On 05/21/16 14:05, Conrad Meyer wrote:
On Sat, May 21, 2016 at 10:52 AM, Pedro F. Giffuni <p...@freebsd.org> wrote:
Author: pfg
Date: Sat May 21 17:52:44 2016
New Revision: 300377
URL: https://svnweb.freebsd.org/changeset/base/300377
Log:
ndis(4): Avoid overflow.
This is a long standing problem: our random() function returns an
unsigned integer but the rand provided by ndis(4) returns an int.
Scale it down.
MFC after: 2 weeks
Modified:
head/sys/compat/ndis/subr_ntoskrnl.c
Modified: head/sys/compat/ndis/subr_ntoskrnl.c
==============================================================================
--- head/sys/compat/ndis/subr_ntoskrnl.c Sat May 21 17:38:43 2016
(r300376)
+++ head/sys/compat/ndis/subr_ntoskrnl.c Sat May 21 17:52:44 2016
(r300377)
@@ -3189,7 +3189,7 @@ static int
rand(void)
{
- return (random());
+ return (random() / 2 + 1);
}
static void
Won't this still return a negative integer in many cases?
random(9) returns u_long, whereas this rand() routine returns 'int'.
Even on architectures where long is the same size as ordinary
integers, the range of possible results of the 'random() / 2 + 1'
expression, before implicit cast to signed, is [1, 2^31] (inclusive).
According to:
sys/libkern/random.c
The result is uniform on [0, 2^31 - 1].
2^31 is not representable by typical signed 32-bit integers, so this
will wrap to INT_MIN. Also, I'm not sure why zero is excluded from
the range.
It is not a good reason but the zero is sometimes inconvenient: if
the value is going to be used as a multiplier in some calculation
it will basically kill the random component.
On architectures where long is larger than ordinary integers, this
expression has no hope of fitting in the non-negative range of a
signed integer.
Why not instead:
return ((u_int)random() / 2);
TBH, I have seen the same conversion formula over and over and I
just repeated it, so I am glad you are asking the question. Hopefully
some else has a better answer? ;).
Pedro.
Pedro.
Best,
Conrad
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