# Wildcard Bits explained

When you are studying Cisco and access-lists you will encounter the so-called Wildcard Bits. Most CCNA students find these very confusing so I’m here to help you and explain to you how they work. Let’s take a look at an example access-list:

``````Router#show access-lists
Standard IP access list 1
10 permit 192.168.1.0, wildcard bits 0.0.0.255
20 permit 192.168.2.0, wildcard bits 0.0.0.255
30 permit 172.16.0.0, wildcard bits 0.0.255.255``````

Access-lists don’t use subnet masks but wildcard bits. This means that in binary, a “0” will be replaced by a “1” and vice versa.

Let me show you some examples:

Subnet mask 255.255.255.0 would be 0.0.0.255 as the wildcard mask. To explain this, I need to show you some binary:

 Bits 128 64 32 16 8 4 2 1 255 1 1 1 1 1 1 1 1

This is the first octet of the subnet mask (255.255.255.0) in binary. As you can see all values have a 1 making the decimal number 255.

 Bits 128 64 32 16 8 4 2 1 0 0 0 0 0 0 0 0 0

This is also the first octet, but now with wildcard bits. If you want the wildcard equivalent, you need to flip the bits. If there’s a 1 you need to change it into a 0. That’s why we now have the decimal number 0.

Let me show you another subnet mask…let’s take 255.255.255.128. What would be the wildcard equivalent of this? We know the 255.255.255.X part, so I’m only showing you the .128 part.

 Bits 128 64 32 16 8 4 2 1 128 1 0 0 0 0 0 0 0

That’s the last octet of our subnet mask. Let’s flip the bits:

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## Forum Replies

1. From your first example,you gave a subnetmask of 255.255.255.0 and you gave the wildcard mask as 0.255.255.255. And from the other explanation in your next example,you 255.255.255.128’s wildcard mask would be 0.0.0.127 of which I understood by your explanation of turning the 1s to 0s. Applying that to the first example of subnetmask 255.255.255.0 with wildcardmask of 0.255.255.255 which you gave, I was thinking the wildcard mask of that should be 0.0.0.255 since the 1s are to be turned to 0s and 0s are to be 1s…pls review it and let me know which is it that is right. Thank you

2. Hi Ali,

You are 100% right, just fixed this typo. Thanks for sharing!

Rene

3. Having trouble with the following example: Not clear why A and C are the correct answers.

A network administrator is configuring ACLs on a Cisco router, to allow traffic from hosts on networks 192.168.146.0, 192.168.147.0, 192.168.148.0, and 192.168.149.0 only. Which two ACL statements, when combined, would you use to accomplish this task? (Choose two)

A. access-list 10 permit ip 192.168.146.0 0.0.1.255
B. access-list 10 permit ip 192.168.147.0 0.0.255.255
C. access-list 10 permit ip 192.168.148.0 0.0.1.255
D. access-list 10 permit ip 192.168.149.0 0.0.255.255

... Continue reading in our forum

4. Having trouble with the following example: Not clear why A and C are the correct answers.

A network administrator is configuring ACLs on a Cisco router, to allow traffic from hosts on networks 192.168.146.0, 192.168.147.0, 192.168.148.0, and 192.168.149.0 only. Which two ACL statements, when combined, would you use to accomplish this task? (Choose two)

A. access-list 10 permit ip 192.168.146.0 0.0.1.255
B. access-list 10 permit ip 192.168.147.0 0.0.255.255
C. access-list 10 permit ip 192.168.148.0 0.0.1.255
D. access-list 10 permit ip 192.168.149.0 0.0.255.255

... Continue reading in our forum

5. Hi Donald,

Let’s do this one in binary so you can see how it works.

146 - 10010010
147 - 10010011

As you can see the first 7 bits are the same, only the 7th bit is different. When a bit is the same we use a 0, when it’s different we use a 1. Our wildcard will be:

00000001 which in decimal is .1 which matches with answer A. We can use the same logic for answer C:

148 - 10010100
149 - 10010101

The first 7 bits are the same, only the 8th bit is different.

I hope this helps!

Rene

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