Code Golf Asked by orlp on January 25, 2021
Write a program or function that prints or returns a string of the alphanumeric characters plus underscore, in any order. To be precise, the following characters need to be output, and no more:
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
When printing to stdout, an optional trailing newline after your output is permitted.
Built-in constants that contain 9 or more of the above characters are disallowed.
Shortest code in bytes wins.
This is a very simple challenge, which I believe will generate some interesting answers nevertheless.
Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.
To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:
# Language Name, N bytes
where N
is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:
# Ruby, <s>104</s> <s>101</s> 96 bytes
If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:
# Perl, 43 + 2 (-p flag) = 45 bytes
You can also make the language name a link which will then show up in the leaderboard snippet:
# [><>](http://esolangs.org/wiki/Fish), 121 bytes
var QUESTION_ID=85666,OVERRIDE_USER=4162;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<hd>s*([^n,]*[^s,]),.*?(d+)(?=[^nd<>]*(?:<(?:s>[^n<>]*</s>|[^n<>]+>)[^nd<>]*)*</hd>)/,OVERRIDE_REG=/^Overrides*header:s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>
New method! Also, I realized that is it pretty much exactly the same as Luis' answer but in Convex, but I came up with this independently.
'{,®W"Oò
Explanation:
'{, Array of chars from NUL to 'z
®W" Regex to match non-word characters
Oò Replace all matches with emtpy string
A,'[,_¬^'_
Explanation:
A, 0-9
'[,_¬^ A-Za-z
'_ _
Correct answer by GamrCorps on January 25, 2021
psh 0a1a
sto 0100 41
sto 011a 61
orr 0201 0201
jpz 0220
sto 0134 30
inc 0215
inc 0216
dec 0201
inc 0205
inc 0206
inc 0209
inc 020a
dec 0202
jpz 0235
jmp 0203
sto 013e 5f
NoRAL is a toy assembly language I found on Esolangs.org. Let's give it a whirl (using a minimally adapted version of the Python implementation): Try it online!
There's also a browser-based interpreter here if you want to step through the code.
Everything in NoRAL is stored in addressable memory, including the output (256 consecutive bytes, displayed as a 32x8 rectangle) and the program itself, which is self-modifying.
# We use two memory locations in the code, 0x0201 and 0x0202, as counters
# Set digit counter to ten (0x0a), letter counter to twenty-six (0x1a)
# The psh instruction pushes something to the stack, which we're not using, so it's
# effectively a no-op
psh 0a1a
# The output section of memory runs from addresses 0x0100 to 0x01ff
# We're going to put A-Z at 0100-0119, a-z at 011a-0133, 0-9 at 0134-013d, and _ at 013e
# Output uppercase and lowercase letter, starting with A (ASCII 0x41) & a (0x61)
# These instructions will be modified later to output the other letters
sto 0100 41
sto 011a 61
# OR digit counter with itself (no-op, but sets the zero flag if counter is zero)
orr 0201 0201
# If the digit counter is now zero, skip next section:
jpc 0220
# Output digit, starting with 0 (ASCII 0x30)
sto 0134 30
# Increment memory location and ASCII code in previous statement
inc 0215
inc 0216
# Decrement digit counter
dec 0201
# Increment letters' memory locations and ASCII codes
inc 0205
inc 0206
inc 0209
inc 020a
# Decrement letter counter
dec 0202
# If letter counter is now zero, skip next instruction:
jpz 0235
# Jump back to "output uppercase and lowercase letter"
jmp 0203
# Output underscore (ASCII 0x5f)
sto 013e 5f
Answered by DLosc on January 25, 2021
(C,127XW)
(C,127XW)
,127 Range(127)
C Cast each integer in the range to a character
( XW) In each character, find all regex matches of `w`
The result is a list of lists, some of which are empty, some of which contain an alphanumeric or underscore character. By default, Pip outputs lists with no separator, so the output of the program is simply the requested characters plus a trailing newline.
Answered by DLosc on January 25, 2021
Full program.
'_',∊⍳¨'zZ9'
⍳¨'zZ9'
"ɩndices" until each of the three characters
∊
ϵnlist (flatten)
'_',
prepend an underscore
Answered by Adám on January 25, 2021
Binary:
00000000: be12 01ad 918a c1cd 293a cde0 f83c 5f75 ........):...<_u
00000010: f2c3 7a61 5a41 3930 5f5f ..zaZA90__
Listing:
BE 0112 MOV SI, OFFSET TBL ; SI to range table
RLOOP:
AD LODSW ; AL = high, AH = low
91 XCHG AX, CX ; move to CX
CLOOP:
8A C1 MOV AL, CL ; move current char to AL for display
CD 29 INT 29H ; write to console
3A CD CMP CL, CH ; high and low char the same yet?
E0 F8 LOOPNZ CLOOP ; if not, decrement and keep looping
3C 5F CMP AL, '_' ; was last char displayed a '_'?
75 F2 JNZ RLOOP ; loop until then
C3 RET ; return to DOS
TBL DB 'zaZA90__' ; handy table of ranges to display
Uses a table with a range of chars to display. Turned out that it wasn't any longer to do it this way than unroll all of the 4 different conditions in code (though I'm positive I can be proven wrong there).
Standalone PC DOS executable (COM) program.
Runtime:
Answered by 640KB on January 25, 2021
<<<${(j::)${:-{0..z}}#[^[:IDENT:]]}
The [:IDENT:]
character class is exactly this. We use empty fallback ${:- }
to a brace expansion, remove all non-[:IDENT:]
characters, and then (j::)
oin.
Answered by GammaFunction on January 25, 2021
_ 9ɧZAɧzaɧ(,
Explanation:
_# Push _
9ɧ# Push range 0-9
ZAɧ# Push range Z-A
zaɧ# Push range z-a
(,#Output
Answered by user85052 on January 25, 2021
Answered by Oliver on January 25, 2021
['a':'z','A':'Z','0':'9','_']
Very simple :)
Or, at the same cost:
char(48+[49:74,17:42,0:9,47])
or
char([97:122,65:90,48:57,95])
Answered by PieCot on January 25, 2021
Answered by JayCe on January 25, 2021
_ci,/(97 65+:!26),95,48+!10
Generates [97..122]++[65..90]++[95]++[48..57]
and maps the numbers to chars.
Answered by uryga on January 25, 2021
"AaZ_"o9:n:?!;1-80.
+}$:@=?}:o1+}::o1
Fairly simple, prints _, loop prints a, A then increments and compares to Z. At the end it'll print 9-0.
Answered by Teal pelican on January 25, 2021
write(*,1)(char(i),i=97,122)
write(*,1)(char(i),i=65,90)
write(*,2)(i,i=0,9)
write(*,1)'_'
1 format(1A,25A,$)
2 format(10I1,$)
end
Just because every challenge is improved by a Fortran entry.
You'd think it would be easy to golf this down, but the required variable declarations get expensive very quickly so I think this is about as short as it's going to get.
If delimiters (n
) were allowed between characters in the output, this could be reduced to 83 bytes:
print*,(char(i),i=97,122)
print*,(char(i),i=65,90)
print*,(i,i=0,9)
print*,'_'
end
Answered by georgewatson on January 25, 2021
Shame we can't return an array or set of the characters, I could've saved quite a few bytes there.
_=>[...new Set(Object.keys(this).join`4Z_`+2**29)].join``
The second line is sorted to make it easy to check all characters are included.
f=
_=>[...new Set(Object.keys(this).join`4Z_`+2**29)].join``
o.innerText=f()+`n`+[...f()].sort().join``
<pre id=o></pre>
Answered by Shaggy on January 25, 2021
•zR.'W'-
•zR.'W'-
• # Push a space
z # Push the character 'z'
R # Create a stack of all characters from space to z.
. # Concatenate them into a single string.
'W'- # Filter out everything that matches the regex 'W' (Non-alpha-numeric characters)
# Implicit print
Answered by ATaco on January 25, 2021
Solution:
("c"$48+(!)75)except":;<=>?@[\]^`"
Example:
q)("c"$48+(!)75)except":;<=>?@[\]^`"
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz"
Notes:
Turns out I cannot read :) Q.an
solves this in 5 bytes. I'll try to golf this new solution.
Answered by streetster on January 25, 2021
PRINT'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_'
Procedural solutions in SQL are just too long, due to the length of keywords. A couple of my best attempts:
76 Bytes, store the alphabet then UPPER
it:
DECLARE @ CHAR(26)='abcdefghijklmnopqrstuvwxyz'PRINT'0123456789_'+@+UPPER(@)
100 Bytes, add all chars from 65-122, then strip out the ones I don't want:
DECLARE @ VARCHAR(99)='0123456789'A:SET @+=CHAR(LEN(@)+55)IF LEN(@)<68GOTO A
PRINT STUFF(@,37,6,'_')
Answered by BradC on January 25, 2021
<[<.-<.->>-]<<-.<[<.->-]9
zZ
The last byte is a literal x1A
(decimal 26). It shows in the "edit" mode of this answer as a tiny arrow, but cannot otherwise be seen for some reason...
I use literals in the source code to provide a starting value and how many times to loop. 26 times for the loop printing Z-A
and z-a
, then subtract and print _
, then use the newline (decimal 10) to print 9
and subtract, looping 10 times.
Output:
ZzYyXxWwVvUuTtSsRrQqPpOoNnMmLlKkJjIiHhGgFfEeDdCcBbAa_9876543210
Answered by mbomb007 on January 25, 2021
Had some coffee and re-read the specs:
[65,90|?chr$(a)+chr$(a+32)]?@0123456789_`
This prints Aan
all the way up to Zzn
, then the literal 0123456789_
.
There are quite a lot of ways QBIC can approach this challenge if we want to print the string in the order provided.
By adding all the requested ranges to Z$
(Z$
gets printed implicitly), 71 bytes
[97,122|Z=Z+chr$(a)][65,90|Z=Z+chr$(b)][48,57|Z=Z+chr$(c)]Z=Z+chr$(95)
Printing each range to screen, with ';` to prevent newlines, 67 bytes
[97,122|?chr$(a)';`][65,90|?chr$(b)';`][48,57|?chr$(c)';`]?chr$(95)
Creates one range, printing only substrings, 70 bytes
[48,122|A=A+chr$(a)]?mid$(A,50,26)+mid$(A,18,26)+left$(A,10)+chr$(95)
Creating one range, but with modified start of that range, 69 bytes
[122|A=A+chr$(a)]?mid$(A,96,26)+mid$(A,64,26)+mid$(A,47,10)+chr$(95)
But sadly the best at 66 bytes, just plainly printing this string:
?@abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_`
Answered by steenbergh on January 25, 2021
using formule for lowercase'ing from Practical Lisp, namely (format t "~(~a~)" "ExAMpleStRINg")
(format t"~a~:*~(~a~)0123456789_"'abcdefghijklmnopqrstuvwxyz)
Ideas for improvement are welcomed.
Answered by user65167 on January 25, 2021
¬ ~Ó×
¬ ~ " Outputs characters in the range " " to `~`
Ó× " Removes all non alphanumeric characters
" Synonym of vim's :s/W//g
Answered by user41805 on January 25, 2021
puts abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
time {append s [incr i]} 9
set i 64;time {set s $s[format %c%c [incr i] [expr $i+32]]} 26
puts _0$s
time {append s [incr i]} 9
time {set s $s[format %c%c [expr [incr i]+55] [expr $i+87]]} 26
puts _0$s
Answered by sergiol on January 25, 2021
FOR I=65TO 90?CHR$(I);CHR$(I+32);
NEXT?1234567890;"_
Answered by 12Me21 on January 25, 2021
In hex:
B030B90A00AA40E2FCB05FAA40B11A5040AAE2FC5834207BF4C3
Input: DI
: pointer to an array of at least 63 bytes. Function outputs the sequence
0123456789_abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ
without any termination, since its length is constant.
Disassembly:
00: B0 30 mov al,'0'
02: B9 0A 00 mov cx,10
_00000005:
05: AA stosb ;*DI++=AL
06: 40 inc ax ;AL++
07: E2 FC loop _00000005 ;Print digits
09: B0 5F mov al,'_' ;0x5f
0B: AA stosb
0C: 40 inc ax ;AX=0x60
_0000000D:
0D: B1 1A mov cl,26
0F: 50 push ax
_00000010:
10: 40 inc ax
11: AA stosb
12: E2 FC loop _00000010 ;Print single-case letters
14: 58 pop ax
15: 34 20 xor al,020 ;Flip "case" bit
17: 7B F4 jnp _0000000D ;Repeat if "not parity", i.e. AX is back to 0x40
19: C3 ret
Answered by meden on January 25, 2021
s="abcdefghijklmnopqrstuvwxyz"print(s..s:upper().."_0123456789")
Answered by Katenkyo on January 25, 2021
Ⓒ…⩥ṻ)ŋ/W⌿
Try it here (ES6 browsers only).
Translated roughly to ES6:
String.fromCharCode(_.range(123)).replace(/W/g,'')
Answered by Mama Fun Roll on January 25, 2021
Answered by Mama Fun Roll on January 25, 2021
()->join(['a':'z','A':'Z',0:9,'_'])
Alternative solution, also 35 bytes:
()->replace(join('0':'z'),r"W","")
Answered by Mama Fun Roll on January 25, 2021
"A}_8_47<='>({a/[email protected]!356);');
Expanded:
" A }
_ 8 _ 4 7
< = ' > ( {
a / _ x . @ .
9 ! 3 5 6
) ; ' ) ;
. . . .
Output:
aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyYzZ1203568479_
Note that there is an unprintable character 0x1A
as the first byte of the program. This also makes the first row of the expanded Hexagon look sort of off. Many thanks to Martin for showing me this trick, as well as for suggesting the algorithm for printing the alphabet!
This prints the alphabet by storing a
and A
on two edges of a hexagon and the number 26 on the edge of the hexagon that touches the joint between the letters. This looks something like this:
A / a
|
26
Then it enters a loops that prints the letters and then increments them, and then decrements the number. After one iteration we would have:
B / b
|
25
And so on. The linear code for the initialisation is: 0x1A " A } a
. The linear code for the loops outside of control flow changes is: ; ) ' ; ) { ( ' =
.
Once the counter reaches zero, we follow a different path to print the numbers and an underscore. Written out linearly this is: x 3 5 6 8 4 7 9 ! ; { @
. This replaces the current memory edge's value with the number 1203568479 (note that x
's ASCII code is 120), which contains all of the decimal digits. We print out this number and then we use a neat feature of Hexagony: we print out the number mod 256 as an ASCII character. This just happens to be 95, or underscore.
Answered by FryAmTheEggman on January 25, 2021
Answered by primo on January 25, 2021
print(("0".."z").grep(~/w/).join())
Sample run:
bash-4.3$ groovy -e 'print(("0".."z").grep(~/w/).join())'
0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz
Answered by manatwork on January 25, 2021
[la1+ddsaPlc>b]dddsb96saC2scx64sa90scx47sa57scx95P
This doesn't feel very elegant.
[ # Open macro definition
la 1+ # Load a, increment
dd # Duplicate a+1 twice
sa # Store a+1 as `a'
P # Print new value of a as a character
lc>b # If c>a, do this again
] # Close macro definition
ddd sb # DDDuplicate and store as `b'
96sa C2sc x # Set lower and upper bounds (`a' and `c') for [a-z] and execute a copy of b
64sa 90sc x # Set bounds for [A-Z] and execute a copy of b
47sa 57sc x # Set bounds for [0-9] and execute a copy of b
95P # Print underscore
Answered by Joe on January 25, 2021
>++++++++[<++++++>-]>+++[<+++>-]<+[<.+>-]++[<+++>-]<+>>++++[<++++++>-]<++[<.+>-]<++++.++>>++++[<++++++>-]<++[<.+>-]
Output: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz
>++++++++[<++++++>-]
Get yourself up to 48
, the start of [0-9]
>+++[<+++>-]<+
Count up 10
[<.+>-]
Then print each character and bump it up
++[<+++>-]<+>
jump up to the start of [A-Z]
>++++[<++++++>-]<++
start a counter of 26
[<.+>-]
and print and increment 26 values
<++++.++>
jump to [a-z]
, stopping in at _
>++++[<++++++>-]<++
counter of 26
[<.+>-]
and print and increment 26 values
there's got to be a way to simplify that counter to 26 and the print happening twice, but I'm not nearly good enough with brainfuck to do so
Answered by Jack B on January 25, 2021
Answered by Lynn on January 25, 2021
target byteorder little;export main;import putchar,isalnum;foreign"C"main(){bits32 v,t;v = 48;T:t=foreign"C"isalnum(v);if (t!=0){foreign"C"putchar(v);}if (v==95){foreign"C"putchar(v);}v=v+1;if (v<123){goto T;}foreign"C"return(0);}
Ungolfed:
target byteorder little;
export main;
import putchar, isalnum;
foreign "C" main(){
bits32 v, tmp;
v = 48;
Top:
tmp = foreign "C" isalnum(v);
if (tmp != 0){
foreign "C" putchar(v);
}
if (v == 95){
foreign "C" putchar(v);
}
v=v+1;
if (v < 123) { goto Top; }
foreign "C" return (0);
}
Answered by kirbyfan64sos on January 25, 2021
echo 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_';
Answered by MonkeyZeus on January 25, 2021
+++[[<+>->++<]>]<<[-<->]<<-.+<<++[->>+.>+.<<<]<--[->>.+<<]
Initializes the tape to 3·2n, and works from there.
+++[[<+>->++<]>] initialize the tape
| 0 | 3 | 6 | 12 | 24 | 48 | 96 | 192 | 128 | 0 | 0 |
^
<<[-<->] subract 128 from 192
| 0 | 3 | 6 | 12 | 24 | 48 | 96 | 64 | 0 | 0 | 0 |
^
<<-.+<<++ ouput '_'; increment 24 twice
| 0 | 3 | 6 | 12 | 26 | 48 | 96 | 64 | 0 | 0 | 0 |
^
[->>+.>+.<<<] output aAbBcC ~ zZ
| 0 | 3 | 6 | 12 | 0 | 48 | 122 | 90 | 0 | 0 | 0 |
^
<--[->>.+<<] decrement 12 twice; output 0 ~ 9
| 0 | 3 | 6 | 0 | 0 | 58 | 122 | 90 | 0 | 0 | 0 |
^
Answered by primo on January 25, 2021
[[D2*(,]B6*(+{n.84*+}dC8*(C4*.9+mr]{nc}d
Answered by kirbyfan64sos on January 25, 2021
_ci95,(97+!26),65+!26
Answered by kirbyfan64sos on January 25, 2021
start in BASIC with PRINT "" AND USR 4e4
org 40000
dump 40000
ld b,"z"
ld a,"_"
rst 16
nchar ld a,b
cp "/"
ret z
cp ":"
jr c,ok ; print numbers
cp "A"
jr c,fnext ; in between ranges
cp "Z"+1
jr c,ok ; print A-Z
cp "a"
jr c,fnext ; in between ranges
ok rst 16
fnext djnz nchar
Hexcode
067A3E5FD778FE2FC8FE3A390CFE
413809FE5B3804FE613801D710E9
Output: _zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA9876543210
Answered by Johan Koelman on January 25, 2021
i;k(){for(;putchar(i%26+"aA0"[i++/26])^57;);puts("_");}
Output:
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
Answered by o79y on January 25, 2021
+++++++++[>+++++>+<<-]>+++.>[<+.>-]<+++++++>>+++++++++++++[<+<+.+.>>-]<<+++++.+>[<+.+.>-]
Details:
+++++++++[>+++++>+<<-]>+++. Goes to '0' while remembering a 9 for the 9 other numbers
[<+.>-] Simply prints for the next 9 characters
<+++++++> Moves it 7 space to the letters
>+++++++++++++ Saves a 13
[<+<+.+.>>-] Prints 2 char at a time while making a second '13' space
<<+++++.+> Moves 5, prints '_' and moves to the lowercases
[<+.+.>-] And again the double print
If I could have commented, I would have to improve others answers. But since I can't, I might as well post my own. As I started writing this the lowest BF one was 96 long.
Answered by Robijoe on January 25, 2021
print''.join(chr(a)for a in range(123)if chr(a).isalnum())+'_'
Explanation:
chr(a)for a in range(123)if chr(a).isalnum() # generates a list iterating through ascii
# symbols, picking just numbers alphabet characters
''.join(...)+'_' # joins a list of items with no spacing;
# appends '_' at the end
My first golfing attempt; thanks to mbomb007 for the hints
Answered by harry on January 25, 2021
(setq a 47)(loop(setq a(+ a 1))(princ(code-char a))(when(and(> a 56)(< a 64))(setq a 64))(when(and(> a 89)(< a 96))(setq a 96))(when(> a 121)(return)))(princ'_)
Answered by CocoaBean on January 25, 2021
Seq.iter(printf"%c"<<char)(95::[48..57]@[65..90]@[97..122])
Output:
_0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
Edit: missed the digits the first time
Edit2, inspired by this Haskell solution this F# snippet is 67 bytes.
Seq.zip"aA0_""zZ9_"|>Seq.iter(fun(x,y)->Seq.iter(printf"%c")[x..y])
Output:
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
Answered by asibahi on January 25, 2021
a=CharacterRange;"_"<>{48~a~57,65~a~90,97~a~122}&
Anonymous function. Takes no input and returns the string "_0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
as output. Just concatenates a few character ranges.
Answered by LegionMammal978 on January 25, 2021
>++++++[-<++++++++>]+++++[<.+.+>-]<+++++++>+++++++++++++[-<.+.+>]<++++.++>+++++++++++++[-<.+.+>]
Explanation:
>++++++[-<++++++++>] Increment to '0'
+++++[<.+.+>-] Print 10 characters (0 to 9)
<+++++++> Increment to lower upper characters
+++++++++++++[-<.+.+>] Print 26 characters (A to Z)
<++++.++> Increment to '_' Print it and move to a
+++++++++++++[-<.+.+>] Print 26 characters (a to z)
EDIT: Most straightforward solution IMHO, still shorter than the others
Answered by Hans on January 25, 2021
x=>[...new Set(Object.keys(this).join`j345689`)].join``
devtools.chrome.enabled
to true
in about:config
. (You can then open it using Ctrl + Shift + J).Answered by user2428118 on January 25, 2021
_.Range(97,26).Concat(_.Range(65,26).Concat(_.Range(48,10))).Push(95).Write("",x=>String.fromCharCode(x))
What this does is gets the char code ranges from a-z, concats with A-Z, then concats 0-9, then pushes the code for _. Write takes a delimiter and joins everything together
Library: https://github.com/mvegh1/Enumerable/blob/master/linq.js
Answered by applejacks01 on January 25, 2021
Have fun hitting enter...
Sub q()
For i = 1 To 123
If i > 47 And i < 58 Or i > 64 And i < 91 Or i > 96 And i < 123 Or i = 95 Then MsgBox Chr(i)
Next
End Sub
Answered by tjb1 on January 25, 2021
10,123,65>'[]^`'^
Answered by primo on January 25, 2021
Just plain object pascal using a set of chars. Writing a full program instead of a procedure shaves off 2 bytes. Removing the program keyword shaves 10 more bytes.
var c:char;begin for c in['a'..'z','A'..'Z','0'..'9','_']do write(c);end.
Answered by hdrz on January 25, 2021
f()->[X||X<-lists:seq(1,$z),re:run([X],"\w")/=nomatch].
f()->S=fun lists:seq/2,[$_|S($0,$9)]++S($A,$Z)++S($a,$z).
Answered by c.P.u1 on January 25, 2021
put |grep /<alnum>/,'0'..'z'
put |grep /w/,'0'..'z'
put # 「print()」 but with trailing newline
# turn the following into a Slip so that there aren't
# any spaces in the output
|
grep
/w/, # match wordchars
'0' .. 'z' # in this Range
( grep
returns a List, which puts spaces between each element when stringified )
Answered by Brad Gilbert b2gills on January 25, 2021
"[]^`""A{"Oix♂c-9ur♂$+Σ
There's probably a shorter way
Explanation:
"[]^`""A{"Oix♂c-9ur♂$+Σ
"A{"O push a list containing the ordinals of "A" and "{"
ix flatten the list, end-exclusive range
♂c convert each ordinal into its corresponding character
"[]^`" - remove non-alphabetic characters except "_"
9ur♂$+ push range(10), convert to strings, concatenate lists
Σ concatenate strings in the list
Answered by user45941 on January 25, 2021
from c in Enumerable.Range(0,123)where char.IsLetterOrDigit((char)c)|c==95 select(char)c
A LINQ expression (OK it is almost C#) where the output is an IEnumerable<char>
. You can try it with LinqPad.
Answered by aloisdg on January 25, 2021
00000000: a854be 2cbc9e 71d597 14bc56 1ad99e 713b .T.,..q....V...q;
0123456789AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz_
Try it online! Check Debug to see the generated binary code.
The binary file above has been generated by assembling the following SASM code.
add 48 ; Set cell 0 to 48 ('0').
fwd 1 ; Advance to cell 1.
add 10 ; Set cell 1 to 10.
jmp ; Set an entry marker and jump to the jnz instruction.
rwd 1 ; Retrocede to cell 0.
put ; Print its content (initially '0').
add 1 ; Increment cell 0 ('0' -> '1', etc.).
fwd 1 ; Advance to cell 1.
sub 1 ; Decrement cell 1.
jnz ; While cell 1 in non-zero, jump to 'rwd 1'.
; This loop will print "0123456789".
rwd 1 ; Retrocede to cell 0, which holds 48 + 10 = 58.
add 7 ; Set cell 0 to 65 ('A').
fwd 1 ; Advance to cell 1.
add 26 ; Set cell 1 to 26.
jmp ; Set an entry marker and jump to the jnz instruction.
rwd 1 ; Retrocede to cell 0.
put ; Print its content (initially 'A').
add 32 ; Add 32 to convert to lowercase ('A' -> 'a', etc.).
put ; Print the cell's content.
sub 31 ; Subtract 31 to switch to the next uppercase letter ('a' -> 'B', etc.).
fwd 1 ; Advance to cell 1.
sub 1 ; Decrement cell 1.
jnz ; While cell 1 in non-zero, jump to 'rwd 1'.
; This loop will print "AaBb...YyZz".
rwd 1 ; Retrocede th cell 0, which holds 65 + 26 = 91.
add 4 ; Set cell 0 to 95 ('_').
put ; Print its content.
Answered by Dennis on January 25, 2021
do(x,y)<-zip"aA0_""zZ9_";[x..y]
The expression zip "aA0_" "zZ9_"
gives the list of endpoints [('a','z'),('A','Z'),('0','9'),('_','_')]
. The do
notation takes each (x,y)
to the inclusive (x,y)->[x..y]
and concatenates the results. Thanks to Anders Kaseorg for two bytes with do
instead of >>=
.
Compare to alternatives:
do(x,y)<-zip"aA0_""zZ9_";[x..y]
zip"aA0_""zZ9_">>= (x,y)->[x..y]
f(x,y)=[x..y];f=<<zip"aA0_""zZ9_"
id=<<zipWith enumFromTo"aA0_""zZ9_"
[c|(a,b)<-zip"aA0_""zZ9_",c<-[a..b]]
f[x,y]=[x..y];f=<<words"az AZ 09 __"
Answered by xnor on January 25, 2021
Another trivial (now non-trivial) solution, but this time is BF!
Saved 14 (!) bytes thanks to @primo.
Saved 4 more bytes thanks to @primo's suggestion to generate the range backwards, and I saved another by incrementing before printing for the lowercase letters.
New (recurrence 4, 71):
+[--[<+++++++>->+<]>-]<<+++<--<-<-----<++++.+>>>[-<<.+<+.>>>]>[-<<.+>>]
Old (values, 114):
-[----->+<]>--->++++++++++>--[----->+<]>-------.++>----[---->+<]>++>++++[->++++++<]>++[-<<.+<<.+>>>>]<<<<<<[-<.+>]
Old (recurrence 1, 103):
++++++++++[[<+>->+<]>+++++++++++++++]<<[<]>>+>++++++++>>----->>-----.++<<<<<[->>>.+>>.+<<<<<]<[->>.+<<]
Old (recurrence 2, 90):
+++++++[[<+>->+<]>>+++[-<+++++++>]<]<<[<]>+++>-->->----->++++.++<<<[->>.+>.+<<<]<[->>.+<<]
Old (recurrence 3, 76):
+[[<+++++++>->+<]>+++]<<[<]>+++>-->->----->++++.++<<<[->>.+>.+<<<]<[->>.+<<]
Assumes 8 bit wrapping cells and wrapping memory. I used Try it online.
All print out _AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz0123456789
First, this part
+[--[<+++++++>->+<]>-]<<
initializes the tape with these values
[91, 70, 49, 21, 7]
^
This works because the recurrence relation I modeled basically is f(x) = 7 * (3 * x + 1)
, backwards. See @primo's Hello, World! Answer for an explanation of what a recurrence relation is.
Then, it's fairly simple to change these values to useful ones. (and print the underscore)
Code: +++<--<-<-----<++++.+
Tape: [96, 65, 48, 26, 10]
^
Then, the simple loops use the values to print the rest of characters. I save 1 byte by having an increment before the print.
>>>[-<<.+<+.>>>]>[-<<.+>>]
I really need to find a shorter sequence generation.
I found a recurrence relation that seems to work well, but there might be a shorter one w/ less hunt and peck.
I used a linear regression calculator to find what should be the shortest possible linear recurrence relation, so I should probably find some other formula if I want to improve.
@primo really improved the recurrence relation a lot, thanks.
Answered by Blue on January 25, 2021
RψΜz@~dpAHd mHLSE
RψΜz@~dpAHd mHLSE
z@~ range from 1 to 126
Μ dpAH chars of
ψ d mHLS filter all that don't match "w+" (LS)
R E join by ""
Answered by Conor O'Brien on January 25, 2021
Answered by Digital Trauma on January 25, 2021
cat(letters,LETTERS,0:9,"_",sep="")
I believe this does not count as letters
and LETTERS
are built-in constants. But this one should be fine:
cat(intToUtf8(c(65:90,97:122)),0:9,"_",sep="")
Answered by djhurio on January 25, 2021
man sh|egrep -o \w|sort -u|tr -d \n
Output on my system is:
_0123456789aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyYzZ
Thanks to Digital Trauma for helpful suggestions.
On some systems you might be able to use ascii
instead of man sh
to save a byte.
Answered by user15259 on January 25, 2021
first code golf attempt using this
>++++++[<++++++++>-]++++++++++[-<.+>]<+++++++>>+++++[<+++++>-]<+[-<.+>]<++++.++>>+++++[<+++++>-]<+[-<.+>]
ungolfed
>++++++[<++++++++>-] // p1=48
++++++++++ // p2=10
[-<.+>]< // print 0-9, p1=58, p2=0
+++++++ // p1=65
>>+++++[<+++++>-]<+ // p2=26
[-<.+>]< // print A-Z, p1=91, p2=0
++++.++ // print underscore, p1=97
>>+++++[<+++++>-]<+ // p2=26
[-<.+>] // print a-z
Answered by Kevin L on January 25, 2021
New version that's much shorter!
<?=preg_replace('/W/','',join(range(' ','z')));
Inspired by TimmyD's solution. Takes a range of all characters from
to z
, joins them into a string, then replaces all characters that match W
(which is any character not specified in this challenge) with nothing.
Old version:
0123456789_<?php for($i=64;++$i<91;)echo chr($i).chr($i+32);
Ungolfed:
0123456789_
<?php
for($i=64;++$i<91;) echo chr($i).chr($i+32);
Anything written outside the <?php
tag is considered plain text. The for loop in the PHP code echoes the uppercase and lowercase of each letter.
Answered by Business Cat on January 25, 2021
Answered by nicael on January 25, 2021
print('_',*filter(str.isalnum,map(chr,range(123))),sep='')
A full program that prints to STDOUT.
The output is: _0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
How it works
map(chr,range(123)) Yield an iterator containing all ascii characters with
code-points in [0,122]...
*filter(str.isalnum,...) ...keep characters if alphanumeric and unpack into tuple...
print('_',...,sep='') ...add underscore and print all characters with no separating
space
If string constants were allowed, the following would have been 45 bytes:
from string import*
print('_'+printable[:62])
Answered by TheBikingViking on January 25, 2021
My first attempt at codegolf!
for(var a='0';a<'~';a++){Console.Write(System.Text.RegularExpressions.Regex.IsMatch(a+"","\w")?a+"":"");}
Thanks a lot to aloisdg, AstroDan, Leaky Nun and Kevin Lau - not Kenny for all the help in comments.
Answered by Daniel on January 25, 2021
String A(char b){String B="_";for(b=48;b++<58;)B+=b;for(b=65;b++<91;)B+=b;for(b=97;b++<123)B+=b;return B;}
Returns _0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
by abusing for-loops.
Making the above function compilable costs 9 bytes, resulting in a 115-byte program:
class a{String A(char b){String B="_";for(b=48;b++<58;)B+=b;for(b=65;b++<91;)B+=b;for(b=97;b++<123)B+=b;return B;}}
The equivalent monolithic program which prints _0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
is 148 bytes long:
interface a{static void main(String[]A){char b;String B="_";for(b=48;b++<58;)B+=b;for(b=65;b++<91;)B+=b;for(b=97;b++<123)B+=b;System.out.print(B);}}
(b,B)->{B="_";for(b=48;b++<58;)B+=b;for(b=65;b++<91;)B+=b;for(b=97;b++<123)B+=b;return B;};
This is a java.util.function.BiFunction<Character, String, String>
.
Answered by dorukayhan on January 25, 2021
()=>{var r="_";for(char c='/';c<'z';)r+=char.IsLetterOrDigit(++c)?c+"":"";return r;};
C# lambda where the output is a string
.
A full string would be 69 bytes...
Code:
()=>{
var r="_";
for(char c='/';c<'z';)
r+=char.IsLetterOrDigit(++c)?c+"":"";
return r;
};
Answered by aloisdg on January 25, 2021
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
Answered by Erik the Outgolfer on January 25, 2021
Saved a byte thanks to randomra!
~.u:95,;48 65 97+i."*10,,~26
Output:
~.u:95,;48 65 97+i."*10,,~26
_0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
I wont provide an explanation per se, but will provide intermediate results.
10,,~26
10 26 26
i. b. 0
1 _ _
* b. 0
0 0 0
i."* b. 0
i."*2 3 4
0 1 0 0
0 1 2 0
0 1 2 3
i. 2
0 1
i. 3
0 1 2
i. 4
0 1 2 3
i."*10,,~26
0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
0 1 2 + i."*10,,~26
0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
48 65 97+i."*10,,~26
48 49 50 51 52 53 54 55 56 57 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48
65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122
;48 65 97+i."*10,,~26
48 49 50 51 52 53 54 55 56 57 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122
95,;48 65 97+i."*10,,~26
95 48 49 50 51 52 53 54 55 56 57 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 ...
u:95,;48 65 97+i."*10,,~26
_01234567890000000000000000ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
~.u:95,;48 65 97+i."*10,,~26
_0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
Answered by Conor O'Brien on January 25, 2021
žyLçá9Ý'_)˜J
Explanation
žyL # push [1..128]
ç # convert to char
á # keep only members of the alphabet
9Ý # push [0..9]
'_ # push underscore
)˜J # add to lists of lists, flatten and join
# implicit output
Edit: Saved 4 bytes thank to Adnan
Answered by Emigna on January 25, 2021
I modified my original alphabet attempt for this latest version. Each character is printed in a loop.
The first line is empty.
;
+T`;w`w_
Output:
_0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
Thanks to Leaky Nun for golfing 4 bytes off my latest attempt.
Answered by mbomb007 on January 25, 2021
r=range
print''.join(map(chr,r(97,123)+r(65,91)+r(48,58)))+'_'
prints:
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
This could probably be golfed more but I'm not sure how!
Answered by Cowabunghole on January 25, 2021
Requires -E
at no extra cost.
say+a.._,A.._,_..9,_
So, my original answer (below) was a bit too boring. The only thing I've managed to come up with is the above, that's exactly the same, but looks a bit more confusing... It's pretty much exactly equivalent to the below:
say a..z,A..Z,0..9,_
I like @msh210's suggestions in the comments, but they're just a bit too long!
Answered by Dom Hastings on January 25, 2021
;.;.;.;.;.;.:,,_abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789.
There's no shorter way to get all the letters, numbers and the underscore than explicitly writing them down. I also need 62 .
commands to concatenate all the chars into a single string. These are generated by starting with a single .
(on the very right), duplicating an concatenating (-> ;.
) 8 times, splitting into 64 single dots again (-> :
) and dropping two of them (-> ,,
).
Answered by nimi on January 25, 2021
-join([char[]](1..127)-match'w')
Constructs a dynamic array 1..127
, casts it as a char
array. That's fed to the -match
operator working on the regex w
, which will return all elements that match (i.e., exactly alphanumeric and underscore). We encapsulate those array elements in a -join
to bundle it up as one string. That's left on the pipeline and output is implicit.
Answered by AdmBorkBork on January 25, 2021
4 bytes off thanks to @FryAmTheEggman and @Dennis!
A,'[,_el^'_
A, e# Push range [0 1 ... 9]
'[, e# Push range of chars from 0 to "Z" ("[" minus 1)
_el e# Duplicate and convert to lowercase
^ e# Symmetric difference. This keeps letters only, both upper- and lower-case
'_ e# Push "_".
e# Implicitly display stack contents, without separators
Answered by Luis Mendo on January 25, 2021
I did it! The lame solution has been defeated by 6 bytes:
fun&123="_"| &91= &97| &58= &65| &n=str(chr n)^ &(n+1);print(&48)
Try it online! Better readable:
fun t 123 = "_"
| t 91 = t 97
| t 58 = t 65
| t n = str(chr(n)) ^ t (n+1);
print(t 48)
Keep reading to see past me whining about not having found this solution yet.
print"_0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
The sad truth so far: I didn't manage to get something shorter than this, and believe me, I've tried.
Straight forward using build-in functions:
print("_"^implode(List.filter Char.isAlphaNum(List.tabulate(123,chr))))
Generate Char list, filter, implode (char list -> string
), add _
, print:
_0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
Uses 71 bytes and is thereby 2 bytes to long to beat the lame solution. As more or less only keywords remain, I'm pretty sure this approach can't be golfed any further.
Let's build our own function!
This approach yielded multiple solutions of which the shortest one
fun&26a=""| &n a=str(chr(n+a))^ &(n+1)a;print(&16 32^ &0 65^ &0 97^"_")
also still needs 71 bytes. At least some a bit more interesting stuff is happening here. Let's name the function f
instead of &
and have closer look:
1 fun f 26 a = ""
2 | f n a = str(chr(n+a)) ^ f (n+1) a
3 ;
4 print(f 16 32 ^ f 0 65 ^ f 0 97 ^ "_")
f n a
returns a string of 26-n
consecutive ascii-chars starting at char number a
. ^
concats two strings. n
is not yet 26, get the current char, convert it into a string and append it to the (recursively build) rest of the string. f
so we can use it afterwards. 26-n? Why not do something more intuitive like
fun f 0 a = "" | f n a = str(chr(n+a)) ^ f (n-1) a; print(f 10 47 ^ f 26 64 ^ f 26 96 ^ "_")
, would nobody ask here ever.
Because on the one hand this would print
9876543210ZYXWVUTSRQPONMLKJIHGFEDCBAzyxwvutsrqponmlkjihgfedcba_
which albeit correct doesn't look very nice. However, more importantly in this case we have one 0
and two 26
and in the other case two 0
and one 26
, which saves 1 byte.
Nevertheless it's still two bytes to go to underbid the infamous solution. At least for this approach remains a tiny bit of hope to achieve this goal, some time, in a brighter future ...
But probably not.
Answered by Laikoni on January 25, 2021
7W:'W'[]YX
7W % Push 2 raised to 7, i.e. 128
: % Range [1 2 ... 128]
'W' % Push string to be used as regex pattern
[] % Push empty array
YX % Regex replace. Uses (and consumes) three inputs: source text, regex pattern,
% target text. The first input (source text) is implicitly converted to char.
% So this replaces non-word characters by nothing.
% Implicitly display
Answered by Luis Mendo on January 25, 2021
∊'w'⎕S'&'⎕UCS⍳255
prints:
0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz
Answered by Adám on January 25, 2021
Characters can be printed in any order? Don't mind if I do!
$><<(?0..?z).grep(/w/)*''
Answered by Value Ink on January 25, 2021
i1122ñYpñvHgJ|éidd@"Í×
This answer is horribly convoluted. I'll post an explanation later.
Hexdump:
00000000: 6916 1631 1b31 3232 f159 7001 f176 4867 i..1.122.Yp..vHg
00000010: 4a7c e969 6464 4022 1bcd d7 J|.idd@"...
Explanation:
Readable:
i<C-v><C-v>1<esc> "Insert the text "<C-v>1"
"<C-v> means "literal"
122ñ ñ "122 times,
Yp "Duplicate this line
<C-a> "And increment the first number on this line
vHgJ "Join every line together
|éi "Insert an 'i' at the beginning of this line
dd "Delete this line
@"<esc> "And execute it as V code.
"That will generate every ascii value from 1-123
Í× "Now remove every non-word character.
Answered by James on January 25, 2021
"_"w9yrcw"A":"Z"ycL@l:Lcw
This prints the following to STDOUT
:
_9876543210abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ
"_"w Write "_"
9y Get the list [0:1:2:3:4:5:6:7:8:9]
rcw Reverse it, concatenate into one number, write
"A":"Z"y Get the list of all uppercase letters
cL Concatenate into a single string L
@l:Lcw Concatenate L to itself lowercased and write
Answered by Fatalize on January 25, 2021
'_':['a'..'z']++['A'..'Z']++['0'..'9']
Nothing to explain here.
Answered by nimi on January 25, 2021
_=>String.fromCharCode(...Array(123).keys()).replace(/W/g,'')
Returns 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz
, so only 6 bytes shorter than a function that returns the string literal. Yes, it sucks.
Answered by Neil on January 25, 2021
->97@"123+65@"91+48@"58+"_"
This showcases the @"
operator well
Non-completing because I finally got aroudn to fixing the @"
operator. The bug was that it was generating a Cheddar range not a JS range so it couldn't properly work
The @"
operator was designed by @CᴏɴᴏʀO'Bʀɪᴇɴ, and what it does is generate a string range from LHS to RHS. When used as an unary operator, it returns the char at the given code point (like python's chr
)
->
97 @" 123 +
65 @" 91 +
48 @" 58 +
"_"
Answered by Downgoat on January 25, 2021
Call f()
without any arguments.
f(n){for(n=128;--n;)isalnum(n)|n==95&&putchar(n);}
Prints
zyxwvutsrqponmlkjihgfedcba_ZYXWVUTSRQPONMLKJIHGFEDCBA9876543210
Answered by owacoder on January 25, 2021
150m.C#P)_+s
Generates ascii 0-150 and filters by alphanumericness and adds _
to the end
Answered by Blue on January 25, 2021
95 48:57 65:90t32+vc
This is a very boring and straightforward answer.
Answered by James on January 25, 2021
s:#"w"0rk|
Finds all characters in U+0000 to U+007B that matches the regex /w/
.
Outputs 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz
.
ssrMc4"0:A[a{_`
basically generates the half-inclusive ranges required: 0-:, A-[, a-{, _-`
.
Answered by Leaky Nun on January 25, 2021
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