46
\$\begingroup\$

This might be a very simple challenge, but I am surprised it hasn't been done on code-golf yet:

Print all Integers from 1 to 10 inclusive in ascending order to standard output.

Your output format can be whatever your language supports. This includes arbitrary separators (commas, semicolons, newlines, combinations of those, etc., but no digits), and prefixes and postfixes (like [...]). However, you may not output any other numbers than 1 through 10. Your program may not take any input. Standard loopholes are disallowed.

This is , so shortest answer in bytes wins!

Leaderboard

var QUESTION_ID=86075,OVERRIDE_USER=42570;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=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;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>

\$\endgroup\$
  • 10
    \$\begingroup\$ Related (duplicate?) \$\endgroup\$ – Luis Mendo Jul 21 '16 at 9:07
  • 17
    \$\begingroup\$ If the only change is hard-coding a single parameter then that falls under the banner of "trivial change", and by the standards of this site still counts as a dupe. \$\endgroup\$ – Peter Taylor Jul 21 '16 at 9:54
  • 10
    \$\begingroup\$ @PeterTaylor The other challenge has a huge problem with the integer limits though. The way it's specified every TC language that doesn't have 64-bit integers needs to implement them. (And that affects quite a lot of languages.) \$\endgroup\$ – Martin Ender Jul 21 '16 at 10:01
  • 18
    \$\begingroup\$ @xnor Quite frankly, I'd rather close the other challenge as a duplicate of this one. The requirement pretty much ruins it. \$\endgroup\$ – Dennis Jul 21 '16 at 14:09
  • 9
    \$\begingroup\$ I can't believe every single of the (currently) 71 answers assumes the base should be decimal… \$\endgroup\$ – Skippy le Grand Gourou Jul 22 '16 at 15:05

227 Answers 227

1
4
5
6 7 8
1
\$\begingroup\$

Clojure, 20 bytes

(print (range 1 11))

Output:

(1 2 3 4 5 6 7 8 9 10)
\$\endgroup\$
1
\$\begingroup\$

C++ : 67 bytes

#include<iostream>
int main(int i){for(;i<11;std::cout<<i++<<",");}

Output:

1 2 3 4 5 6 7 8 9 10

\$\endgroup\$
1
\$\begingroup\$

K, 5 Bytes

    1+!10
1 2 3 4 5 6 7 8 9 10

Explanation;

!2    --> 0 1
!5    --> 0 1 2 3 4
!10   --> 0 1 2 3 4 5 6 7 8 9
1+!10 --> 1 2 3 4 5 6 7 8 9 10
\$\endgroup\$
1
\$\begingroup\$

C#, 54 bytes

n=>{for(int i=1;i<11;)System.Console.Write(i+++" ");};
\$\endgroup\$
  • \$\begingroup\$ Only one byte shorter than n=>{System.Console.Write("1,2,3,4,5,6,7,8,9,10");};. Silly C#! Edit: Actually, if I take out the spaces VS added, this comes in at 51... \$\endgroup\$ – BMac Oct 19 '16 at 0:57
  • \$\begingroup\$ @BMac Yeah but solutions like that are no fun! And with the above I could pass I through as an argument and force it to 1 saving 4 bytes and probably some other changes :) On second thoughts that's disallowed so no I can't do that \$\endgroup\$ – TheLethalCoder Oct 19 '16 at 8:20
1
\$\begingroup\$

RETURN, 7 bytes

1{11
}.

Try it here.

\$\endgroup\$
1
\$\begingroup\$

Batch, 35 bytes

@for /l %%i in (1,1,10)do @echo %%i

Hardcoding would have saved 10 bytes...

\$\endgroup\$
  • \$\begingroup\$ you can cut the CR/LF to save two bytes. \$\endgroup\$ – peter ferrie Mar 16 '18 at 4:15
  • \$\begingroup\$ @peterferrie But there is no CR/LF... \$\endgroup\$ – Neil Mar 16 '18 at 8:48
1
\$\begingroup\$

Fourier, 12 bytes

Prints a leading newline

10(10aX^o~X)

Try it online!

\$\endgroup\$
1
\$\begingroup\$

TI-BASIC, 9 bytes

seq(X,X,1,10

TI-BASIC is tokenized, so seq( is represented as 1 byte, as are all the other characters. The seq function is actually more powerful: the first X is an expression, and the second X is the variable that is used in the expression using the values 1 to 10, instead of using the predefined variable for X. For example, the squares of the numbers from 1 to 10 would be seq(X²,X,1,10.

\$\endgroup\$
1
\$\begingroup\$

Emmet (HTML) - 6 bytes

You'll have to excuse me, I'm new to code golf but I think I understand the concept.

{$}*10

Output:

12345678910

Alternatively, if it's required for the numbers to be seperate, adding a p infront of the braces will put it into <p> tags, like so:

<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
<p>5</p>
<p>6</p>
<p>7</p>
<p>8</p>
<p>9</p>
<p>10</p>

It also requires a tab to make it "go", I've left that out of the byte count.

Please let me know if I've stuffed up somewhere. Thanks!

\$\endgroup\$
1
\$\begingroup\$

Silicon, 3 bytes

(Silicon uses CP037, so 3 bytes, not 4.)

0Â\

Explanation:

0Â\
  \     Push a list with the numbers in the range...
0       Zero
 Â      Ten
\$\endgroup\$
1
\$\begingroup\$

Seriously/Actually, 3 bytes

9uR

Try it online: Seriously, Actually

Explanation:

9uR
9u   push 9, increment (10)
  R  range(1, 11) ([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
\$\endgroup\$
1
\$\begingroup\$

GNU sed, 22 21 bytes

c1 2 3 4 5 6 7 8 9 10

With coreutils, the code is only 7 bytes long!

eseq 10

Adding to the diversity of languages used so far, I present a sed solution. The consensus is that sed is exempt from the "no input" rule, since the script doesn't start without.

Run:

echo | sed -f script.sed
\$\endgroup\$
1
\$\begingroup\$

APL, 3 bytes

⍳10

Explanation:

⍳   range
10   10
\$\endgroup\$
  • \$\begingroup\$ Of course, your ⎕IO must be 1. \$\endgroup\$ – Zacharý Nov 15 '16 at 0:53
  • \$\begingroup\$ @ZacharyT Which is its default value \$\endgroup\$ – user41805 Mar 25 '17 at 19:48
  • \$\begingroup\$ Depending on the APL you use. \$\endgroup\$ – Zacharý Mar 26 '17 at 23:18
1
\$\begingroup\$

k, 5 bytes

1+!10

Explanation:

1+ //Projection of +, add 1 to the argument
!10 // "til" 10 - i.e. generate a list of numbers from 0 to n-1

Output:

1 2 3 4 5 6 7 8 9 10
\$\endgroup\$
  • \$\begingroup\$ You can also do 1_!11, which seems faster (at least in kdb+).. count 0..10 then drop the first element, rather than adding 1 to each item of a 10 item list. \$\endgroup\$ – streetster Jun 15 '17 at 22:34
1
\$\begingroup\$

Java 7, 50 bytes

void m{for(int i=1;i<11;System.out.println(i++));}
\$\endgroup\$
  • \$\begingroup\$ Welcome to Programming Puzzles & Code Golf! Just so you know, the downvote was cast automatically by the Community user when your answer was edited. I consider this a bug. I'd upvote, but I don't know Java and have no idea how to test your code. Could you maybe include a link to an online interpreter? \$\endgroup\$ – Dennis Oct 18 '16 at 16:38
1
\$\begingroup\$

Lithp, 64 bytes

((import "lists")(def f #::((each (seq 1 10) #N::((print N))))))

Fairly simple, but unfortunately fairly verbose. I'm counting the import because modules don't automatically load (ie, each and seq are from the lists module, and need to be imported manually.)

To use:

(
    (import "lists")(def f #::((each (seq 1 10) #N::((print N)))))
    (f)
)

Alternate Answer, 68 bytes, recursive and no modules

(def x #::((def y #N::((print N)(if (< N 10) ((y (+ N 1))))))(y 1)))

Defines a recursive function y which calls itself until N is 10.

I've made this a little more readable here:

(
    (def x #::(
        (def y #N::(
            (print N)
            (if (< N 10) (
                (y (+ N 1))
            ))
        ))
        (y 1)
    ))
    (x)
)

Sadly my language is a bit verbose, but Lisp-like language tend to do that. I'm more interested in ensuring the language can handle everything I'd want to throw at it.

\$\endgroup\$
1
\$\begingroup\$

DARTMOUTH BASIC,42 BYTES

EH, WHY NOT?

1 FOR I=1 TO 10
2 PRINT I
3 NEXT I
4 END
\$\endgroup\$
  • \$\begingroup\$ Do you need the 4 END with DARTMOUTH BASIC? Does the symbolic listing not end when there is nothing left to interpret? \$\endgroup\$ – Shaun Bebbers Jun 11 '19 at 15:48
1
\$\begingroup\$

ASMD, 5 bytes

T(i{p

Body must be at least 30 characters; you entered 20.

\$\endgroup\$
1
\$\begingroup\$

APL with any ⎕IO, 9 bytes.

1-⎕IO-⍳10

APL with ⎕IO=0, 5 bytes.

1+⍳10
\$\endgroup\$
1
\$\begingroup\$

Japt, 3 bytes

1oB

This is very simple: o creates a semi-inclusive range between two values, and B is pre-defined to 11. Thus, this creates the range [1..11), or [1,2,3,4,5,6,7,8,9,10], which is automatically sent to STDOUT.

Test it online!

\$\endgroup\$
  • \$\begingroup\$ Why not just ? (EDIT: Should have checked the date first!) \$\endgroup\$ – Shaggy Feb 20 '18 at 15:32
1
\$\begingroup\$

MATLAB, 4 Bytes

1:10

Output:

 1     2     3     4     5     6     7     8     9    10

The colon operator acts as a range function in Matlab, working from the preceding number to second one, with a default step of 1. (1:3) returns [1,2,3]

As for printing, MatLab Auto prints any line not terminated with a semicolon; Printing a line actually saves space!

\$\endgroup\$
1
\$\begingroup\$

Brainf***, 57 Bytes

++++[>++++<-]>[>++>+++>+++<<<-]+++++++++[>>+.<.<-]>>>+.-.

This is my first attempt at a program in this language. I think it's pretty optomized

\$\endgroup\$
  • \$\begingroup\$ Are you sure? I'm afraid this in best case outputs the characters with codes 1..10 (“␁␂␃␄␅␆␇␈␉␊”), not the numbers 1..10. \$\endgroup\$ – manatwork Nov 16 '16 at 20:19
  • \$\begingroup\$ Yeah, so that's my bad. The Esoteric IDE that I'm using has it print out 1,2,3...10 \$\endgroup\$ – bioweasel Nov 16 '16 at 20:22
  • 1
    \$\begingroup\$ Okay, I've redone it. Does that look better? \$\endgroup\$ – bioweasel Nov 16 '16 at 21:29
  • \$\begingroup\$ Yepp, except the . immediately after the last loop, which outputs an unnecessary ␀ character. \$\endgroup\$ – manatwork Nov 17 '16 at 11:17
  • \$\begingroup\$ Ha. Forgot about that. I was using that for troubleshooting. Thanks for the help! \$\endgroup\$ – bioweasel Nov 17 '16 at 15:18
1
\$\begingroup\$

Brainfuck, 59 Bytes

+++++[>++++++++++>++<<-]>-<+++++++++[>.+>.<<-]>---------.-.
\$\endgroup\$
1
\$\begingroup\$

Cubix, 12 bytes

\;;u>)ONo-?@

Test it online! I will add an explanation within the next few hours.

\$\endgroup\$
1
\$\begingroup\$

Pushy, 3 bytes

TR_

(non-competing as the language postdates the challenge)

It's extremely simple:

T  % Push 10
R  % Generate range (1 to 10, including both endpoints)
_  % Output representation of stack (1 2 3 4 5..)
\$\endgroup\$
1
\$\begingroup\$

VIM, using Bash and coreutils: 9 bytes

:!seq 10
\$\endgroup\$
  • 2
    \$\begingroup\$ Technically this is 10 bytes because you need to hit <CR> for the command to work. \$\endgroup\$ – James Nov 16 '16 at 21:31
  • 1
    \$\begingroup\$ @DJMcMayhem Good point. However, it does not need <CR> for the command to work if the command is passed to vim from the shell vim -c ':r!seq 10'. Of course one has to hit <CR> to execute the command from the shell, but this is then 'technically' not part of the vim command. Slightly different, but one could even argue that a C program without a newline wouldn't be valid either link. \$\endgroup\$ – ttq Dec 6 '16 at 11:18
  • \$\begingroup\$ Actually, if you were to take that approach, you wouldn't even need the colon, so it could be 8. I think technically that would be a vimscript answer (or ex?), rather than vim. You could post that as a separate answer I suppose \$\endgroup\$ – James Dec 6 '16 at 23:00
  • \$\begingroup\$ It can be even reduced to 7 bytes, if you only care about displaying the numbers 1 to 10. Then the r can be omitted too. \$\endgroup\$ – ttq Dec 7 '16 at 12:37
1
\$\begingroup\$

SmileBASIC, 19 bytes

FOR I=1TO 10?I
NEXT

Nothing to see here

\$\endgroup\$
1
\$\begingroup\$

Batch: 36 bytes

for /l %%i in (1,1,10) do (echo %%i)

Breakdown:

for: for operation in batch. Similar to C.

/l: option for the above command

%%i: define %%i, or %i in CMD, just like how you would define i in a for loop in C

in (1,1,10): pretty much "in (start, step, increment)", or in C " for (start, increment, step)".

do: well, run the code after this each time %%i is between 1-10.

(echo %%i): print %%i which is going from 1 to 10
\$\endgroup\$
1
\$\begingroup\$

Minecraft 26

say "1 2 3 4 5 6 7 8 9 10"

I know there is no special clue in it but I don't know.

Real answer:

scoreboard objective add a dummy

Repeating

give @p wool
stats entity @p set AffectedItems a @p
scoreboard player set @p a 0
clear @p wool 0 0
tellraw @p {"selector":"@p","objective":"a"}
\$\endgroup\$
  • \$\begingroup\$ IIRC you don't need the quotes when using the say command. But this is not relevant, because your answer with the say command is invalid anyway, because hardcoding the output is a standard loophole everywhere other than at [kolmogorov-complexity] challenges. See meta.codegolf.stackexchange.com/a/1063/29672 \$\endgroup\$ – CocoaBean Jul 23 '16 at 11:22
  • \$\begingroup\$ @CocoaBean This IS kolmogorov-complexity \$\endgroup\$ – SuperJedi224 Jan 26 '17 at 18:54
1
\$\begingroup\$

Q/KDB+ 8 Bytes

1+til 10

Explanation:

til 10

Outputs list of numbers 0 to 9

1+

Increments each number in the list by one

Output:1 2 3 4 5 6 7 8 9 10

\$\endgroup\$
  • \$\begingroup\$ 1 byte shorter using k shorthand 1+(!)10 \$\endgroup\$ – streetster Jun 15 '17 at 22:26
1
4
5
6 7 8

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.