As you should be aware in the wildcard mask a binary 0 is a bit that must match and a binary 1 is a bit that may vary.
So in your mask of 0.0.0.32 the first 3 octets must match exactly (it must be 10.1.1). The mask of the fourth octet has a single binary 1. It may help to write out the 4 octet in binary ( 00100000). So for this mask there are exactly 2 values of the address that will match the mask. These values are 10.1.1.0 and 10.1.1.32.
As your comment indicates this mask is quite unusual in an access list. It is much more common to have the mask be the inverse of common subnet masks (such as the mask .31 which is the inverse of mask 224).
We are pleased to announce availability of Beta software for 16.6.3.
16.6.3 will be the second rebuild on the 16.6 release train targeted
towards Catalyst 9500/9400/9300/3850/3650 switching platforms. We are
looking for early feedback from customers befor...
Introduction Featured Speakers Luis Espejel is the Telecommunications
Manager of IENova, an Oil & Gas company. Currently he works with Cisco
IOS® and Cisco IOS XE platforms, and NX to some extent. He has also
worked as a Senior Engineer with the Routing P...
In this session you can learn more about Layer 3 multicast and the best
practices to identify possible threats and take security measures. It
provides an overview of basic multicast, the best security practices for
use of this technology, and recommendati...