Sorting, how should I cout the round and pass? - c++

Question

Our professor told us to make our program cout the round and pass of the sorting algorithm.. What should I do then?
https://qph.is.quoracdn.net/main-qimg-21f7faf7e69c3317661f5f959008e031?convert_to_webp=true
Like this one ^
Bubble sort code
#include <stdio.h>
int main()
{
int array[100], n, c, d, swap;
printf("Enter number of elements\n");
scanf("%d", &n);
printf("Enter %d integers\n", n);
for (c = 0; c < n; c++)
scanf("%d", &array[c]);
for (c = 0 ; c < ( n - 1 ); c++)
{
for (d = 0 ; d < n - c - 1; d++)
{
if (array[d] > array[d+1]) /* For decreasing order use < */
{
swap = array[d];
array[d] = array[d+1];
array[d+1] = swap;
}
}
}
printf("Sorted list in ascending order:\n");
for ( c = 0 ; c < n ; c++ )
printf("%d\n", array[c]);
return 0;
}
Is this right??? --->
#include<stdio.h>
int main()
{
int a[100],c[100],b,n,i,temp;
printf("\t\t\tBUBBLE SORT\n");
printf("Enter total numbers : ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("\nEnter %d number : ",i+1);
scanf("%d",&a[i]);
c[i]=a[i];
}
printf("\n");
for(i=0;i<n;i++)
{
for(int j=0;j<n-1;j++)
{
b=0;
if(a[j]<a[j+1])
{ b++;
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
if(b!=0)
{
for(int k=0;k<n;k++)
{
printf("%d ",a[k]);
}
printf("\n");
}
}
}
printf("\nAssending order By Bubble sort\n\t");
for(i=0;i<n;i++)
{
printf("%d ",c[i]);
}
return 0;
}

Answer 1

This is just simple step by step bubble sort problem.
just make two functions as bubble_asc(), bubble_desc()
and call it from the main function().
int main() {
...
bubble_asc();
bubble_desc();
...
}
void bubble_asc() {
...
do step by step bubble sort with ascending order here.
for(~~~~~) {
for(~~~~~~) {
if(array[a] < array[b])
....
}
}
print arrays here..
}
void bubble_desc() {
...
do step by step bubble sort with descending order here.
for(~~~~~) {
for(~~~~~~) {
if(array[a] > array[b])
....
}
}
print arrays here..
}