Question

Quoting from Kernighan and Ritchie’s ‘The C Programming Language’ Page 16 –

#include<stdio.h>

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

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

getchar();
return 0;
}

“The type char is specifically meant for storing such character
data, but any integer type can be used. We used int for a subtle but
important reason. The problem is distinguishing the end of the input
from valid data. The solution is that getchar returns a distinctive
value when there is no more input, a value that cannot be confused
with any real character. This value is called EOF, for “end of
file”. We must declare c to be a type big enough to hold any value
that getchar returns. We can’t use char since c must be big
enough to hold EOF in addition to any possible char. Therefore we
use int.”.

I looked up in stdio.h, it says #define EOF (-1)

The book conclusively states that char cannot be used whereas this program “works just fine” (See EDIT) with c as char data type as well. What is going on? Can anyone explain in terms of bits and signed values?

EDIT:
As Oli mentioned in the answer, the program cannot distinguish between EOF and 255. So it will not work fine. I want to know what’s happening – Are you saying that when we do the comparison c!=EOF, the EOF value gets cast to a char value = 255 (11111111 in binary; i.e. the bits 0 through 7 of EOF when written in 2’s complement notation)?

Answer 1

Your program doesn’t work fine; it won’t be able to distinguish between EOF and 255.

The reason it appears to work correctly is because char is probably signed on your platform, so it’s still capable of representing -1.