Japt, 15 bytes

Port of m90's JS solution, until I manage to come up with something shorter.

;Bg7^Mm8|#ÿ^3îU

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posted 29 days ago

CC BY-SA 4.0

27d ago

Shaggy‭

2131 reputation 4 116 216 166

JavaScript, 43 bytes

s=>"ABCDE"[Math.log2(s.slice(0,3)^255|8)^7]

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Taking the first 3 characters of the string is always enough to include the first octet, while including at most one full stop.

Before the bitwise XOR, the substring is converted to a Number (if a full stop is present, it is interpreted as a decimal point) and then to a 32-bit integer (rounding towards zero), thus becoming the value of the first octet.

XORing with 255 produces 128–255 for class A, 64–127 for class B, 32–63 for class C, 16–31 for class D, and 0–15 for class E.

ORing with 8 narrows those ranges; in particular, class E is reduced to 8–15, which fits the pattern of the original ranges for the other classes.

Taking the base-2 logarithm, and (implicitly) rounding towards zero, produces 7, 6, 5, 4, 3 for classes A, B, C, D, E; XORing with 7 changes those to 0, 1, 2, 3, 4, which is finally used as an index into the string "ABCDE" to produce the result.

posted almost 3 years ago

CC BY-SA 4.0

m90‭

271 reputation 0 10 27 1

Japt, 26 25 bytes

;Bg4mUqL În10,2 ùT8 qT Îl

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Long and unwieldy, I'm rusty. Looks at the first 0 in the bitstring of the first octet.

posted almost 3 years ago

CC BY-SA 4.0

3y ago

Quintec‭

2306 reputation 17 46 242 48

Python 3, 56 bytes

lambda x:chr(65+f'{int(float(x[:3]))&~8:08b}'.find('0'))

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Takes in the ip address as a string and returns the classification letter.

Explanation:

• x[:3] extracts the first 3 characters
• int(float(…)) converts str to int, passing through float to handle cases like 1.2.3.4 -> 1.2 -> 1
• …&~8 sets the 8 bit (fourth to last) of the integer to 0
• f'{…:08b}' converts the integer to a binary string, left-padded to 8 characters (e.g. 192 becomes '11000000' and 117 becomes '01110101')
• ….find('0') finds the index of the first '0' character
• 65+… adds the character code for 'A'
• chr(…) converts the sum from the previous step to a character

posted almost 3 years ago

CC BY-SA 4.0

3y ago

dino‭

121 reputation 0 10 12 31

Vyxal, 24 bytes

\.€hEb:L8-0wẋf$J0ḟkA5Ẏ$i

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This is absolutely abysmal but whatever. A lot of bytes are taken up by trying to pad the list with zeroes (ẋ was buggy or something so molding to 0 8ẋ didn't work).

Explanation:

• \.€ splits on . and h gets the first of these part/octet
• E turns that into an integer and b turns it into binary (list of 0/1)
• : duplicates that list on the stack
  • The first copy is used to create a list of extra 0s to pad
    • L gets the length of the list and 8- subtracts 8 from it
    • 0w is the singleton list [0]
    • ẋ then repeats the singleton list 8 - len(binary) times
    • f flattens that list to get [0, 0, 0, 0, 0, 0, 0, 0]
  • $ swaps this pad and the original list and J prepends the pad to the original
• 0ḟ finds the index of the first 0
• kA is the uppercase alphabet and 5Ẏ gives the first 5 letters (ABCDE), our classes
• $ swaps before i indexes into the string of classes using the index of 0

posted almost 3 years ago

CC BY-SA 4.0

3y ago

user‭

1546 reputation 9 59 174 89

C (gcc), 64 62 bytes

Saved 2 bytes thanks to m90 in the comments.

i;f(char*s){i=atoi(s)>>4;return'A'+(i&8?i&4?i&2?i%2+3:2:1:0);}

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posted almost 3 years ago

CC BY-SA 4.0

3y ago

celtschk‭

1364 reputation 28 48 176 73

J, 46 char

'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.',

Sample runs

    'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.', '127.255.255.255'
A
    'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.', '191.255.255.255'
B
    'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.', '223.255.255.255'
C
    'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.', '239.255.255.255'
D
    'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.', '255.255.255.255'
E

posted over 2 years ago

CC0

torres‭

101 reputation 0 17 10 28