Bubble Sort

Pseudocode:

for j = 0 to n -2 // no need till n-2 as it will validate already properly aligned values(see code)
    if a[j] > a[j+1]
        swap (a[j], a[j+1])
 After the above pass, the largest element will come at the end.

Perform the above pass for each of the elements in the array.

Code
#include <stdio.h>

#define NELEMS(x) (sizeof (x) / sizeof (x[0]))

void display(int arr[], int n)
{
    int i;
    for (i = 0; i < n; ++i)
        printf("%d ", arr[i]);

    printf("\n");
}

void swap(int *x, int *y)
{
    int t = *x;
    *x = *y;
    *y = t;
}

void bubble(int arr[], int n)
{
    int i, j;
    int swapped = 0;
    for (i = 0; i < n - 1; ++i) { // no need of <n as bigger will find its way anyway
        swapped = 0;
        for (j = 0; j < n - i - 1; ++j) { // j < n-2 also works, but this is not needed
            if (arr[j] > arr[j + 1]) {
                swap(arr + j, arr + j + 1);
                swapped = 1;
                display(arr, 8);
            }
        }
        if (swapped == 0) break; // if no swap happened, no need to continue any further
    }
}

int main()
{
    int arr[] = { 17, 21, 19, 15, 5, 99, 1, 0 };
    /* int arr[] = { 17, 21, 19, 98, 99, 100, 101, 102 }; // almost sorted (usage of swapped flag helps here) */

    printf("Unsorted: ");
    display(arr, NELEMS(arr));

    bubble(arr, NELEMS(arr));

    printf("Sorted: ");
    display(arr, NELEMS(arr));
}

Output:

Unsorted: 17 21 19 15 5 99 1 0

17 19 21 15 5 99 1 0

17 19 15 21 5 99 1 0

17 19 15 5 21 99 1 0

17 19 15 5 21 1 99 0

17 19 15 5 21 1 0 99

17 15 19 5 21 1 0 99

17 15 5 19 21 1 0 99

17 15 5 19 1 21 0 99

17 15 5 19 1 0 21 99

15 17 5 19 1 0 21 99

15 5 17 19 1 0 21 99

15 5 17 1 19 0 21 99

15 5 17 1 0 19 21 99

5 15 17 1 0 19 21 99

5 15 1 17 0 19 21 99

5 15 1 0 17 19 21 99

5 1 15 0 17 19 21 99

5 1 0 15 17 19 21 99

1 5 0 15 17 19 21 99

1 0 5 15 17 19 21 99

0 1 5 15 17 19 21 99

Sorted: 0 1 5 15 17 19 21 99

Time Complexity: O(n^n)