Why the output of this code is 9?

Question

int n,f,s;
n=18;
while (n>0) {
f=n%10;
n=n/10;
s+=f;
}
cout<<s;

Answer 1

Debugging tip for beginners: Don't give yourself a headache! Just print each cycle of the loop like so

#include <iostream>
using namespace std;

int main()
{
int n,f,s = 0;
n=18;
while (n > 0) {
cout << "Cycle: " << n << ", ";
f = n % 10;
cout << f << ", ";
n = n / 10;
cout << n << ", ";
s += f;
cout << s << "\n";
}
cout << s;
}

Output:
Cycle: 18, 8, 1, 8
Cycle: 1, 1, 0, 9
9

Answer 2

● After 1st cycle of loop : f=8 , n=1 (as int/int gives int & decimal part is removed) , s=8(adding value of f to it)
● After 2nd cycle of loop : f=1 , n=0 , s=9 (adding f to 8)

Answer 3

first iteration (while)

f = 8
n = 1
s = 8

second (last) iteration:

f = 1
n = 0
s = 1 + 8 or s = 9

remember % takes the remainder of int division.

and s+=f
have the same effect compared with
s = s + f

Answer 4

While loop runs 2 times

For first time:
n=18
f=n%10 = 18%10 = 8
n=n/10 = 18/10 = 1
s=s+f = 0+8 = 8

For the second time:
n=1
f=n%10 = 1%10 = 1
n=n/10 =1/10 = 0
s=s+f = 8+1 = 9

Answer 5

The code op posted can potentially cause problems if s isn't initialised to zero. Since you're adding something to s, and it isn't initialised, it's going to use the value stored at the 4 bytes it is given, which is going to ultimately return some garbage value.

Answer 6

first loop :
f=18%10=8
n=18/10=1
s=0+8=8
s=8 store
second loop:
1>0 true
f=1%10=1
n=1/10=0
s=8+1=9. s=9 stored
third loop:
0>0 false

output:
9

Answer 7

Because s != 0! s it’s nil(or null, I don’t know). It’s right code:
#include <iostream>
using namespace std;
int main()
{
int n=18,f,s=0;
while(n>0)
{
f=n%10;
n/=10;
s+=f;
}
cout<<s;
}