Usage of int instead of char

Question

I've got a copy of the book "The C Programming Language", and in one of the first chapters it shows a short example of a code that copies characters from stdin to stdout. Here is the code:

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

The writer then said, he used int instead of char because char isn't big enough to hold EOF. But when I tried it with char, it worked in the same way, even with ctrl+z. Stack says it’s a duplicate, so I ask shortly:

Why using ‘char’ is wrong?

Answer 1

If you will write for example

char c;
c = getchar();
while (c != EOF) {
//...

when in the condition c != EOF the value of the object c is promoted to the type int and two integers are compared.

The problem with declaring the variable c as having the type char is that the type char can behave either as the type signed char or unsigned char (depending on a compiler option). If the type char behaves as the type unsigned char then the expression c != EOF will always evaluate to logical true.

Pay attention to that according to the C Standard EOF is defined the following way

EOF which expands to an integer constant expression, with type int and a negative value, that is returned by several functions to indicate end-of-file, that is, no more input from a stream;

So after this assignment

c = getchar();

when c is declared as having the type char and the type char behaves as the type unsigned char then the value of c will be a positive value after the integer promotion and hence will not be equal to a negative value.

To simulate the situation just declare the variable c as having the type unsigned char

unsigned char c;