60
\$\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\$
17
  • 10
    \$\begingroup\$ Related (duplicate?) \$\endgroup\$
    – Luis Mendo
    Commented Jul 21, 2016 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\$ Commented Jul 21, 2016 at 9:54
  • 11
    \$\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\$ Commented Jul 21, 2016 at 10:01
  • 20
    \$\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
    Commented Jul 21, 2016 at 14:09
  • 11
    \$\begingroup\$ I can't believe every single of the (currently) 71 answers assumes the base should be decimal… \$\endgroup\$ Commented Jul 22, 2016 at 15:05

273 Answers 273

1
6 7 8
9
10
0
\$\begingroup\$

LibreLogo, 20 bytes

Code:

print set range 1 11

Output:

enter image description here

\$\endgroup\$
0
\$\begingroup\$

Triangular, 15 bytes

S]([email protected]/AS%<

Formats into this triangle:

    S
   ] (
  l . U
 @ . . i
/ A S % <
  1. The first command, S, stashes the top of stack (0) in memory. (This is only so that the U will work the first time.)
  2. ( opens a loop.
  3. U pulls memory onto stack.
  4. i increments the stack value.
  5. < directs the IP left, hitting %, which prints the top of stack as an integer.
  6. S stashes the top of stack in memory.
  7. A pushes 10 to the stack.
  8. / directs the IP up-right, hitting @, which prints the top of stack (10) as ASCII (\n).
  9. l checks if the second-to-last stack value (the counter value) is less than 10.
  10. If so, jump back to (. Otherwise, exit the program.
\$\endgroup\$
1
  • \$\begingroup\$ 14 \$\endgroup\$
    – squid
    Commented Sep 16, 2019 at 13:17
0
\$\begingroup\$

tcl, 23

time {puts [incr i]} 10

demo

\$\endgroup\$
0
\$\begingroup\$

><>, 10 bytes

llnao:9=?;

Try it online!

Explanation:

This program pushes the length of the stack instead of trying to increment the top of the stack, because l is shorter than 1+, and you can do it to an empty stack.

l             Push the length of the stack. The first time through pushes 0.
 ln           Print the length of the stack. The first time through prints 1.
   ao         Print a linefeed. (ASCII value 10)
     :9=      If the top of the stack is 9 (i.e. we just printed 10), push 1. 
              Otherwise, push 0.
        ?;    If the top of the stack is non-0, end the program.
              
              Wrap to the start and Repeat. At the end of the loop, we end up
              with one more value on the stack than we started with.
          
\$\endgroup\$
1
  • \$\begingroup\$ Whoops, I didn't see the other ><> answer that also uses the length of the stack as an incrementer. I'm going to leave this up, but you should really go upvote them if you're going to upvote me. It's even the same byte count. \$\endgroup\$ Commented Jun 19, 2017 at 18:03
0
\$\begingroup\$

8th, 18 bytes

( . cr ) 1 10 loop
\$\endgroup\$
0
\$\begingroup\$

TLDCode, 12 10 Bytes

='1'p9{c+}

Clears the stack of the default input, pushes 1 to the stack, parses it from a string to a number, reapeats c+ 9 times, which copies the top of the stack and pushes to the stop of the stack, then incremetnts it. Implicitly prints the stack with _ as the seperator.

Output

1_2_3_4_5_6_7_8_9_10

Previous attempt (12 bytes)

'1'pe9{n+e}=

Pushes 1 to the stack, parses it from a string to a number and prints, then repeats n+e 9 times, which prints a new line, increments the top of the stack and then prints the top. Then it clears the stack to avoid the implicit printing of the default input.

Output

1
2
3
4
5
6
7
8
9
10
\$\endgroup\$
0
\$\begingroup\$

Firebird, 113 bytes

The compacted select, used to calculate the score is below:

WITH RECURSIVE T AS (SELECT 1 AS I FROM RDB$DATABASE UNION ALL SELECT 1+I FROM T WHERE I < 10 ) SELECT * FROM T

The formated select:

WITH RECURSIVE T AS (
  SELECT 1 AS I
  FROM RDB$DATABASE
  UNION ALL
  SELECT 1+I
  FROM T
  WHERE I < 10
)
SELECT *
FROM T

Short explanation: I start with a select on a system table from firebird, witch always has only one record. Then, using recursion on that projection, I add the column "I" until it reaches 10.

\$\endgroup\$
4
  • \$\begingroup\$ Welcome to PPCG! Seems like a decent first post, but I think that you could improve it by giving a short explanation for how it works. \$\endgroup\$ Commented Jun 22, 2017 at 19:56
  • 1
    \$\begingroup\$ Thanks. Yes, I should have add a short explanation on my answer. Already updated it. \$\endgroup\$ Commented Jun 22, 2017 at 20:00
  • \$\begingroup\$ Can you remove some of the spaces/newlines? \$\endgroup\$ Commented Jun 22, 2017 at 20:02
  • \$\begingroup\$ @CalculatorFeline Yes I Can! Will do that now, thanks! \$\endgroup\$ Commented Jun 22, 2017 at 20:04
0
\$\begingroup\$

Python 2.7, 17 bytes

print range(1,11)

Why 11? The range() function is non-inclusive for the end argument.

repl.it

\$\endgroup\$
2
  • \$\begingroup\$ Duplicate of TuukkaX's Python2 answer posted 11 months earlier. (With sufficient privileges you could see that there are already 3 completely or partially identical deleted answers too.) \$\endgroup\$
    – manatwork
    Commented Jun 23, 2017 at 7:51
  • \$\begingroup\$ @manatwork Oops. Sorry. \$\endgroup\$
    – Joseph
    Commented Jun 23, 2017 at 16:16
0
\$\begingroup\$

Swift 3, 20 bytes

print(Array(1...10))
\$\endgroup\$
1
  • \$\begingroup\$ Welcome on Programming Puzzle & Code Golf. About the downvote on your answer: it was automatically applied by the system when @Toto edited your answer from the "First Post" review queue; but no one actually downvoted it. Happy golfing! \$\endgroup\$
    – Dada
    Commented Jun 23, 2017 at 8:44
0
\$\begingroup\$

MY (noncompeting), 3 bytes

Here's the hex:

0A 49 27

Finally, a reasonable solution. Explanation:

0A - Push 10 to the stack
49 - Pop n; Push [1 ... n]
27 - Pop n; Output n
\$\endgroup\$
0
\$\begingroup\$

Python 3, 24 bytes

print(list(range(1,11)))

Try it online!

\$\endgroup\$
0
\$\begingroup\$

Python 2, 19 Bytes

print range(11)[1:]

Try it online!

\$\endgroup\$
0
\$\begingroup\$

anyfix, 3 bytes

⁴RK

Note that TIO says that this is 5 bytes but that is because of the unicode character ; in the codepage of anyfix (which is the same as the Jelly codepage), this solution encodes to 3 bytes.

Explanation

⁴   10
 R  Range [1..10]
  K Join on spaces

Alternatively, ⁴RY separates by newlines. R⁴K and R⁴Y also work because one of the key features of anyfix is that it works with prefix, infix, and postfix (though K and Y will work regardless of what's on the stack because if the top of the stack is not a list or does not exist, then it will use the entire stack as a list). Alternatively, R⁴ or ⁴R and the @l interpreter flag will make it output [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] because @l makes it output the stack as its string representation in Python.

Try it online!

\$\endgroup\$
0
0
\$\begingroup\$

AnyDice, 10 bytes

output d10

Try it here.

This program creates and outputs a single 10 sided die. This lists out the numbers on the die (1-10) next to a bar graph of the probabilities in the standard output form.

\$\endgroup\$
0
\$\begingroup\$

Eukleides, 26 bytes

for i=1 to 10
print i
end

Beats hardcoding by 2 bytes.

\$\endgroup\$
0
\$\begingroup\$

><>, 17 Bytes

1>:nao1\
;\?(b:+/

Pretty basic:

1                       | Push 1 to the stack
 >                      | Make sure pointer is going the right direction
  :n                    | Duplicate the number and print it
    ao                  | Print a new line
      1\                | Increment, redirect pointer
      +/                |-┘ 
   (b:                  | Check that the result is less than 11
;\?                     | If it is, continue looping, else end.
\$\endgroup\$
0
\$\begingroup\$

Implicit, 10 bytes

(;.@%<10)&

Try it online!

(            do..while loop
 ;            pop (nothing initially, n<10 after that)
  .           increment (0 initially, n after that)
   @          output ASCII value of n for delimiter
    %         output n as integer
     <10      push n<10
        )    loop while top of stack truthy (n<10)
         &   exit (no implicit output)

Output (? denotes an unprintable):

?1?2?3?4?5?6?7?8    9
10
\$\endgroup\$
0
\$\begingroup\$

Bitwise, 208 191 175 bytes

-17 bytes using the registers 3 and 4 instead of function frame registers
-16 bytes using position-based jumps instead of LABELs

.A:
AND $1 $2 3
XOR $1 $2 $1
SL 3 &1 $2
JMP &-4 $2
RET $1
.G:
NOT $1 3
A $2 3 4
A 4 &1 4
SR 4 &31 4
AND 4 &1 4
RET 4
.L:
G $2 $1 3
RET 3
A 1 &1 1
OUTI 1 &1
L 1 &10 2
JMP &-4 2

Try it online!

Explanation:

.A:          # subroutine: arg3 = arg1 + arg2
.G:          # subroutine: arg3 = arg1 > arg2
.L:          # subroutine: arg3 = arg1 < arg2
A 1 &1 1     # add l1 (literal 1) to r1 (register 1)
OUTI 1 &1 -1 # output r1 as integer
L 1 &10 2    # set r2 to r1 < 10
JMP &-4 2    # jump 3 lines back in the code if r2 is nonzero

Output:

1 2 3 4 5 6 7 8 9 10 
\$\endgroup\$
0
\$\begingroup\$

FALSE, 18 17 bytes

0[1+$11-][$."
"]#

Explanation:

0 {push 0}
{conditional function}
[
 1+ {add 1}
 $11- {check if 11}
]
{loop function}
[
 $. {print number}
 "
" {print line break}
]
# {while loop (run second function while first returns true)}
\$\endgroup\$
0
\$\begingroup\$

Excel VBA, 20 Bytes

Anonymous VBE immediates window function that takes no input and outputs to the VBE Immediates window

For i=1To 10:?i:Next
\$\endgroup\$
0
\$\begingroup\$

uBASIC, 20 bytes

Anonymous function that takes no input and outputs to the console

0Forx=1To10:?x:NextX

Try it online!

\$\endgroup\$
0
\$\begingroup\$

Python3 - 19 Characters

print(*range(1,11))

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

\$\endgroup\$
0
\$\begingroup\$

Rust, 42 bytes

fn main(){for i in 1..11{print!("{},",i)}}
\$\endgroup\$
0
\$\begingroup\$

jamal, 17 characters

{@for i/1..10/i }

Sample run:

bash-4.4$ jamal.pl number.jam
1 2 3 4 5 6 7 8 9 10 
\$\endgroup\$
0
\$\begingroup\$

Javascript (potentially non-competing)

As a function (6 bytes):

_=>110

Full program (10 bytes):

alert`110`

This prints all number from 1 to 10 inclusive in binary...
If you closely read the question, there is never any mention about mandatory separators (thus allowing 123456...) nor the output's base...

\$\endgroup\$
3
  • \$\begingroup\$ Afaik, it alerts "110" literally... \$\endgroup\$
    – CreaZyp154
    Commented Feb 16, 2022 at 7:48
  • \$\begingroup\$ @CreaZyp154 1 10 is 1 to 10 in binary \$\endgroup\$
    – Brian H.
    Commented Feb 16, 2022 at 12:43
  • \$\begingroup\$ I'm confused ngl... \$\endgroup\$
    – CreaZyp154
    Commented Feb 16, 2022 at 13:34
0
\$\begingroup\$

Rattle, 6 bytes

[+p]10

Try it Online!

Loop 10 times while incrementing and printing each time

\$\endgroup\$
0
\$\begingroup\$

Pascal, 55 B

This complete program requires a processor compliant with (at least) ISO standard 7185 “Standard Pascal”. All numerical values supplied to the built‑in write/writeLn routines have an implementation‑defined minimum printing width. Considering an implementation defines this width to be 8, the program

program p(output);begin write(1,2,3,4,5,6,7,8,9,10)end.

prints

       1       2       3       4       5       6       7       8       9      10

If you are a user of the FreePascal Compiler, you need to insert a {$mode ISO} compiler directive comment or supply the ‑MISO parameter to the fpc compiler command. Otherwise the built‑in minimum width is 1, which just concatenates all numbers without any intervening space.

\$\endgroup\$
0
\$\begingroup\$

Swift, 18 bytes

print((1...10)+[])

Self-explanatory. This literally just creates a ClosedRange<Int> from 1 to 10, and then converts it to an [Int] by concatenating an empty array literal. (If we don't do this bit, it'll just output it as 1...10 instead of [1, 2, 3, 4, 5, 6, 7, 8, 9, 10].

\$\endgroup\$
0
\$\begingroup\$

APL language

⍳10
1 2 3 4 5 6 7 8 9 
\$\endgroup\$
2
  • \$\begingroup\$ Already posted by Adám then also posted by Niclas M about 7 and half years ago. \$\endgroup\$
    – manatwork
    Commented Apr 1 at 10:49
  • \$\begingroup\$ I'm exploring APL, Language it was very interesting .by solving this problems improve my skills in APL \$\endgroup\$
    – Hapisnake
    Commented Apr 9 at 2:11
0
\$\begingroup\$

Alice, 18 bytes

a!ar/ Q \;?&/ O @

Try it online!

I'll try to reorganize the code to make it denser and lower the byte count, so far no luck.

a!             # ‎⁡Push 10, pop it and put it on the tape
  ar           # ‎⁢Create a range from 0 to 10 on the stack
    /Q\;       # ‎⁣Reverse the stack and pop the 0
        ?      # ‎⁤Push the 10 from the tape onto the stack
         &/O   # ‎⁢⁡Pop 10 from the stack and repeat "pop and print" ten times
            @  # ‎⁢⁢Bye
💎

Created with the help of Luminespire.

\$\endgroup\$
1
6 7 8
9
10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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