On 6/6/2012 10:31 PM, Dave McGuire wrote:
On 06/06/2012 10:11 PM, Joe Ferraro wrote:
This isn't the thread for it, I realize.... the product slick states
that the switch does BGP, HSRP, OSPF, VRRP, etc.. again, not my area of
expertise... is this still considered a switch?
Hmm, it may actually be ok then. I've not worked with any of the 49xx
series. Supporting those dynamic routing protocols certainly suggests
that it knows how to route! ;)
Hey Brian, have you ever worked with those hybrid router/switch boxen?
Ok, I stand somewhat corrected. IP SERVICES can do tunnels on switches. Interesting.
My 3560g can't do decnet routing, however. (not that i need it to)
Just FYI, the 4948 is a BAD ASS switch.
It's basically a single sup Cat 4500 with a single line card crammed into a 1U case.
When you say hybrid router/switch boxes your talking about the ISR stuff like George was talking about recently, yes?
I now have THREE of those in the forms of an 1811w, and 1841 and the recent acquisition of a 2851.
The ones George has are the bigger ones, the 3000 series I believe.
Interesting beasts.
-brian
On 2012-06-07 18:18, Bob Armstrong wrote:
I was meaning as in non-root.
Ah, well, but that's good because it means kernel hacks are not required.
Right. Basically, you need to interface to ethernet. That you do the same way as my bridge, for instance. All the rest is just massaging the data in the packet, and eventually passing data to the next higher/lower level. The one semi-sensitive thing is that there are timers that needs to be kept, and things that needs to happen at times. But I think they all have such tolerances that you will be good even running it all in userland.
Johnny
On 2012-06-07 18:11, Kari Uusim ki wrote:
On 7.6.2012 5:05, Joe Ferraro wrote:
On Wed, Jun 6, 2012 at 7:13 PM, Sampsa Laine <sampsa at mac.com
<mailto:sampsa at mac.com>> wrote:
I've got a Windows NT 4.0 VM up occasionally too..
Sampsa
boy that gives me bad memories of long days ... perhaps you should've
gone for NT 3.50 while you were at it... I do have a Win 3.11 box
running... perhaps there's a decnet stack for it?!!?
There is a DECnet stack for DOS and Windows in the Pathworks V5 or V6
client kit. Those versions were meant to be used on Windows'es before
Windows NT4. Then the Pathworks32 kit was released and that could be
used on WNT4.0 and Windows 2000. Maybe on XP as well. Don't remember for
sure.
Some pretty recent version of Pathworks32 works on Windows XP anyway.
Johnny
On 7.6.2012 5:05, Joe Ferraro wrote:
On Wed, Jun 6, 2012 at 7:13 PM, Sampsa Laine <sampsa at mac.com
<mailto:sampsa at mac.com>> wrote:
I've got a Windows NT 4.0 VM up occasionally too..
Sampsa
boy that gives me bad memories of long days ... perhaps you should've
gone for NT 3.50 while you were at it... I do have a Win 3.11 box
running... perhaps there's a decnet stack for it?!!?
There is a DECnet stack for DOS and Windows in the Pathworks V5 or V6 client kit. Those versions were meant to be used on Windows'es before Windows NT4. Then the Pathworks32 kit was released and that could be used on WNT4.0 and Windows 2000. Maybe on XP as well. Don't remember for sure.
On 2012-06-07 17:53, Marc Chametzky wrote:
Marc, are you sure ? It means that the assignment gets evaluated as a
boolean expression??
Yes... you've got "port == DPORT" with two equal signs. That's not an
assignment. That is going to evaluate to see whether it's true or false.
Since DPORT = 0 by default, then it's evaluating "port == 0" which is
true, thus its value is 1.
So, you've got a statement which is:
1;
Essentially, it's a no-op.
With DPORT = 0, it doesn't make any difference, save for a compiler
warning. If you were to set DPORT to some other number, though, the
assignment wouldn't take place as would otherwise be expected.
The assignment never took place, as it should have, since there was no assignment.
If you changed DPORT to something else, you'd just get a
if (port == 0) 0;
which is just as meaningless as the result of DPORT was set to 0, which gave
if (port == 0) 1;
Johnny
On 2012-06-07 17:50, hvlems at zonnet.nl wrote:
Marc, are you sure ? It means that the assignment gets evaluated as a boolean expression??
No, he's right.
The *original* code (not what I wrote further down), is (was):
if (port == 0) port == DPORT;
Johnny
-----Original Message-----
From: Marc Chametzky<marc at bluevine.net>
Sender: owner-hecnet at Update.UU.SE
Date: Thu, 07 Jun 2012 11:44:31
To:<hecnet at Update.UU.SE>
Reply-To: hecnet at Update.UU.SESubject: Re: [HECnet] New bridge.c feedback
1171 cc: WARNING File = bridge.c, Line = 684
The indicated expression has no effect.
if (port == 0) port == DPORT;
^
Apart from putting in a conditional?
DPORT is a constant that I intended for users that compile to be able
to change, if they want to.
By default is't zero, thus the compiler sees:
if (port == 0) port = 0;
No, that's not what the compiler is seeing. Instead, it's seeing this:
if (port == 0) 1;
That's because you have "port == DPORT" (a conditional, not an
assignment statement).
which is pretty meaningless, and thus something to warn about. But how
should I change the code to avoid that? I could (of course) add a
check to the if, but that makes it look more complicated, and silly.
Oh well, checking if DPORT have been defined to something non-zero is
a pretty reasonable solution I think.
--Marc
On 2012-06-07 17:44, Marc Chametzky wrote:
1171 cc: WARNING File = bridge.c, Line = 684
The indicated expression has no effect.
if (port == 0) port == DPORT;
^
Apart from putting in a conditional?
DPORT is a constant that I intended for users that compile to be able
to change, if they want to.
By default is't zero, thus the compiler sees:
if (port == 0) port = 0;
No, that's not what the compiler is seeing. Instead, it's seeing this:
if (port == 0) 1;
That's because you have "port == DPORT" (a conditional, not an
assignment statement).
Jeeeez. When the f* did that sneak in there??? I'm way too tired today (obviously). That was totally meaningless. (Maybe just my brain, knowing what it should say just managed to ignore that extra equal sign.)
Thanks for pointing out the obvious. Fixed now.
Johnny
Marc, are you sure ? It means that the assignment gets evaluated as a boolean expression??
Yes... you've got "port == DPORT" with two equal signs. That's not an assignment. That is going to evaluate to see whether it's true or false. Since DPORT = 0 by default, then it's evaluating "port == 0" which is true, thus its value is 1.
So, you've got a statement which is:
1;
Essentially, it's a no-op.
With DPORT = 0, it doesn't make any difference, save for a compiler warning. If you were to set DPORT to some other number, though, the assignment wouldn't take place as would otherwise be expected.
--Marc
