C Programming (Beginner help) relating to getchar() putchar()

Question

Im new to C programming and wanted to clear, what may seem to be, a silly doubt...

Code:

#define EOF 0
main()
{ 
    int c;

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

this should just return the value I enter...but accordingly...shouldnt it terminate when I enter 0? If not...what exactly does the statement under 'while' signify? Any help would be greatly appreciated guys :)

Answer 1

The getchar function returns the ASCII value entered (48 for zero, or '0'), or a value called EOF in the header file <stdio.h> (normally -1).

So if you want to stop either on EOF (the proper EOF, not the one you defined) or if the user writes a zero, then this will work much better:

#include <stdio.h>

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

    while (c != EOF && c != '0')
    {
        putchar(c);
        c = getchar();
    }

    return 0;
}

Answer 2

EOF means when you hit, for instance, ctrl + d in linux, it sends EOF signal, it's not necessarily a 0.

Answer 3

EOFis certainly not 0 (nor is it '0', the char literal equaling 48). It's not char at all. stdio.h (which you should be #include-ing!) defines it as -1. It indicates that whatever file you are reading has no more data in it. It's not a byte in the stream, rather the way that the io library indicates that it has finished.

The whole reason getchar needs to return an int is so that the end-of-file indicator cannot possibly be a valid byte in the file.

Other functions indicate that the EOF has been reached differently: fgets() returns NULL. read() which returns the number of bytes read returns 0, indicating that it could read nothing.

Answer 4

When you enter a 0, it's not actually a zero, just the ASCII character 0. It has a numerical value of 48, which can be verified by printf("%d\n", '0');, which prints 48. An EOF is an actual zero, that is, it's representation is essentially zero.