On Sun, 9 Oct 2005, Richard A Downing wrote:
Yes, Google gave me a lot of hits on it too, but I didn't understand a them :-)
I don't understand how to find the connection between the ioctl visible to the program, and the code that gets called in the kernel, I was hoping somebody would jump in. Of course, I'm assuming that the C library (dietlibc, in this case), passes that ioctl to the kernel.
Taking your suggestion, I commented the ioctl out. This produces a working (?) fgetty. However, I now have the problem that the bash that results from logging in has job control off. Starts: -bash: job control disabled in this shell I think this is because it's been started without the tty being the controlling tty.
Yes. I didn't mean you to not call the ioctl, just ignore any error code (or perhaps call 'logger' to log it on syslog if you are up to it).
I also tried mingetty, from which fgetty is derived, and got the same problem - no job control. And gogetty, same thing. Clearly there is something here that I don't yet understand.
Does regular agetty provide job control when linked against dietlibc (if it can be linked against it) ? From memory, dietlibc is by far the smallest of the C libraries, which means some things have to go. Sorry if you are actually using a full dietlibc system, but it isn't clear from this thread.
Ken -- das eine Mal als Tragödie, das andere Mal als Farce
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