Question

This example is from the K&R book

#include<stdio.h>


main()
{
    long nc;

    nc = 0;
    while(getchar() != EOF)
        ++nc;
    printf("%ld\n", nc);
}

Could you explain me why it works that way. Thanks.

^Z^Z doesn’t work either (unless it’s in the beginning of a line)

Answer 1

Traditional UNIX interpretation of tty EOF character is to make blocking read return after reading whatever is buffered inside a cooked tty line buffer. In the start of a new line, it means read returning 0 (reading zero bytes), and incidentally, 0-sized read is how the end of file condition on ordinary files is detected.

That’s why the first EOF in the middle of a line just forces the beginning of the line to be read, not making C runtime library detect an end of file. Two EOF characters in a row produce 0-sized read, because the second one forces an empty buffer to be read by an application.

$ cat
foo[press ^D]foo <=== after ^D, input printed back before EOL, despite cooked mode. No EOF detected
foo[press ^D]foo[press ^D] <=== after first ^D, input printed back, and on second ^D, cat detects EOF

$ cat
Some first line<CR> <=== input
Some first line <=== the line is read and printed
[press ^D] <=== at line start, ^D forces 0-sized read to happen, cat detects EOF

I assume that your C runtime library imitates the semantics described above (there is no special handling of ^Z at the level of kernel32 calls, let alone system calls, on Windows). That’s why it would probably detect EOF after ^Z^Z even in the middle of an input line.