
if (counter > 0)
{
printf("Repeated 8's");
}
else
{
printf("No 8's found");
}
return 0;
}
In this example, counter would have the number of occurrences of 8s in your number. Just display it in the printf and it's done.
EDIT : here's a solution using an array:
#include <stdio.h>
int main(void)
{
int digit = 0;
long n;
int arrayNumber[10] = {0};
printf("Enter number: ");
scanf("%ld", &n);
while (n>0)
{
digit = n % 10;
arrayNumber[digit]++;
n /= 10;
}
if (arrayNumber[8] > 0)
{
printf("Repeated 8's");
}
else
{
printf("No 8's found");
}
return 0;
}
This way, you would know occurrence of every number from 0 to 9 in your integer. I'd also point out that you need to define what is a repeated number. If it's when there's at least 2 occurrence, you need to change arrayNumber[8] > 0 by arrayNumber[8] > 1