IPv4 address structure in Computer Networks - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to show the number of bits in an IPv4 address.

Computer Networks

ipv4_bits = [1]

Drag options to blanks, or click blank then click option'

A128

B64

C32

D16

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing IPv4 with IPv6 bit length.

Choosing 64 or 128 bits which are for other protocols.

An IPv4 address is made up of 32 bits, which are divided into four 8-bit sections called octets.

2fill in blank

medium

Complete the code to represent an IPv4 address as four decimal numbers separated by dots.

Computer Networks

ipv4_address = "[1].168.1.1"

Drag options to blanks, or click blank then click option'

A192

B256

C300

D500

Attempts:

3 left

💡 Hint

Common Mistakes

Using numbers greater than 255 in an IPv4 address.

Confusing decimal and binary representations.

Each part of an IPv4 address is a decimal number between 0 and 255. 192 is a common starting number for private networks.

3fill in blank

hard

Fix the error in the code that tries to split an IPv4 address into its parts.

Computer Networks

parts = ipv4_address.[1](".")

Drag options to blanks, or click blank then click option'

Ajoin

Bsplit

Creplace

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using join which combines strings instead of splitting.

Using replace which changes characters but does not split.

The split method breaks a string into parts based on a separator. Here, it splits the IPv4 address into its four parts using the dot.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each octet string to its integer value.

Computer Networks

octet_values = {octet: int([1]) for octet in ipv4_address.[2](".")}

Drag options to blanks, or click blank then click option'

Aoctet

Bipv4_address

Csplit

Djoin

Attempts:

3 left

💡 Hint

Common Mistakes

Using join instead of split.

Trying to convert the whole address string instead of each octet.

The comprehension converts each octet string to an integer. The split method divides the address into octets, and int(octet) converts each to a number.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each octet to True if it is greater than 127, otherwise False.

Computer Networks

octet_flags = {octet: int(octet) [1] 127 for octet in ipv4_address.[2]("[3]")}

Drag options to blanks, or click blank then click option'

A>

Bsplit

C.

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' for the comparison.

Using wrong separator in split method.

This comprehension checks if each octet number is greater than 127. The split method divides the address by dots, and the comparison operator > tests the value.

Practice

1. What is the correct structure of an IPv4 address?

easy

A. Three numbers separated by dots, each from 0 to 255

B. Six numbers separated by colons, each from 0 to 65535

C. Four numbers separated by dots, each from 0 to 255

D. Four numbers separated by commas, each from 0 to 255

Solution

1. Step 1: Understand IPv4 format

   An IPv4 address consists of four parts called octets, separated by dots.

2. Step 2: Check the range of each octet

   Each octet ranges from 0 to 255, which fits in 8 bits.

3. Final Answer:

   Four numbers separated by dots, each from 0 to 255 -> Option C

4. Quick Check:

   IPv4 = 4 octets (0-255) [OK]

Hint: IPv4 always has 4 dot-separated numbers 0-255 [OK]

Common Mistakes:

• Confusing IPv4 with IPv6 format
• Using commas instead of dots
• Incorrect number of parts

2. Which of the following is a valid IPv4 address?

easy

A. 10.0.0.1

B. 192.168.1.256

C. 172.16.300.5

D. 255.255.255

Solution

1. Step 1: Check each octet's value

   Valid IPv4 octets must be between 0 and 255 inclusive.

2. Step 2: Validate each option

   A has all octets 0-255 and 4 parts (valid). B: 192.168.1.256 has 256 (invalid), C has 300 (invalid), D has only 3 parts (invalid). Only A is valid.

3. Final Answer:

   10.0.0.1 -> Option A

4. Quick Check:

   All octets 0-255 and 4 parts = valid IPv4 [OK]

Hint: Check each number is 0-255 and count 4 parts [OK]

Common Mistakes:

• Allowing numbers above 255
• Ignoring missing octets
• Confusing IPv4 with IPv6

3. What is the decimal value of the first octet in the IPv4 address 11000000.10101000.00000001.00000001?

medium

A. 255

B. 168

C. 1

D. 192

Solution

1. Step 1: Convert first octet from binary to decimal

   Binary 11000000 equals 128+64=192.

2. Step 2: Confirm decimal value

   The first octet decimal value is 192.

3. Final Answer:

   192 -> Option D

4. Quick Check:

   11000000 binary = 192 decimal [OK]

Hint: Convert binary octet to decimal by adding powers of 2 [OK]

Common Mistakes:

• Misreading binary digits
• Adding wrong powers of two
• Confusing octet positions

4. Identify the error in this IPv4 address: 256.100.50.25

medium

A. Too many octets

B. The first octet is out of range

C. Octets must be separated by commas

D. The last octet is missing

Solution

1. Step 1: Check each octet's value range

   IPv4 octets must be between 0 and 255 inclusive.

2. Step 2: Identify invalid octet

   The first octet is 256, which is outside the valid range.

3. Final Answer:

   The first octet is out of range -> Option B

4. Quick Check:

   Octet > 255 means invalid IPv4 [OK]

Hint: Octets must be 0-255; 256 is invalid [OK]

Common Mistakes:

• Thinking 256 is valid
• Confusing separators
• Ignoring octet count

5. You need to assign IP addresses to 300 devices in a network. Which of these IPv4 address ranges can you use without running out of addresses?

hard

A. 172.16.0.0 to 172.16.1.255

B. 10.0.0.0 to 10.0.0.127

C. 192.168.1.0 to 192.168.1.255

D. 192.168.0.0 to 192.168.0.100

Solution

1. Step 1: Calculate number of addresses in each range

   Range A: 256 addresses (0-255), B: 128 addresses, C: 512 addresses (two full /24 blocks), D: 101 addresses.

2. Step 2: Compare with required 300 devices

   Only range C has enough addresses (512) to cover 300 devices.

3. Final Answer:

   172.16.0.0 to 172.16.1.255 -> Option A

4. Quick Check:

   More than 300 addresses needed, only C fits [OK]

Hint: Count addresses: 256 per block; need ≥300 [OK]

Common Mistakes:

• Assuming 256 addresses cover 300 devices
• Ignoring address range size
• Confusing IP blocks