#include <iostream>
#include <math.h>
using namespace std;

// converting from decimal to binary

int main()
{
    int n, bit, ans = 0, i = 0;
    cout << "enter a number: ";
    cin >> n;
    while (n > 0)
    {
        bit = n & 1;
        ans = round(bit * pow(10, i)) + ans; /* we used 'round' bcoz pow(10, 2) give 99.99999
                                                it differs from compiler to compiler  */
        n = n >> 1;
        i++;
    }

    cout << ans;
    return 0;
}

I am unable to understand that in while(n>0), n will be stored as binary form or decimal form.
That is if n=5, so whether while loop check for (5>0) or (101>0).

Can anyone explain what is happening here?

I am new to this platform, please don’t delete my question. My earlier questions are also gets deleted due enough dislikes. I am here to learn and am still learning.

>Solution :

"Decimal" and "Binary" are text representations of a value. n is an int, not text, so it holds the value 5. If I have 5 apples on a desk, then there is no "decimal" or "binary" involved. THere’s just 5 apples. Same with int.

cin >> n;
while (n > 0)

This loop continues because 5 > 0.

n = n >> 1;

This is just a fancy way of writing n = n / 2, so then n becomes 2. Since 2 > 0, the loop continues. On the third run, n becomes 1, and since 1 > 0, the loop continues a third time. On the fourth run, n becomes 0, and 0 == 0, so the while loop exits.