From mboxrd@z Thu Jan  1 00:00:00 1970
From: "Nils M Holm" <nmh@t3x.org>
Date: Sat, 27 Jun 2015 23:19:43 +0200
To: Fans of the OS Plan 9 from Bell Labs <9fans@9fans.net>
Message-ID: <20150627211943.GA5035@ananda.local>
References: <20150627203011.GB2121@ananda.local>
	<CAOw7k5gd8NLYukayhHNh_aOAPDc_QEy293mJEqmk2fCJ0rfnJg@mail.gmail.com>
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In-Reply-To: <CAOw7k5gd8NLYukayhHNh_aOAPDc_QEy293mJEqmk2fCJ0rfnJg@mail.gmail.com>
Subject: Re: [9fans] Help with interrupting fgetc()
Topicbox-Message-UUID: 5c4dd558-ead9-11e9-9d60-3106f5b1d025

On 2015-06-27T21:58:47+0100, Charles Forsyth wrote:
> if interrupted, fgetc returns EOF because the underlying read system call
> returns -1 (with error string "interrupted").
> System calls are interrupted by a note (see notify(2)). Your loop will
> therefore stop and exit.
>
> Also, see /sys/doc/comp.ms for some other details of the Plan 9 C
> environment. For instance, main is
>       void main(int, char**);
> and you need an explicit call to exits [sic] at the end of main, not a
> return 0 (or you'll get an error status "main"
> returned to the shell).

Thanks! I updated my code accordingly, and it works fine now. It seems
like a clearerr(stdin) is necessary to restore the original behavior of
fgetc() after an interrupted system call. Would the below code be
acceptable?

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#include <u.h>
#include <libc.h>
#include <stdio.h>

volatile int intr = 0;

void n(void *x, char *s) {
	if (!strcmp(s, "interrupt")) {
		print("oopsie!\n");
		intr = 1;
		noted(NCONT);
	}
	else {
		noted(NDFLT);
	}
}

void main(int argc, char **argv) {
	int	c;

	notify(n);
	c = fgetc(stdin);
	while (c != EOF || intr) {
		if (intr)
			clearerr(stdin);
		else
			fputc(c, stdout);
		intr = 0;
		c = fgetc(stdin);
	}
	exits(NULL);
}
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--
Nils M Holm  < n m h @ t 3 x . o r g >  www.t3x.org