[PATCH] i386: fix infinite loop with singlestep int80 syscalls

[PATCH] i386: fix infinite loop with singlestep int80 syscalls

The commit 635cf99a80f4ebee59d70eb64bb85ce829e4591f introduced a
regression.  Executing a ptrace single step after certain int80
accesses will infinitely loop and never advance the PC.

The TIF_SINGLESTEP check should be done on the return from the syscall
and not before it.

The new test case is below:

/* Test whether singlestep through an int80 syscall works.
 */
#define _GNU_SOURCE
#include <stdio.h>
#include <unistd.h>
#include <fcntl.h>
#include <sys/ptrace.h>
#include <sys/wait.h>
#include <sys/mman.h>
#include <asm/user.h>
#include <string.h>

static int child, status;
static struct user_regs_struct regs;

static void do_child()
{
	char str[80] = "child: int80 test\n";

	ptrace(PTRACE_TRACEME, 0, 0, 0);
	kill(getpid(), SIGUSR1);
	write(fileno(stdout),str,strlen(str));
	asm ("int $0x80" : : "a" (20)); /* getpid */
}

static void do_parent()
{
	unsigned long eip, expected = 0;
again:
	waitpid(child, &status, 0);
	if (WIFEXITED(status) || WIFSIGNALED(status))
		return;

	if (WIFSTOPPED(status)) {
		ptrace(PTRACE_GETREGS, child, 0, &regs);
		eip = regs.eip;
		if (expected)
			fprintf(stderr, "child stop @ %08lx, expected %08lx %s\n",
					eip, expected,
					eip == expected ? "" : " <== ERROR");

		if (*(unsigned short *)eip == 0x80cd) {
			fprintf(stderr, "int 0x80 at %08x\n", (unsigned int)eip);
			expected = eip + 2;
		} else
			expected = 0;

		ptrace(PTRACE_SINGLESTEP, child, NULL, NULL);
	}
	goto again;
}

int main(int argc, char * const argv[])
{
	child = fork();
	if (child)
		do_parent();
	else
		do_child();
	return 0;
}

Signed-off-by: Jason Wessel <jason.wessel@windriver.com>
Cc: Jeremy Fitzhardinge <jeremy@goop.org>
Cc: Chuck Ebbert <cebbert@redhat.com>
Signed-off-by: Chris Wright <chrisw@sous-sol.org>
Signed-off-by: Greg Kroah-Hartman <gregkh@suse.de>