Why doesn't this array save the scanned numbers?

This is my code. It is supposed to scan numbers until 0 is entered. Then it should print the numbers. But all it prints is zero.

#include <stdio.h>

int main()
{
    int x[100];
    int n;
    int i;
    for(i = 0; ; i++)
    {
        while(x[i] != 0)
        {
            scanf("%d", &x[i]);
            n++;
        }
        if(x[i] == 0)
            {
                break;
            }
    }
    for(i = 0; i < n;i++)
    {
        printf("%d\n", x[i]);
    }

}

I was expecting to see the exact numbers that I entered.

>Solution :

The reason for the early exit is because here: while(x[i] != 0) can fail if the uninitialized value (x[i]) is 0. Though it might not be 0. But this logic is severely flawed in a couple other ways.

• There is really no need to do a while loop since the for loop is already iterating through the array.

• You should really have a terminating condition in the loop since, you could iterate passed 100 and write outside the bounds of the array.

• n is uninitialized, so it could actually start counting at any number like 12456 or something =]

Here is what I think you want:

#include <stdio.h>

int main()
{
    int x[100];
    int n = 0;  // init this!
    int i;
    for(i = 0; i < 100; i++) // added terminating condition here
    {    
       scanf("%d", &x[i]);
       n++;
       if(x[i] == 0)
       {
           break;
       }
    }
    for(i = 0; i < n;i++)
    {
        printf("%d\n", x[i]);
    }
}

Also, you should really be checking the return value of scanf. I’ll leave that one to you =].