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On Sun, 17 Nov 2002 14:33:54 -0500, Michael H. Warfield wrote:

> > > > Actually that is a class B address.
> > > >
> > > > The first octet of a class A is 1-126 (127 reserved for loop
> > > > back)
> > > >                      class B is 128-191
> > > >                    class C is 192-223
> > > >
> > > > since 172 is between the ranges of 128-191 that would make it
> > > > class B
> > > >
> > > > Class B subnet 255.255.0.0 or /16
> > >
> > > The step from Class B to /16 is beyond me. If memory serves
> > > correctly, the Class B subnet in RFC1918 is 172.16.0.0/12
> > > which would be netmask 255.240.0.0.
> 
> > I want to know how do you know the the netmask is 255.240.0.0 ?
> 
>       He doesn't.  He's wrong.  An old Class B was a /16 by
>       definition.
> What he's quoting is the definition for the 16 Class B addresses which
> were specified for private address space usage.  The formed a /12
> band. He has not answered the original question and has provided
> erronious information about the old Class B addresses based on a
> missreading an missunderstanding of the Private Address Space RFC
> 1918.

Wow, what a nice way of replying to a message where I explicitly
wrote "if memory serves correctly". ;) Consider helping more often
instead of lurking in the background until someone makes a small
mistake. It's so easy to dig out RFC1918 and look up the section
which mentions the 16 contiguous class B network numbers 172.16.0.0
- - 172.31.255.255 in Class B (128.0.0.0 - 191.255.255.255). Bundled
together they still don't have a netmask of 255.255.0.0, but
255.240.0.0.

Btw, you didn't answer the question either. It was:

> What is the subnet-mask of IP "172.16.0.1" - "172.16.0.253"

Which would be impossible, because it is missing 172.16.0.254.
Leaving that minor detail aside, the answer would be a netmask of
255.255.255.0, because actually it is an 8-bit subnet of class B.

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