EOF adds to counter, no idea why

Question

I have been stuck trying to understand why triggering eof using ctrl-D adds to a counter in a for loop.

Here is my code:

#include <stdio.h>
int main()
{
    double nc;
    for (nc = 0; getchar() != EOF; nc++){
       getchar();
    }
    printf("%.0f\n", nc);
    return 0;
}

My outcome is :

0
1
2
3
4
5
6
7
8

The 8 is what's given to me when I use ctrl-D after inputting 7. Is there a reason why triggering the eof causes the code to run another complete loop? I thought an empty buffer will return nothing.

Answer 1

Here is something you might miss. Take piece of your code:

for (nc = 0; getchar() != EOF; nc++)
    getchar();

There are two getchar(). Let's call them, getchar1() and getchar2(). Your input should be like this:

0\n
.
.
.
7\n
EOF

getchar1() catch the sequence of digits and EOF. getchar2() always catch the newline ('\n'). And the count of you go through for loop body is 8 (0 to 7).

Hope it helpful for you.