svn commit: r300377 - head/sys/compat/ndis

[Pedro Giffuni]

On 05/21/16 14:05, Conrad Meyer wrote:
> On Sat, May 21, 2016 at 10:52 AM, Pedro F. Giffuni <pfg at 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
>