> From: Peter Hurley > [ +Jiri Slaby who doesn't read linux-serial ;) ] > > On Thu, 2013-03-21 at 23:32 +0000, Craig McQueen wrote: > > > > > From: Peter Hurley [mailto:peter@xxxxxxxxxxxxxxxxxx] On Thu, > > > 2013-03-21 at 20:38 +0000, Grant Edwards wrote: > > > > On 2013-03-21, Craig McQueen > > > > <craig.mcqueen@xxxxxxxxxxxxxxxxxxxxxx> > > > wrote: > > > > > It sounds as though people have done pseudo-ttys with HW > > > handshaking > > > > > support--eg tty0tty project. However I'd rather implement this > > > > > function in the kernel pseudo-terminal driver itself. Is there > > > > > any reason not to do that? > > > > > > > > No reason other than you and I are the only two people who care > > > > about it. :) > > > > > > Assuming you're leaning toward an in-kernel solution, why not just > > > implement a new tty driver that behaves like a local serial port? > > > > The pseudo-tty already provides most of the functionality I want, so > I > > don't want to reinvent the wheel. I want to use it to simulate a > modem > > device. Various other programs could benefit from an enhanced > > pseudo-tty, so they also don't have to implement their own kernel > > drivers--e.g.: > > I should have been more specific: I didn't mean necessarily start from > scratch. As a starting point you could just dup pty.c, rip out the BSD > legacy support, and rename the driver/tty device base names. > > Whatever that was would behave just like ptm/pts. I'm a little fuzzy about this... If I do this, how would userland programs create pty master/slave device pairs? Could it work with the API of the UNIX 98 style pseudo-tty in 'man 7 pty'? That is: posix_openpt() grantpt() unlockpt() ptsname() I see posix_openpt(flags) is essentially equivalent to open("/dev/ptmx", flags), so maybe if I made my own driver, posix_openpt(flags) would be replaced by open("/dev/my-driver-ptmx", flags), and the other function calls could stay the same. Is that right? Thanks, Craig McQueen ��.n��������+%������w��{.n�����{��ǫ����{ay� ʇڙ���f���h������_�(�階�ݢj"��������G����?���&��
