On 03/19/2013 10:48:53 PM, Wang Dongsheng-B40534 wrote:

=20
=20

quoted

-----Original Message-----
From: Wood Scott-B07421
Sent: Wednesday, March 20, 2013 6:55 AM
To: Wang Dongsheng-B40534
Cc: Wood Scott-B07421; Gala Kumar-B11780; =20

linuxppc-dev@lists.ozlabs.org;

quoted

Zhao Chenhui-B35336; Li Yang-R58472
Subject: Re: [PATCH 3/3] powerpc/fsl: add MPIC timer wakeup support

On 03/19/2013 01:25:42 AM, Wang Dongsheng-B40534 wrote:

quoted

quoted

-----Original Message-----
From: Wood Scott-B07421
Sent: Tuesday, March 19, 2013 8:31 AM
To: Wang Dongsheng-B40534
Cc: Gala Kumar-B11780; linuxppc-dev@lists.ozlabs.org; Wang

Dongsheng-

quoted

B40534; Zhao Chenhui-B35336; Li Yang-R58472
Subject: Re: [PATCH 3/3] powerpc/fsl: add MPIC timer wakeup =20

support

quoted

quoted

quoted

On 03/08/2013 01:38:47 AM, Wang Dongsheng wrote:

quoted

+static ssize_t fsl_timer_wakeup_store(struct device *dev,
+				struct device_attribute *attr,
+				const char *buf,
+				size_t count)
+{
+	struct timeval interval;
+	int ret;
+
+	interval.tv_usec =3D 0;
+	if (kstrtol(buf, 0, &interval.tv_sec))
+		return -EINVAL;

I don't think the buffer will NUL-terminated...  Ordinarily

there'll be

quoted

an LF terminator, but you can't rely on that (many other sysfs

attributes

quoted

seem to, though...).

I think we don't need to care about LF terminator.
The kstrtol--> _kstrtoull has been done.

My point is, what happens if userspace passes in a buffer that has =20

no

quoted

terminator of any sort?  kstrtol will continue reading beyond the =20

end of

quoted

the buffer.

Do not care about terminator.

kstrtol() obviously *does* because it doesn't take the buffer length as =20
a parameter.

kstrtol--> _kstrtoull--> _parse_integer
=20
_kstrtoull(...) {
	...
	rv =3D _parse_integer(s, base, &_res);
	if (rv & KSTRTOX_OVERFLOW)
		return -ERANGE;
	rv &=3D ~KSTRTOX_OVERFLOW;
	if (rv =3D=3D 0)
		return -EINVAL;
	s +=3D rv;
=20
	if (*s =3D=3D '\n')
		s++;
	if (*s)
		return -EINVAL;
	...
}
=20
_parse_integer(...) {
	...
	while (*s) {
		if ('0' <=3D *s && *s <=3D '9')
			val =3D *s - '0';
		else if ('a' <=3D _tolower(*s) && _tolower(*s) <=3D 'f')
			val =3D _tolower(*s) - 'a' + 10;
		else
			break;	//this will break out to convert.

Really?  How do you know that the next byte after the buffer isn't a =20
valid hex digit?  How do you even know that we won't take a fault =20
accessing it?

quoted

Echoing a nonzero value wouldn't just be to cancel, it would be to =20

set a

quoted

new timer after cancelling the old.

If you think this way is better, I can change.

I do.

But why should do it?
Explicitly stop the timer (echo 0) before reuse it is more reasonable =20
for me.

It's an unnecessary restriction, and eliminating it doesn't make =20
anything simpler.

-Scott=