29
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Challenge

Write a program or function that converts a 32 bit binary number to its quad-dotted decimal notation (often used for representing IPv4)

Quad-dotted decimal

A quad-dotted decimal is formed like so:

  1. Split the binary representation into its 4 individual bytes
  2. Convert each byte into denary
  3. place "." between each number

Example:

input:  10001011111100010111110001111110
step 1: 10001011 11110001 01111100 01111110
step 2:    139      241      124      126
step 3: 139.241.124.126

I/O examples

input --> output

10001011111100010111110001111110 --> 139.241.124.126
00000000000000000000000000000000 --> 0.0.0.0
01111111000000000000000000000001 --> 127.0.0.1
11000000101010000000000111111111 --> 192.168.1.255

Rules

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  • 6
    \$\begingroup\$ Input will always be a 32 bit binary number but your examples are binary lists. Is that also an acceptable input form? \$\endgroup\$ – Adám Jan 10 at 12:49
  • 2
    \$\begingroup\$ Is it allowed to use a string as input instead of a list of binary digits? \$\endgroup\$ – Galen Ivanov Jan 10 at 13:31
  • \$\begingroup\$ I have added a brief description of how the qaud-dotted notation is formed @LuisMendo. \$\endgroup\$ – mabel Jan 10 at 13:31
  • \$\begingroup\$ @Adám this is actually how I meant for the challenge to be interpreted, but I guess I couldn't find the right word. I have added it as an acceptable input form, but left the 32 bit binary number as valid input also. \$\endgroup\$ – mabel Jan 10 at 13:33
  • 2
    \$\begingroup\$ I'm not clear, can we take in an actual number as input, like 2130706433 to give 127.0.0.1? \$\endgroup\$ – xnor Jan 11 at 4:00

43 Answers 43

10
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05AB1E, 6 bytes

4äC'.ý

Try it online or verify all test cases.

Explanation:

4ä       # Convert the (implicit) input-string into 4 equal-sized parts
  C      # Convert each part from binary to an integer
   '.ý  '# Join this list by "."
         # (after which the result is output implicitly)
| improve this answer | |
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19
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x86-16 machine code, IBM PC DOS, 54 47 45 bytes

Binary:

00000000: be82 00b3 04b1 08ac d0d8 d0d4 e2f9 8ac4  ................
00000010: 41d4 0a50 8ac4 84c0 75f6 580c 30b4 0ecd  A..P....u.X.0...
00000020: 10e2 f74b 7406 b02e cd10 ebd9 c3         ...Kt........

Build and test BIN2IP.COM using xxd -r from above.

Unassembled listing:

BE 0082     MOV  SI, 82H        ; command line input address
B3 04       MOV  BL, 4          ; loop 4 bytes 
        BYTE_LOOP:
B1 08       MOV  CL, 8          ; loop 8 bits 
        BIT_LOOP: 
AC          LODSB               ; load next bit char into AL 
D0 D8       RCR  AL, 1          ; put LSB of char into CF 
D0 D4       RCL  AH, 1          ; put CF into LSB of byte value, then shift left
E2 F9       LOOP BIT_LOOP       ; continue bit loop 
8A C4       MOV  AL, AH         ; put byte result into AL
        GET_DIGIT: 
D4 0A       AAM                 ; byte divide by 10, AH = AL / 10, AL = AL % 10 
50          PUSH AX             ; save remainder in AL on stack 
8A C4       MOV  AL, AH         ; put quotient back into AL 
41          INC  CX             ; increment decimal digit count 
D4 C0       TEST AL, AL         ; quotient = 0? 
75 F6       JNZ  GET_DIGIT      ; if not, continue looping 
        PRINT_DIGIT: 
58          POP  AX             ; restore digit in AL 
0C 30       OR   AL, '0'        ; ASCII convert              
B4 0E       MOV  AH, 0EH        ; BIOS write char function   
CD 10       INT  10H            ; write to console 
E2 F7       LOOP PRINT_DIGIT    ; loop until done 
4B          DEC  BX             ; is last byte? 
74 06       JZ   END_LOOP       ; if so, don't display a '.' 
B0 2E       MOV  AL, '.'        ; otherwise display '.'
CD 10       INT  10H            ; write to console 
        END_LOOP:
75 D7       JNZ  BYTE_LOOP      ; continue byte loop 
C3          RET                 ; exit to DOS 

Output:

enter image description here

A standalone PC DOS executable. Input is command line, output to console.

Notes:

The "interesting part" (converting binary string to bytes) is about 15 bytes, whereas the rest of code is writing the itoa() function to convert binary bytes into a decimal string representation for display.

  • -2 bytes eliminating unnecessary PUSH/POP thx to @PeterCordes!
| improve this answer | |
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  • \$\begingroup\$ dec cx / jz end_loop could be re-arranged using a loop instruction that either falls through to a ret or jumps forward to mov al/int 10h/jmp byte_loop, for a net saving of I think 1 byte, from replacing dec cx/jcc with loop. Or maybe use DI for that loop counter instead of CX, avoiding push/pop? \$\endgroup\$ – Peter Cordes Jan 11 at 20:13
  • \$\begingroup\$ BTW, several size-neutral changes that would make it more efficient (especially on modern CPUs): shr al,1 to shift the low bit into CF, not RCR. You don't need an input dependency on FLAGS. (Or hmm, possibly even scasb with AL='1' instead of loading, but that would invert). adc ah,ah can be 1 uop, RCL isn't. Also, test reg,reg/jcc is more efficient than or reg,reg, and the same size. I think the inefficient or idiom comes from 8080 ORA A, which didn't have an opcode that only set flags. \$\endgroup\$ – Peter Cordes Jan 11 at 20:24
  • \$\begingroup\$ The question allows input as a 32-bit binary integer so you could just write a function that takes that (in memory pointed to by DS:SI if you want), instead of converting from a base 2 string. Also, instead of push/pop for digit-order, you could maybe store backwards into a buffer, creating a $-terminated string, and when you're done you have a pointer to the first byte. How do I print an integer in Assembly Level Programming without printf from the c library? shows the algorithm (and using a Linux write system call on it). \$\endgroup\$ – Peter Cordes Jan 11 at 20:27
  • 1
    \$\begingroup\$ @PeterCordes yes, I just tried and that's what's happening. I was doing two different LOOPs on the "inner" loop, and since I can count on CX being 0 at the end of the first, I used it as the counter for decimal digits and looped again. Unfortunately DEC / JNZ is 1 byte more than LOOP so my gain there offsets the BX outer loop gain. So, yes, BX would seem to win but it was a great idea and fun exercise to go through! \$\endgroup\$ – 640KB Jan 11 at 21:10
  • 1
    \$\begingroup\$ @PeterCordes I experimented with writing in reverse to a string buffer, and was only able to get it down to 48 bytes. Here's the scratchpad (godbolt.org/z/6853--) if you have any thoughts or ideas. \$\endgroup\$ – 640KB Jan 13 at 17:22
9
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C (gcc), 48 bytes

i;f(n){for(i=4;i--;)printf(".%hhu"+i/3,n>>i*8);}

Takes as input a 32-bit integer.

Thanks to ceilingcat and gastropner for getting this answer where it is now!

Try it online!

| improve this answer | |
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  • \$\begingroup\$ Missing semicolon in for loop. 48 bytes \$\endgroup\$ – girobuz Jan 12 at 6:38
7
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Jelly, 10 6 bytes

s8Ḅj“.

Try it online!

s            Slice input list
 8           into size 8 chunks
  Ḅ          Convert from binary list to integer
   j“.       Join with dots as the separator
             Implicit output
| improve this answer | |
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  • \$\begingroup\$ If you use (character) rather than (start list of characters) this would be a Link rather than a full program (although submission may want to ba a full-program anyway, since the Link result is not just a list of characters, so we may prefer the implicit printing) \$\endgroup\$ – Jonathan Allan Jan 10 at 14:52
  • \$\begingroup\$ Are you counting as a single byte? \$\endgroup\$ – chrylis -cautiouslyoptimistic- Jan 11 at 7:13
  • 1
    \$\begingroup\$ @chrylis-onstrike- Jelly uses a custom code page to encode its characters. So, in this instance, yes, it's one byte. \$\endgroup\$ – AdmBorkBork Jan 12 at 15:59
5
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PHP, 26 bytes

<?=long2ip(bindec($argn));

Try it online!

OR taking input as a integer as an anonymous function :

PHP, 7 bytes

long2ip

Try it online!

| improve this answer | |
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  • \$\begingroup\$ No need for bindec, $argn will be an integer. \$\endgroup\$ – Kerkouch Jan 18 at 3:23
  • \$\begingroup\$ @kerkouch yes, that's true. I think that option was allowed after I posted the original answer. That simplifies this quite a bit. :) \$\endgroup\$ – 640KB Jan 18 at 14:05
4
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PowerShell, 46 bytes

""+[IPAddress]"$([Convert]::ToInt64($args,2))"

Try it online!

First takes input $args binary string and [System.Convert]s it into Int64. Uses the .NET type call [System.Net.Ipaddress] to parse that Int64 into an IPAddress object, then coerces the .IPAddressToString() method by prepending ""+.

| improve this answer | |
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  • \$\begingroup\$ Unfortunately on systems that are LittleEndian (like TIO and most computers... I think?) this parses the binary backwards so the IP in your TIO link has the quads in reverse order. \$\endgroup\$ – Malivil Jan 10 at 15:45
  • 1
    \$\begingroup\$ @Malivil Dang, that's a side-effect of removing the "$( )" that I golfed. I'll put that back in. Thanks for the bug catch! \$\endgroup\$ – AdmBorkBork Jan 10 at 16:06
4
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Python 2, 47 bytes

f=lambda n,k=-2:k*`n`or f(n>>8,k+1)+'.'+`n%256`

Try it online!


Python 3, 46 bytes

lambda n:('.%d'*4%(*n.to_bytes(4,"big"),))[1:]

Try it online!

Shaving a byte from David Foerster's to_bytes solution using string formatting.

| improve this answer | |
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3
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MathGolf, 6 bytes

8/å'.u

Try it online.

Explanation:

8/       # Split the (implicit) input-string into parts of size 8
  å      # Convert each part from a binary-string to an integer
   '.u  '# Join by "."
         # (after which the entire stack joined together is output implicitly)
| improve this answer | |
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3
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APL (Dyalog Unicode), 20 19 bytesSBCS

-1 thanks to Kritixi Lithos.

Full program. Prompts for 32-bit integer, optionally as list of bits.

' '⎕R'.'⍕256|83⎕DR⎕

Try it online!

 console prompt for numeric input

83⎕DR interpret the bits of that data as 8-bit integers (internal Data Representation type 3)

256| convert to unsigned integers (lit. 256-mod of that)

 stringify (makes space-separated string)

' '⎕R'.'Replace spaces with dots

| improve this answer | |
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  • \$\begingroup\$ Using ⎕r on 'ing the entire list saves a byte \$\endgroup\$ – user41805 Jan 12 at 7:57
  • \$\begingroup\$ @KritixiLithos How? \$\endgroup\$ – Adám Jan 12 at 8:16
  • \$\begingroup\$ ' '⎕r'.'⍕256|... tio.run/… \$\endgroup\$ – user41805 Jan 12 at 8:20
  • \$\begingroup\$ @KritixiLithos Oh right, for some reason I thought it was nested and would give too many spaces. \$\endgroup\$ – Adám Jan 12 at 8:21
3
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JavaScript (V8), 49 44 bytes

-5 bytes thanks to Arnauld

s=>s.match(/.{8}/g).map(x=>'0b'+x|0).join`.`

Try it online!

| improve this answer | |
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3
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C#7+, 117 73 bytes

s=>string.Join('.',(new System.Net.IPAddress(s)+"").Split('.').Reverse())

Accepts the bit representation of an IP address to transform it into a four-number notation, converts it to a string array, reverses the elements to account for the endianness differences of machines, then joins them together again with a dot.

Try it online!

  • Reduced to 73 bytes by utilizing the string hack and chaining (c/o @Expired Data)

Initial answer, 117 bytes

string I(long b){var x=new System.Net.IPAddress(b).ToString().Split('.');Array.Reverse(x);return string.Join(".",x);}

Use it like:

void Main()
{
    Console.WriteLine(I(0b_10001011111100010111110001111110));
}

public string I(long b)
{
    var x = new System.Net.IPAddress(b).ToString().Split('.');
    Array.Reverse(x);
    return string.Join(".", x);
}
| improve this answer | |
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  • 1
    \$\begingroup\$ by using lambdas, chaining and a hack for ToString() you can reduce it to 73 bytes \$\endgroup\$ – Expired Data Jan 13 at 11:47
  • \$\begingroup\$ Thanks, @ExpiredData, totally forgot the string hack. \$\endgroup\$ – terrible-coder Jan 13 at 15:59
2
\$\begingroup\$

Japt, 7 bytes

ò8 mÍq.

Try it here

| improve this answer | |
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2
\$\begingroup\$

k4, 18 17 bytes

saved a byte by expressing in composed form instead of as a lambda:

("."/:$2/:'4 0N#)

original explanation: output is string, quad-dot decimal not supported by k

{"."/:$2/:'4 0N#x}

{                } /lambda with implicit arg x
           4 0N#x  /cut x into 4 pieces
       2/:'        /convert each piece to decimal
      $            /stringify
 "."/:             /join with .

called on 2 binaries:

{"."/:$2/:'4 0N#x}'(10001011111100010111110001111110b;11000000101010000000000111111111b)
("139.241.124.126";"192.168.1.255")
| improve this answer | |
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2
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Java 9, 97 94 92 81 bytes

s->{for(int i=0;i<32;)System.out.print((i>0?".":"")+Long.parseLong(s,i,i+=8,2));}

-2 bytes thanks to @AZTECCO.
-11 bytes thanks to @Holger by combining the Long.parseLong(s.substring(i,i+=8),2) into Long.parseLong(s,i,i+=8,2).

Try it online.

Explanation:

s->{                   // Method with String parameter and no return-type
  for(int i=0;i<32;)   //  Loop `i` in the range [0, 32):
    System.out.print(  //   Print:
      (i>0?            //    If `i` is larger than 0 (so it's not the first iteration):
        "."            //     Print a dot
       :               //    Else:
        "")            //     Print nothing instead
      +                //    Appended with:
       Long.parseLong(s,i,i+=8,2));}
                       //     A substring of the input `s` from index `i` to `i+8`,
                       //     (and increase `i` by 8 for the next loop iteration)
                       //     Converted from binary-String to integer
| improve this answer | |
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  • 1
    \$\begingroup\$ Since Java 9, you can use Long.parseLong(s,i,i+=8,2) which will not only improve it from a golfing point of view, even the performance will be better as it saves the substring operations. \$\endgroup\$ – Holger Jan 13 at 15:12
  • \$\begingroup\$ @Holger Oh, I didn't knew that. Thanks! :) \$\endgroup\$ – Kevin Cruijssen Jan 13 at 16:56
2
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Excel. 96 bytes

=BIN2DEC(LEFT(A1,8))&"."&BIN2DEC(MID(A1,9,8))&"."&BIN2DEC(MID(A1,17,8))&"."&BIN2DEC(RIGHT(A1,8))
| improve this answer | |
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2
\$\begingroup\$

Python 3 (125 bytes)

def f(x):
    y=''
    for j in range(4):
        y+=str(int(x[j*8:j*8+8],2))
        if j<4:
            y+="."
    return y

Try it online

| improve this answer | |
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  • 1
    \$\begingroup\$ Hi. Small problem in that this outputs an extra "." at the end. Easily fixed by changing j<4 to j<3. Also we normally count bytes and not characters and according to TIO this is 125 bytes. \$\endgroup\$ – ElPedro Jan 11 at 8:28
  • 1
    \$\begingroup\$ Here are a few hints for 80 bytes Try it online! \$\endgroup\$ – ElPedro Jan 11 at 8:50
  • \$\begingroup\$ Thank you @ElPedro for the knowledge. \$\endgroup\$ – Merin Nakarmi Jan 15 at 17:43
  • \$\begingroup\$ My pleasure, however there are a couple more golfs that I didn't mention. I'll leave those for you to find. Since OP has said that trailing space is alowed then you can save at least 6 more. Have fun :) \$\endgroup\$ – ElPedro Jan 15 at 21:59
  • \$\begingroup\$ BTW, even if you don't use the hints, you should really update the answer to correct the identified problem and please feel free to post my suggestions as an improved answer. \$\endgroup\$ – ElPedro Jan 15 at 22:00
1
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Red, 33 bytes

func[n][to 1.1.1 debase/base n 2]

Try it online!

Takes the input as strings.

| improve this answer | |
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1
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Burlesque, 13 bytes

8co{b2}]m'.IC

Try it online!

8co    #Break into chunks 8 long
{b2}]m #Read each chunk as base-2 and turn to string
'.IC   #Intercalate "." between each and collapse
| improve this answer | |
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1
\$\begingroup\$

Stax, 8 bytes

Ç∩0&→Ö¡ 

Run and debug it at staxlang.xyz!

Unpacked (9 bytes) and explanation:

8/{|Bm'.*
8/           Split into length-8 chunks. 4M would work just as well.
  {|Bm       Convert each chunk to decimal
      '.*    Join with .
| improve this answer | |
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1
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Python 2, 81 \$\cdots\$ 58 55 bytes

lambda s:'.'.join(`int(s[i:i+8],2)`for i in(0,8,16,24))

Try it online!

Saved 3 bytes thanks to ElPedro!!!

Lambda function that takes a string of 32 "0"s and "1"s.

| improve this answer | |
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  • \$\begingroup\$ Ha, just came up wth the same thing! If you switch to Python 2 you can use backticks instead of str() to save 3 more Try it online! \$\endgroup\$ – ElPedro Jan 11 at 7:50
  • \$\begingroup\$ @ElPedro Awesome, great minds think alike! Good idea switching to 2 for str() to backticks save. :-) \$\endgroup\$ – Noodle9 Jan 11 at 8:31
1
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C (clang), 66 61 bytes

i;g(*m){for(i=32;i--;)*++m+=i%8?*m*2:!printf(".%d"+i/24,*m);}

Try it online!

Input as an array of integers (bits)

Adds current number shifted to the next. Every 8 bits it prints instead of adding.

Saved 5 thanks to @gastropner and @ceilingcat

| improve this answer | |
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1
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CJam, 10 bytes

l~8/2fb'.*

Try it online!

Explanation

l~    e# Read a line and evaluate it. Pushes list to the stack
8/    e# Split into sublists of 8 elements each. Gives list of sublists
2fb   e# Map "base conversion" with extra parameter 2 over the list of sublists
'.*   e# Join sublists with character ".". Implicitly display
| improve this answer | |
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1
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Python 3, 47 bytes

lambda n:".".join(map(str,n.to_bytes(4,"big")))

Try it online!

| improve this answer | |
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1
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PowerShell, 45 44 bytes

$args|%{$r+=+$r+"$_"}
[ipaddress]::Parse($r)

Try it online!


PowerShell, 45 bytes

Pure PowerShell. It does not use external libs.

($args|%{($r=2*$r%256+"$_")[++$i%8]})-join'.'

Try it online!

Unrolled and commented:

$bytes = $args|%{           # $args is array on character 48 or 49 (bits)
    $r = 2 * $r             # one bit shift left
                            # the first operand is integer, so Powershell converts the second operand to an integer
    $r = $r % 256           # bitwise and 0xFF
    $digit = "$_"           # convert a char 48, 49 to string "0" or "1" respectively
    $r = $r + $digit        # add a digit
                            # the first operand is integer, so Powershell converts the second operand to an integer
    # now $r is a byte containing 8 bits to the left of the current one

    $index = ++$i % 8       # 1,2,3,4,5,6,7,0, 1,2,3,4,5,6,7,0, ...
    ($r)[$index]            # represent $r as an array; take an element of this array
                            # index 0 will give $r, other indexes will give $null
                            # Powershell outputs non $null values only
                            # Compare to `Wrtie-Output ($r)[$index]`
}
# now $bytes is array of not $null elements
Write-Output ($bytes -join '.')
| improve this answer | |
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  • 1
    \$\begingroup\$ Witchcraft! I can't even begin to understand how the second one works -- can you add an explanation? \$\endgroup\$ – AdmBorkBork Jan 12 at 16:09
  • \$\begingroup\$ Thanks, O' Valley of Plenty. I've added explanation. \$\endgroup\$ – mazzy Jan 13 at 8:17
1
\$\begingroup\$

Perl 5, 30 bytes

s/.{8}/oct("0b$&").'.'/ge;chop

Try it online!

Search-replaces with regexp that takes eight bits (0 or 1) at a time and converts them to their decimal representation with . placed after each, but chops off the last . char. Using a function named oct here seems counter-intuitive since the input string isn't octal. But when the given string starts with 0b the rest is read as the binary string it is.

| improve this answer | |
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1
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Ruby, 37 34 bytes

->b{(-3..0).map{|w|255&b<<w*8}*?.}

Try it online!

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Perl 6, 28 27 bytes

-1 byte thanks to Jo King

{chop S:g/.**8/{:2(~$/)}./}

Try it online!

| improve this answer | |
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1
\$\begingroup\$

J, 18 bytes

' .'rplc~&":_8#.\]

Try it online!

Old answer

-2 thanks to Kritixi Lithos

' .'rplc~&":2#.4 8$]

Try it online!

My first answer in a non-esolang! (kinda). The way this answer works is pretty simple. Let's look first at a non-tacit form of this expression:

(":#.(4 8 $ n))rplc' .'

Assuming that (for instance) n is:

n =: 1 0 0 0 1 0 1 1 1 1 1 1 0 0 0 1 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 

The expression 4 8 $ n is equal to:

1 0 0 0 1 0 1 1
1 1 1 1 0 0 0 1
0 1 1 1 1 1 0 0
0 1 1 1 1 1 1 0

Then, the #. verb is applied over the matrix, yielding the following result:

139 241 124 126

The resulting list is stringified using ":, and using rplc every space in the string representation of list is swapped to a dot, yielding the final form:

139.241.124.126
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ [:#. can become 2#., and you can also remove the cap in rplc~[:": with rplc~&":. Also 2#.4 8$] can become shorter with the adverb infix jsoftware.com/help/dictionary/d430.htm \$\endgroup\$ – user41805 Jan 16 at 20:26
  • \$\begingroup\$ I can't spot the the possibility of using adverb infix in the code \$\endgroup\$ – Kamila Szewczyk Jan 17 at 9:04
1
\$\begingroup\$

REXX, 91 bytes

PARSE ARG WITH 1 A 9 B 17 C 25 D
SAY X2D(B2X(A))'.'X2D(B2X(B))'.'X2D(B2X(C))'.'X2D(B2X(D))

Online REXX interpreter

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ If REXX had a function that would go directly from binary to decimal I could chop 20 bytes off of this. \$\endgroup\$ – abricker Jan 17 at 23:24
1
\$\begingroup\$

Forth (gforth), 130 bytes.

Requires a Forth that starts in decimal mode, works with gforth.

: p . 8 emit ." ." ;
: d dup 255 and swap ;      
: r 8 rshift d ;
: q 32 2 base ! word number drop d r r r drop decimal p p p . ;

Try it online!

Usage: q 10001011111100010111110001111110 [enter]

Deobfuscated version (or how I would really do it)

\ Forth program to convert a binary IP address to dotted decimal notation.

decimal
: binary   2 base ! ;

\ Get the binary string and convert to a number.
: getbin   32 binary word number drop ;

\ Shift and mask the byte we are interested in. Put all 4 on the stack.
hex
: mask   rshift dup ff and ;
: quad4   dup ff and swap ;
: quad   8 mask swap ;
: 3more   quad quad quad ;

\ Print a quad, backspace over it's trailing space, print a dot.
: .quad   . 8 emit ." ." ;

\ Print all 4 quads in decimal.
: .4quads   decimal .quad .quad .quad . ;

\ Get binary number, chop it up into 4 quads, print in decimal.
: qdot   getbin quad4 3more drop .4quads ;
| improve this answer | |
\$\endgroup\$

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