Why can't I find the value of EOF in C?

Question

I'm reading the book "The C Programming Language" and there is an exercise that asked to verify that the expression getchar() != EOF is returning 1 or 0. Now the original code before I was asked to do that was:

int main()
{
    int c;
    c = getchar();

    while (c != EOF)
    {
        putchar(c);
        c = getchar();
    }  
}

So I thought changing it to:

int main()
{
    int c;
    c = getchar();

    while (c != EOF)
    {
        printf("the value of EOF is: %d", c);
        printf(", and the char you typed was: ");

        putchar(c);
        c = getchar();
    }
}

And the answer in the book is:

int main()
{
  printf("Press a key\n\n");
  printf("The expression getchar() != EOF evaluates to %d\n", getchar() != EOF);
}

Could you please explain to me why my way didn't work?

Answer 1

Because if c is EOF, the while loop terminates (or won't even start, if it is already EOF on the first character typed). The condition for running another iteration of the loop is that c is NOT EOF.

Answer 2

To display the value of EOF

#include <stdio.h>
int main()
{
   printf("EOF on my system is %d\n", EOF);
   return 0;
}

EOF is defined in stdio.h normally as -1