Could anyone please explain this code to me

Question

Here is a code, C, that I dont understand.
---------------------

#include<stdio.h>
#include<math.h>

int main (void)
{
int c = 0;
long long v = 88888;

while (v > 0)
{

v = v / 10;
c++;
}

printf("%lli", c);
}

----------------------

When I run the code , the 'printf' format string returns '5' as a the equivalent of the variable 'c'. And 'c' from the onset was declared as 0.

How come the 'c++' in the while loop changes to 5?
How does the 'v' variable pass its result to 'c'as 4 instead of 8888.8?

Thank you.

Answer 1

The while loop goes on if v > 0. Since v is declared as 88888, it is > 0.
Now, v gets divided by 10 and is 8888 (you declared it as int) and in the next line c gets incremented by 1 and is now 1. The loop goes on until v stops being greater than 0.

So:
Initially: v = 88888, c = 0
1st loop: v = 8888, c = 1
2nd loop: v = 888, c = 2
3rd loop: v = 88, c = 3
4th loop: v = 8, c = 4
5th loop: v = 0, c = 5
loop ends, as v is no longer > 0

Answer 2

v = v/10 is integer division since we don't have a floating point type. This basically drops the number in the ones position off of the value of v with each iteration of the loop until v = 0 and the loop stops.

c++ is the same as c = c + 1
So the value is incremented with each iteration of the loop. Thus, equaling 5 when the loop stops and exits.

Answer 3

I am really greatful for this, thank you. Namit Jain Kuba Siekierzyński ChaoticDawg

Answer 4

ChaoticDawg But v will always remain positive

Answer 5

Kuba Siekierzyński Ohhk👏👏👍👍👌👌