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Exam 200-201 topic 1 question 163 discussion

Actual exam question from Cisco's 200-201
Question #: 163
Topic #: 1
[All 200-201 Questions]

Which regular expression is needed to capture the IP address 192.168.20.232?

  • A. ^(?:[0-9]{1,3}\.){3}[0-9]{1,3}
  • B. ^(?:[0-9]{1,3}\.)*
  • C. ^)?:[0-9]{1,3}\.){1,4}
  • D. ^([0-9].{3})
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Suggested Answer: A 🗳️

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Pwned
Highly Voted 3 years ago
A is correct... but this is the correct form ^(?:[0-9]{1,3}\.){3}[0-9]{1,3}
upvoted 12 times
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anonymous1966
Highly Voted 2 years, 3 months ago
Selected Answer: A
Correct options are: A) ^(?:[0-9]{1,3}.){3}[0-9]{1,3} B) ^(?:[0-9]f1,3}.){1,4} C) ^(?:[0-9]{1,3}.) D) ^([0-9]-{3}) Test and verify the correct option A at https://regex101.com/
upvoted 8 times
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RoBery
Most Recent 9 months, 3 weeks ago
A. ^(?:[0-9]{1,3}\.){3}[0-9]{1,3} Explanation: ^: Asserts the start of the string. (?:[0-9]{1,3}\.){3}: This non-capturing group matches three occurrences of digits 1 to 3 in length followed by a dot (.), representing the first three octets. [0-9]{1,3}: Matches the fourth octet, consisting of digits 1 to 3 in length. $: Asserts the end of the string. This regular expression is designed to match the pattern of an IPv4 address and captures the IP address 192.168.20.232.
upvoted 1 times
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Eng_ahmedyoussef
2 years ago
Selected Answer: A
A. ^(?:[0-9]{1,3}\.){3}[0-9]{1,3}
upvoted 2 times
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