Convert a 32 bit binary IPv4 address to its quad-dotted notation

Question

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:

Split the binary representation into its 4 individual bytes
Convert each byte into denary
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

Input will always be a 32 bit binary number or a binary list or string
Standard I/O rules apply
No standard loopholes
This is code-golf, so the shortest answer in bytes wins

Input will always be a 32 bit binary number but your examples are binary lists. Is that also an acceptable input form? — Adám
– Adám, Commented Jan 10, 2020 at 12:49
Is it allowed to use a string as input instead of a list of binary digits? — Galen Ivanov
– Galen Ivanov, Commented Jan 10, 2020 at 13:31
I have added a brief description of how the qaud-dotted notation is formed @LuisMendo. — mabel
– mabel, Commented Jan 10, 2020 at 13:31
@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. — mabel
– mabel, Commented Jan 10, 2020 at 13:33
I'm not clear, can we take in an actual number as input, like 2130706433 to give 127.0.0.1? — xnor
– xnor, Commented Jan 11, 2020 at 4:00

Community · Accepted Answer · 2020-06-17 09:04:33Z

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:

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!

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? — Peter Cordes
– Peter Cordes, Commented Jan 11, 2020 at 20:13
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. — Peter Cordes
– Peter Cordes, Commented Jan 11, 2020 at 20:24
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). — Peter Cordes
– Peter Cordes, Commented Jan 11, 2020 at 20:27
@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! — 640KB
– 640KB, Commented Jan 11, 2020 at 21:10
@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. — 640KB
– 640KB, Commented Jan 13, 2020 at 17:22

Community · Accepted Answer · 2020-06-17 09:04:33Z

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)

S.S. Anne · Accepted Answer · 2020-01-12 21:48:05Z

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!

\$\begingroup\$ Missing semicolon in for loop. 48 bytes \$\endgroup\$

girobuz
– girobuz

2020-01-12 06:38:53 +00:00
Commented Jan 12, 2020 at 6:38 — girobuz
– girobuz, Commented Jan 12, 2020 at 6:38

AdmBorkBork · Accepted Answer · 2020-01-10 14:21:30Z

7

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

edited Jan 10, 2020 at 14:21

answered Jan 10, 2020 at 14:15

AdmBorkBork

43.7k5 gold badges107 silver badges288 bronze badges

\$\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
– Jonathan Allan

2020-01-10 14:52:44 +00:00
Commented Jan 10, 2020 at 14:52
\$\begingroup\$ Are you counting Ḅ as a single byte? \$\endgroup\$

chrylis -cautiouslyoptimistic-
– chrylis -cautiouslyoptimistic-

2020-01-11 07:13:49 +00:00
Commented Jan 11, 2020 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
– AdmBorkBork

2020-01-12 15:59:10 +00:00
Commented Jan 12, 2020 at 15:59

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xnor · Accepted Answer · 2020-01-12 06:56:04Z

Python 2, 47 bytes

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

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Python 3, 46 bytes

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

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

Community · Accepted Answer · 2020-06-17 09:04:33Z

5

PHP, 26 bytes

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

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OR taking input as a integer as an anonymous function :

PHP, 7 bytes

long2ip

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 10, 2020 at 16:53

640KB

12.4k2 gold badges38 silver badges61 bronze badges

\$\begingroup\$ No need for bindec, $argn will be an integer. \$\endgroup\$

Kerkouch
– Kerkouch

2020-01-18 03:23:15 +00:00
Commented Jan 18, 2020 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
– 640KB

2020-01-18 14:05:37 +00:00
Commented Jan 18, 2020 at 14:05

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AdmBorkBork · Accepted Answer · 2020-01-10 16:06:34Z

4

PowerShell, 46 bytes

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

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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 ""+.

edited Jan 10, 2020 at 16:06

answered Jan 10, 2020 at 13:56

AdmBorkBork

43.7k5 gold badges107 silver badges288 bronze badges

\$\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
– Malivil

2020-01-10 15:45:13 +00:00
Commented Jan 10, 2020 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
– AdmBorkBork

2020-01-10 16:06:19 +00:00
Commented Jan 10, 2020 at 16:06

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

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)

user41805 · Accepted Answer · 2020-01-12 08:35:55Z

APL (Dyalog Unicode), 20 19 bytes^SBCS

-1 thanks to Kritixi Lithos.

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

' '⎕R'.'⍕256|83⎕DR⎕

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⎕ 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

@KritixiLithos Oh right, for some reason I thought it was nested and would give too many spaces. — Adám
– Adám, Commented Jan 12, 2020 at 8:21

Expired Data · Accepted Answer · 2020-01-13 10:08:41Z

3

JavaScript (V8), 49 44 bytes

-5 bytes thanks to Arnauld

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

Try it online!

edited Jan 13, 2020 at 10:08

answered Jan 10, 2020 at 15:51

Expired Data

4,0111 gold badge12 silver badges34 bronze badges

1

\$\begingroup\$ 44 bytes \$\endgroup\$

Arnauld
– Arnauld

2020-01-12 10:37:00 +00:00
Commented Jan 12, 2020 at 10:37

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Expired Data · Accepted Answer · 2020-01-13 16:18:41Z

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); }

by using lambdas, chaining and a hack for ToString() you can reduce it to 73 bytes — Expired Data
– Expired Data, Commented Jan 13, 2020 at 11:47

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Japt, 7 bytes

ò8 mÍq.

Try it here

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 10, 2020 at 19:24

Shaggy

45k4 gold badges39 silver badges95 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

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")

Community · Accepted Answer · 2020-06-17 09:04:33Z

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

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. — Holger
– Holger, Commented Jan 13, 2020 at 15:12

Wernisch · Accepted Answer · 2020-01-16 17:08:56Z

Excel. 96 bytes

=BIN2DEC(LEFT(A1,8))&"."&BIN2DEC(MID(A1,9,8))&"."&BIN2DEC(MID(A1,17,8))&"."&BIN2DEC(RIGHT(A1,8))

lyxal · Accepted Answer · 2020-01-17 09:04:22Z

2

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

edited Jan 17, 2020 at 9:04

lyxal♦

35.6k2 gold badges69 silver badges148 bronze badges

answered Jan 10, 2020 at 22:30

Merin Nakarmi

2571 silver badge6 bronze badges

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
– ElPedro

2020-01-11 08:28:36 +00:00
Commented Jan 11, 2020 at 8:28
1

\$\begingroup\$ Here are a few hints for 80 bytes Try it online! \$\endgroup\$

ElPedro
– ElPedro

2020-01-11 08:50:19 +00:00
Commented Jan 11, 2020 at 8:50
\$\begingroup\$ Thank you @ElPedro for the knowledge. \$\endgroup\$

Merin Nakarmi
– Merin Nakarmi

2020-01-15 17:43:24 +00:00
Commented Jan 15, 2020 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
– ElPedro

2020-01-15 21:59:53 +00:00
Commented Jan 15, 2020 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
– ElPedro

2020-01-15 22:00:17 +00:00
Commented Jan 15, 2020 at 22:00

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Kamila Szewczyk · Accepted Answer · 2020-01-17 12:33:26Z

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

[:#. 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 — user41805
– user41805, Commented Jan 16, 2020 at 20:26
I can't spot the the possibility of using adverb infix in the code — Kamila Szewczyk
– Kamila Szewczyk, Commented Jan 17, 2020 at 9:04

Galen Ivanov · Accepted Answer · 2020-01-10 13:35:25Z

1

Red, 33 bytes

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

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Takes the input as strings.

edited Jan 10, 2020 at 13:35

answered Jan 10, 2020 at 13:25

Galen Ivanov

21.5k3 gold badges26 silver badges62 bronze badges

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DeathIncarnate · Accepted Answer · 2020-01-10 14:09:56Z

Burlesque, 13 bytes

8co{b2}]m'.IC

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8co #Break into chunks 8 long {b2}]m #Read each chunk as base-2 and turn to string '.IC #Intercalate "." between each and collapse

Khuldraeseth na'Barya · Accepted Answer · 2020-01-10 14:28:35Z

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 .

Noodle9 · Accepted Answer · 2020-01-11 08:26:50Z

1

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.

edited Jan 11, 2020 at 8:26

answered Jan 10, 2020 at 14:05

Noodle9

20.4k3 gold badges23 silver badges47 bronze badges

\$\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
– ElPedro

2020-01-11 07:50:01 +00:00
Commented Jan 11, 2020 at 7:50
\$\begingroup\$ @ElPedro Awesome, great minds think alike! Good idea switching to 2 for str() to backticks save. :-) \$\endgroup\$

Noodle9
– Noodle9

2020-01-11 08:31:38 +00:00
Commented Jan 11, 2020 at 8:31

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AZTECCO · Accepted Answer · 2020-01-11 08:54:22Z

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

\$\begingroup\$ 63 bytes \$\endgroup\$

gastropner
– gastropner

2020-01-11 01:51:34 +00:00
Commented Jan 11, 2020 at 1:51 — gastropner
– gastropner, Commented Jan 11, 2020 at 1:51

Luis Mendo · Accepted Answer · 2020-01-11 18:48:38Z

CJam, 10 bytes

l~8/2fb'.*

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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

David Foerster · Accepted Answer · 2020-01-12 00:27:21Z

1

Python 3, 47 bytes

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

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answered Jan 12, 2020 at 0:27

David Foerster

4942 silver badges12 bronze badges

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mazzy · Accepted Answer · 2020-01-13 05:39:50Z

PowerShell, 45 44 bytes

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

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PowerShell, 45 bytes

Pure PowerShell. It does not use external libs.

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

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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 '.')

Witchcraft! I can't even begin to understand how the second one works -- can you add an explanation? — AdmBorkBork
– AdmBorkBork, Commented Jan 12, 2020 at 16:09

Kjetil S · Accepted Answer · 2020-01-13 10:44:53Z

Perl 5, 30 bytes

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

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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.

G B · Accepted Answer · 2020-01-14 08:09:03Z

1

Ruby, 37 34 bytes

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

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edited Jan 14, 2020 at 8:09

answered Jan 10, 2020 at 16:43

G B

23.4k1 gold badge24 silver badges55 bronze badges

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nwellnhof · Accepted Answer · 2020-01-14 12:43:43Z

1

Perl 6, 28 27 bytes

-1 byte thanks to Jo King

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

Try it online!

edited Jan 14, 2020 at 12:43

answered Jan 10, 2020 at 16:37

nwellnhof

10.6k1 gold badge20 silver badges38 bronze badges

\$\begingroup\$ Using chop is one byte shorter \$\endgroup\$

Jo King
– Jo King

2020-01-14 01:39:16 +00:00
Commented Jan 14, 2020 at 1:39

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abricker · Accepted Answer · 2020-01-17 23:27:35Z

1

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

edited Jan 17, 2020 at 23:27

answered Jan 16, 2020 at 11:53

abricker

613 bronze badges

\$\begingroup\$ If REXX had a function that would go directly from binary to decimal I could chop 20 bytes off of this. \$\endgroup\$

abricker
– abricker

2020-01-17 23:24:37 +00:00
Commented Jan 17, 2020 at 23:24

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mabel · Accepted Answer · 2020-01-18 01:14:28Z

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 ;

Convert a 32 bit binary IPv4 address to its quad-dotted notation

Challenge

Quad-dotted decimal

I/O examples

Rules

49 Answers 49

x86-16 machine code, IBM PC DOS, 54 47 45 bytes

05AB1E, 6 bytes

C (gcc), 48 bytes

Jelly, 10 6 bytes

Python 2, 47 bytes

Python 3, 46 bytes

PHP, 26 bytes

OR taking input as a integer as an anonymous function :

PHP, 7 bytes

PowerShell, 46 bytes

MathGolf, 6 bytes

APL (Dyalog Unicode), 20 19 bytesSBCS

JavaScript (V8), 49 44 bytes

C#7+, 117 73 bytes

Initial answer, 117 bytes

Japt, 7 bytes

k4, 18 17 bytes

Java 9, 97 94 92 81 bytes

Excel. 96 bytes

Python 3 (125 bytes)

J, 18 bytes

Old answer

Red, 33 bytes

Burlesque, 13 bytes

Stax, 8 bytes

Unpacked (9 bytes) and explanation:

Python 2, 81 \$\cdots\$ 58 55 bytes

C (clang), 66 61 bytes

CJam, 10 bytes

Explanation

Python 3, 47 bytes

PowerShell, 45 44 bytes

PowerShell, 45 bytes

Perl 5, 30 bytes

Ruby, 37 34 bytes

Perl 6, 28 27 bytes

REXX, 91 bytes

Forth (gforth), 130 bytes.

Linked

Related

Hot Network Questions

APL (Dyalog Unicode), 20 19 bytes^SBCS

Python 2, ^{81 \$\cdots\$ 58} 55 bytes