我需要一个UUID。您的工作是生成一个。

规范的UUID（通用唯一标识符）是一个32位十六进制数字，在某些点上插入了连字符。程序应输出32位十六进制数字（128位），格式为 xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx（8-4-4-4-12位数），其中x是一个随机十六进制数。假设您的语言的PRNG是完美的，则所有有效输出必须具有相同的生成概率。

TL; DR

生成以下形式的32个随机十六进制数字 8-4-4-4-12数字数字。最短的代码获胜。

编辑：必须为十六进制。始终仅生成小数是无效的。编辑2：没有内置。这些不是GUID，只是通用的十六进制数字。

输出示例：

ab13901d-5e93-1c7d-49c7-f1d67ef09198
7f7314ca-3504-3860-236b-cface7891277
dbf88932-70c7-9ae7-b9a4-f3df1740fc9c
c3f5e449-6d8c-afe3-acc9-47ef50e7e7ae
e9a77b51-6e20-79bd-3ee9-1566a95d9ef7
7b10e43c-3c57-48ed-a72a-f2b838d8374b

输入和标准漏洞是不允许的。

这是代码高尔夫球，因此最短的代码获胜。另外，请随时进行澄清。

Pyth，20个字节

j\-msm.HO16*4hdj83 3

示范。

[1, 0, 0, 0, 2]在基数3中编码为83，然后加1并乘以4得到每个段的长度。然后使十六进制数字并连字符。

朱莉娅，80字节

h=hex(rand(Uint128),32)
print(h[1:8]"-"h[9:12]"-"h[13:16]"-"h[17:20]"-"h[21:32])

生成一个随机的128位整数，以填充为32位数字的字符串的形式获取其十六进制表示形式，并将其划分为带破折号的段。

感谢ConfusedMr_C和kvill的帮助！

CJam，26 25字节

8 4__C]{{Gmr"%x"e%}*'-}/;

在CJam解释器中在线尝试。

怎么运行的

8 4__C]{              }/   For each I in [8 4 4 4 12]:
        {         }*         Do I times:
         Gmr                   Pseudo-randomly select an integer between 0 and 15.
            "%x"e%             Apply hexadecimal string formatting.
                    '-       Push a hyphen-minus.
                        ;  Discard the last hyphen-minus.

PowerShell，77 69 67字节

((8,4,4,4,12)|%{((1..$_)|%{'{0:X}'-f(random(16))})-Join""})-Join"-"

编辑：多余的括号：

((8,4,4,4,12)|%{((1..$_)|%{('{0:X}'-f(random(16)))})-Join""})-Join"-"

编辑：能够从原始内容中删除结尾的.Trim（“-”）：

(((8,4,4,4,12)|%{((1..$_)|%{('{0:X}'-f(random(16)))})+"-"})-Join"").Trim("-")

考虑到标志（-f和-Join）的性质，使用某些空格可能会更清楚。我仍然想丢掉最后的Trim（“-”）：

(((8,4,4,4,12)|%{((1..$_)|%{('{0:X}' -f (random(16)))}) + "-"}) -Join "").Trim("-")

或者，使用内置功能（例如上述C＃答案）

'{0}'-f[System.Guid]::NewGuid()

但是，即使它只有31个字节，它似乎还是有点捷径-y。

Python 2，86 84字节

from random import*;print'-'.join('%%0%ix'%i%randint(0,16**i-1)for i in[8,4,4,4,12])

这将链接字符串格式化程序，以使Python为每个段唯一地格式化十六进制数字。

取消高尔夫：

import random

final = []
for i in [8, 4, 4, 4, 12]:               # Iterate through every segment
    max = (16 ** i) - 1                  # This is the largest number that can be
                                         # represented in i hex digits
    number = random.randint(0, max)      # Choose our random segment
    format_string = '%0' + str(i) + 'x'  # Build a format string to pad it with zeroes
    final.append(format_string % number) # Add it to the list

print '-'.join(final)                    # Join every segment with a hyphen and print

这可能需要一些改进，但我感到很自豪。

Perl 5，43个字节

@Xcali节省了2个字节！

printf"%04x"."-"x/[2-5]/,rand 2**16for 1..8

在线尝试！

PHP，69 72 75字节

foreach([8,4,4,4,12]as$c)$r[]=rand(".1e$c","1e$c");echo join('-',$r);

这不会输出十六进制数字（a，...f）。它们是允许的，但问题正文不是必需的。

没有数字组以 0（也不是必需的）。

编辑：由于@IsmaelMiguel而节省了3个字节

C＃，65个字节

using System;class C{void Main(){Console.Write(Guid.NewGuid());}}

编辑：是的！C＃比另一种语言（Java除外）短：)

高夫，86

BEGIN{for(srand();j++<32;printf(j~"^9|13|17|21"?"-":E)"%c",x+(x>10?87:48))x=rand()*16}

您可以每秒使用一次，以生成唯一的随机“ UUID”。这是因为srand()，如果未指定参数，则使用自纪元以来的系统时间（以秒为单位）。

for n in `seq 100` do awk 'BEGIN{for(srand();j++<32;printf(j~"^9|13|17|21"?"-":E)"%c",x+(x>10?87:48))x=rand()*16}'; sleep 1; done

我认为awk部分相当优雅。

BEGIN{
    srand()
    for(;j++<32;) {
        x=rand()*16
        x+=(x>10?87:48)
        printf "%c", x
        if(j~"^8|12|16|20")printf "-"
    }
}

如果您希望使用它的频率超过每秒一次，可以像这样在bash中调用它。请注意，awk部分也已更改。

echo `awk 'BEGIN{for(srand('$RANDOM');j++<32;printf(j~"^9|13|17|21"?"-":E)"%c",x+(x>10?87:48))x=rand()*16}'`

将echo在其中添加，以便每次都打印新行。

K5，35个字节

"-"/(0,8+4*!4)_32?`c$(48+!10),65+!6

要生成十六进制字母，我需要`c$从数字（48+!10）和前6个大写字母（65+!6）的列表中生成一个字符串（）。产生相同长度数字的另一种方法是,/$!10。

使用生成的字符串“ 0123456789ABCDEF”，其余的操作很简单。从该集合（32?）中选择32个随机值，在（）_处0 8 12 16 20通过计算将结果字符串切成（）(0,8+4*!4)，然后用破折号（"-"/）连接结果字符串片段。

实际上：

  "-"/(0,8+4*!4)_32?`c$(48+!10),65+!6
"9550E114-A8DA-9533-1B67-5E1857F355E1"

R，63字节

x=sample(c(0:9,letters[1:6]),36,1);x[0:3*5+9]='-';cat(x,sep='')

在线尝试！

该代码首先构建一个36个字符的随机字符串，然后放置四个连字符。它将UUID输出到stdout。

JavaScript，ES6、106字节

"8-4-4-4-12".replace(/\d+/g, m => {t=0;for(i=0; i<m; i++) {t+=(Math.random()*16|0).toString(16)}return t})

使用正则表达式替换。将格式字符串视为生成十六进制字符的计数。尽我所能吊起；尽可能省略分号。

Perl 6、53字节

显而易见的一个：

say join '-',(0..9,'a'..'f').flat.roll(32).rotor(8,4,4,4,12)».join # 67

使用转换Perl 5示例会printf导致代码短一些。

printf ($_='%04x')~"$_-"x 4~$_ x 3,(0..^4⁸).roll(8) # 53

Kotlin, 175 bytes

fun main(a:Array<String>){
fun f()="0123456789abcdef".get((Math.random()*16).toInt())
var s=""
for(i in listOf(8,4,4,4,12)){
for(j in 1..i)
s+=f()
if(i!=12)s+="-"}
println(s)}

Try it online!

My first ever Kotlin program & PPCG submission

APL (Dyalog Unicode), 115 78 bytes

a←⊣,'-',⊢
H←⊃∘(⎕D,819⌶⎕A)¨16∘⊥⍣¯1
(H 8?16)a(H 4?16)a(H 4?16)a(H 4?16)a H 12?16

Try it online!

This is my first APL submission. Huge thanks to @Adám for bearing with me at the PPCG's APL chat and for the hexadecimal conversion function.

Thanks to @Zacharý for 1 byte

Edited to fix byte count.

Japt, 32 bytes

[8,4,4,4,12]m@MqG**X sG ù0X} q"-

Try it online!

MATLAB/Octave , 95 bytes

a='-';b=strcat(dec2hex(randi(16,32,1)-1)');[b(1:8) a b(9:12) a b(13:16) a b(17:20) a b(21:32)]

Perl, 51 bytes

say"xx-x-x-x-xxx"=~s/x/sprintf"%04x",rand 65536/reg

Requires perl5 >= 5.10 I think. For the /r modifier and for say().

J, 42 39 37 27 bytes

echo'-'(8+5*i.4)},hfd?36#16

Try it online!

C++, 194 193 221 210 201 bytes

+7 bytes thanks to Zacharý ( detected a - that should not be at the end )

#include<iostream>
#include<random>
#include<ctime>
#define L(a)for(int i=0;i<a;++i)std::cout<<"0123456789abcdef"[rand()%16];
#define P(a)printf("-");L(a)
void t(){srand(time(0));L(8)P(4)P(4)P(4)P(12)}

If someone has a way to get a different value every execution without changing srand and without including <ctime>, that would be great

Befunge-93, 97 bytes

v>4448v,+<    <
0*    :  >59*0^
62v0-1_$:|>*6+^
>^>41v < @^99<
v<*2\_$:54+` |
?0>+\1-^ v*68<>
>1^

Try it online!

I'm sure this can be shrunk, but this is my first try :)

Bash, 67 bytes

for l in 4 2 2 2 6;{ o+=`xxd -p -l$l</dev/random`-;}
echo ${o::-1}

JavaScript REPL, 79 bytes

'66-6-6-6-666'.replace(/6/g,_=>(Math.random().toString(16)+'00000').slice(2,6))

Try it online!

Math.random may return 0. Adding 5 zeros make the slicing get 4 0s

Forth (gforth), 91 89 bytes

include random.fs
hex
: f 0 4 4 4 8 20 0 do dup i = if + ." -" then 10 random 1 .r loop ;

Try it online!

Explanation

Changes the base to hexadecimal, then outputs numbers/segments of the appropriate length with dashes at specified intervals

Code Explanation

include random.fs          \ include the random module
hex                        \ set the base to hexadecimal
: f                        \ start a new word definition
  0 4 4 4 8                \ enter the intervals to place dashes
  20 0 do                  \ start a counted loop from 0 to 0x20 (32 in decimal)
    dup i =                \ check if we are on a character that needs a dash
    if                     \ if we are
      +                    \ calculate the next character that gets a dash
      ." -"                \ output a dash
    then                   \ end the if block
    f random               \ get a random number between 0x0 and 0xf
    1 .r                   \ output it right-aligned in 1-character space
  loop                     \ end the loop
;                          \ end the word definition

C (gcc),  94   91  86 bytes

main(i){srand(&i);i=803912;for(;i--%16||(i/=16)&&printf("-");printf("%x",rand()%16));}

Try it online!

I would have liked to suggest this version in a comment to Max Yekhlakov (his answer), but unfortunately I do not have the 50 needed reputation points yet, so here is my answer.

803912 is C4448 in hexadecimal, it describes how the output should be formatted (12-4-4-4-8), it is reversed because least significant digits will be read first.
 

Edits:

• saved 3 bytes thanks to Jonathan Frech
• saved 5 more bytes by replacing srand(time(0)) with srand(&i)

C (gcc), 143 110 103 96 94 bytes

Golfed down to 94 bytes thanks to ceilingcat and Jonathan Frech.

(*P)()="\xf\x31À";*z=L"\10\4\4\4\14";main(n){for(;*z;*++z&&putchar(45))for(n=*z;n--;printf("%x",P()&15));}

Try it online!

Explanation:

/*
  P is a pointer to a function.
  The string literal contains actual machine code of the function:

  0F 31     rdtsc
  C3        ret

  0xc3 is the first byte of the UTF-8 representation of the character À
*/
(*P)() = "\xf\61À";

// encode uuid chunk lengths as literal characters
// we use wide characters with 'L' prefix because
// sizeof(wchar_t)==sizeof(int) for 64-bit gcc C on TIO
// so z is actually a zero-terminated string of ints
*z = L"\8\4\4\4\14"

main (n)
{
    for (
        ; 

        // loop until we reach the trailing zero
        *z;

        // increase the pointer and dereference it
        *++z 
             // and output a hyphen, if the pointer does not point at zero
             && putchar(45) 
    )
        // output a random hex string with length pointed at by z
        for (n = *z; n--; printf ("%x", P()&15));
}

Java with Ten Foot Laser Pole v. 1.06, 126 bytes

String u(){return sj224.tflp.util.StringUtil.replace("aa-a-a-a-aaa","a",s->String.format("%04x",(int)(Math.random()*65536)));}

Tested with version 1.06 of the library, but this should work with any version 1.04 or newer.

Jelly, 17 bytes

ØhWẋ36X€”-“µÇŒð‘¦

Try it online!

SmileBASIC, 65 62 bytes

DEF G H?"-";:END
DEF H?HEX$(RND(65536),4);
END H G G G G H H H

I created a function to print 4 random hex digits: DEF H?HEX$(RND(65536),4);:END as well as 4 digits with a - after them: DEF G:H?"-";:END. Then it just has to call these functions a bunch of times.

Chip, 109 + 6 = 115 bytes

Requires flags -wc36, causing +6 bytes

!ZZZZZZZZZZZZZZZZZZZZZZ
,-----.,+vv--^----^---z
?]]]--R\acd
?xx+-)\\b
?x+x-)\\c
?^xx\--\d
`-xx]v~\e
f*`)'`-\g

Try it online!

Generates 4 random bits (the four ?'s) and converts to hex digits:

• 0x0 - 0x9 => 0 - 9
• 0xa - 0xe => b - f
• 0xf => a

...a bit unconventional, but it saved me some bytes at no expense to the distribution of outcomes.