Why two EOF needed as input?

Question

When I run the code below, I use three inputs (in Ubuntu terminal):

1. abc(Ctrl+D)(Ctrl+D)
2. abc(Ctrl+D)(Enter)(Ctrl+D)
3. abc(Enter)(Ctrl+D)

The code reacts well in all cases. My question is: why in 1) and 2) I need two EOF?

#include <iostream>

int main()
{
  int character;

  while((character=std::cin.get())!=EOF){}

  std::cout << std::endl << character << std::endl;

}

Answer 1

That's how the "EOF" character works (in "canonical" mode input, which is the default). It's actually never sent to the application, so it would be more accurate to call it the EOF signal.

The EOF character (normally Ctrl-D) causes the current line to be returned to the application program immediately. That's very similar to the behaviour of the EOL character (Enter), but unlike EOL, the EOF character is not included in the line.

If the EOF character is typed at the beginning of a line, then zero bytes are returned to the application program (since the EOF character is not sent). But if a read system call returns 0 bytes, that is considered an end-of-file indication. So at the beginning of a line, an EOF will be treated as terminating input; anywhere else, it will merely terminate the line and so you need two of them to terminate input.

For more details, see the .Posix terminal interface specification.

Answer 2

You don't have "two EOF". Bash is putting the tty in raw mode, and interpreting ^D differently depending on context. If you type ^D after a newline, bash closes the input stream on the foreground process. If you type a few characters first, bash requires you to type ^D twice before doing so. (The first ^D is treated like 'delete')