Thread: C program on arrays - Explination

Hey,
i have a code here that scans 10 integers and prints them without they're double.
If i print 1234356792 my output will be 12345679
The code below shows how to do it but i need someone to explain me a loop that's there.

Code:

#include <stdio.h>


int main()
{
    int arr[10];
    int i, j, seen;
    
    printf("Enter 10 integers:\n");
    
    for (i=0; i<10 ; i++)
    {
        scanf("%d", &arr[i]);
    }
    printf("Set(array) is:");
    
    for (i=0; i<10; i++)
    {
        seen = 0;
        for (j=0; j<i; j++)           // What does that mean?
        {
            if (arr[i]==arr[j])
            {
                seen=1;             // what the 1 means?
                break;
            }
        }
        if(!seen)
            printf("%d", arr[i]);
    }
    printf("\n");
    return 0;
}

appreciate any help

I can illustrate one problem. Scanf will read for exactly what you tell it to. So. How many integers is this?

123456789 1 2 3 4 5 6 7 8 9

If you said anything besides 10, you're wrong.

Code:

for (j=0; j<i; j++)           // What does that mean?

Does your book have a section on for loops? You should read it.

Code:

                seen=1;             // what the 1 means?

True, in this later Boolean expression:

Code:

if(!seen)

See also <stdbool.h>

thanks for you reply.
for the scanf it scans 10 integers so 10 is an integer also it could scan it of course

sick appreciate the help!