Why does scanf returns control back to the program on pressing Enter key?

Question

I wrote the following program.

void main()
{
   int   *piarrNumber1   = (int *) calloc(1, sizeof(int));
   int   iUserInput      = 0;

   scanf(\"%d\", &iUse         


        

Answer 1

Details are operating system specific (since standard C99 does not know about terminals).

I assume you are using Linux.

First, stdio(3) is buffering the standard input stream and most other FILE* streams. You might try to change that with setvbuf(3), but this affects output buffering only.

More importantly, when stdin (actually the file descriptor used by it, i.e. STDIN_FILENO which is generally the value of fileno(stdin)) is a terminal (see isatty(3) to test that), the linux kernel is usually line-buffering the terminal (so called cooked mode) - at least to handle the backspace key. You might change that by switching the tty to raw mode (as every editor like emacs or vim or nano would do). See this question. But you should reset the cooked mode before your program exits.

So in normal cases, two levels of buffering happen: in the kernel for the line discipline of the terminal, and in the libc for the buffering of stdin

Read the tty demystified page and the Text Terminal HowTo

In practice, if you want sophisticated terminal input, use some library like ncurses or readline (don't bother using just termios)

See also stty(1) & termios(3) & tty_ioctl(4); read about ANSI escape codes.

Notice that this line buferring at two levels (libc and kernel) is specific to ttys. When stdin is a pipe(7) (as in echo foo | yourprogram) or a file (as in yourprogram < yourinputfile.txt) things are different.

In short, ttys are hard to understand, because they mimic complex and arcane hardware devices of the 1950s-1970s era.