this post was submitted on 15 Nov 2023
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[–] [email protected] 312 points 1 year ago (45 children)

Ok. This covers every ipv6 and ipv4 address.

"^\s*((([0-9A-Fa-f]{1,4}:){7}([0-9A-Fa-f]{1,4}|:))|(([0-9A-Fa-f]{1,4}:){6}(:[0-9A-Fa-f]{1,4}|((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])(.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])){3})|:))|(([0-9A-Fa-f]{1,4}:){5}(((:[0-9A-Fa-f]{1,4}){1,2})|:((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])(.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])){3})|:))|(([0-9A-Fa-f]{1,4}:){4}(((:[0-9A-Fa-f]{1,4}){1,3})|((:[0-9A-Fa-f]{1,4})?:((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])(.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])){3}))|:))|(([0-9A-Fa-f]{1,4}:){3}(((:[0-9A-Fa-f]{1,4}){1,4})|((:[0-9A-Fa-f]{1,4}){0,2}:((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])(.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])){3}))|:))|(([0-9A-Fa-f]{1,4}:){2}(((:[0-9A-Fa-f]{1,4}){1,5})|((:[0-9A-Fa-f]{1,4}){0,3}:((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])(.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])){3}))|:))|(([0-9A-Fa-f]{1,4}:){1}(((:[0-9A-Fa-f]{1,4}){1,6})|((:[0-9A-Fa-f]{1,4}){0,4}:((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])(.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])){3}))|:))|(:(((:[0-9A-Fa-f]{1,4}){1,7})|((:[0-9A-Fa-f]{1,4}){0,5}:((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])(.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])){3}))|:)))(%.+)?\s*$"

[–] [email protected] 83 points 1 year ago* (last edited 1 year ago) (9 children)

Please don't. Use regex to find something that looks like an IP then build a real parser. This is madness, its's extremely hard to read and a mistake is almost impossible to spot. Not to mention that it's slow.

Just parse [0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3} using regex (for v4) and then have some code check that all the octets are valid (and store the IP as a u32).

[–] [email protected] 11 points 1 year ago (8 children)

And dupe check. 0.0.0.0 and 000.000.000.000 may both be valid, but they resolve the same

[–] [email protected] 3 points 1 year ago* (last edited 1 year ago) (3 children)

Definitely, tho if you store it as a u32 that is fixed magically. Because 1.2.3.4 and 1.02.003.04 both map to the same number.

What I mean by storing it as a u32 is to convert it to a number, similar to how the IP gets sent over the wire, so for v4:

octet[3] | octet[2] << 8 | octet[1] << 16 | octet[0] << 24

or in more human terms:

(fourth octet) + (third octet * 256) + (second octet * 256^2) + (first octet * 256^3)
[–] [email protected] 2 points 1 year ago

True enough for database or dictionary storage, but a lot of times things get implemented in arrays where you still wind up with two copies of the same uint32.

[–] [email protected] 2 points 1 year ago (1 children)

Because 1.2.3.4 and 1.02.003.04 both map to the same number.

But 10.20.30.40 and 010.020.030.040 map to different numbers. It's often best to reject IPv4 addresses with leading zeroes to avoid the decimal vs. octal ambiguity.

[–] [email protected] 2 points 1 year ago (1 children)

I don't know why anyone would write their IPs in octal, but fair point

[–] [email protected] 2 points 1 year ago

It's not about how people write them, it's how parsers parse them. IPv4 has been around since 1982, and most parsers interpret leading zeros as octal.

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