Match an IPv4 octet

A pattern for an IP address needs a piece that accepts 0 to 255 and nothing above, and [0-9]{1,3} accepts 999.

import re

octet = r"(25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])"
rx = re.compile(rf"{octet}")
ok = [n for n in range(0, 300) if rx.fullmatch(str(n))]
print(len(ok), ok[0], ok[-1])
print(bool(rx.fullmatch("007")), bool(rx.fullmatch("256")))

ip = re.compile(rf"{octet}(\.{octet}){{3}}")
for s in ["10.0.0.1", "255.255.255.255", "256.1.1.1", "1.2.3", "01.2.3.4"]:
    print(f"{s:16} {bool(ip.fullmatch(s))}")

256 0 255
False False
10.0.0.1         True
255.255.255.255  True
256.1.1.1        False
1.2.3            False
01.2.3.4         False

The three alternatives cover 250 to 255, 200 to 249 and 100 to 199, and the last one covers 0 to 99 with an optional tens digit. Python's fullmatch anchors the pattern, as in the re documentation. The count of 256 matches among the numbers up to 299 shows that nothing above 255 is accepted, and the second pattern repeats the octet three more times with dots.

Order matters when the pattern is not anchored: with the short form [1-9]?[0-9] first, a search in 255 would find only 25. The pattern rejects leading zeros such as 007, which some tools read as octal. For anything beyond a quick filter use the ipaddress module in Python or inet_pton in C, which handle IPv6 and edge cases.