Regex for an IPv4 address

An IPv4 address is four numbers from 0 to 255 separated by dots. A regex can’t compare numbers, so it spells out which digit strings make 0 to 255, then repeats that piece four times with dots between.

How the octet works

25[0-5]    250 to 255
2[0-4]\d   200 to 249
1\d\d      100 to 199
[1-9]?\d   0 to 99, with no leading zero

(?:…\.){3} takes three octets each followed by a dot, and the last octet stands alone. The groups are non-capturing because nothing needs to be pulled out; if you want the four numbers, split the string on dots after the test passes.

What it accepts

The tester below checks one address per line with g and m; for a single value, drop both flags.

192.168.0.1        matches
10.0.0.255         matches
255.255.255.255    matches
0.0.0.0            matches
256.1.1.1          no: 256 is out of range
192.168.01.1       no: leading zero
1.2.3              no: three parts
1.2.3.4.5          no: five parts

Why not \d{1,3}

The short version, ^(?:\d{1,3}\.){3}\d{1,3}$, also accepts 999.999.999.999. It’s fine for picking addresses out of a log you trust, and wrong for checking input.

Leading zeros are rejected on purpose. Some parsers, including the C library’s inet_aton, read 010 as octal, which is 8, so 192.168.010.1 can mean a different machine from the one the person typed.

What it doesn’t know

It checks the format, not the meaning: it can’t tell a private range such as 10.0.0.0/8 from a public address, and it knows nothing of IPv6. In Node, net.isIPv4(text) does the same check without a regex, and net.isIP(text) returns 4, 6 or 0.

For an address with a prefix length, such as 10.0.0.0/8, add an optional suffix before the final $:

(?:\/(?:3[0-2]|[12]?\d))?

That takes /0 to /32. The CIDR & Subnet Calculator shows the range a prefix covers.