Print numbers from 1 to 10

Question

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 code-golf, so shortest answer in bytes wins!

Leaderboard

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Answer 1

Piet, 40 Codels

Original (10x4):

Scaled:

Code runs in a tight loop where the output is printed, followed by 32 ASCII. Previously printed number is compared to 9, and the output is used for both incrementing the counter and redirecting the code execution flow. After 10 increments, the code will get stuck in an infinite loop, performing invalid ops.

Try it online!

Answer 2

Python 3, 24 19 bytes

print(*range(1,11))

Explanation:

       range(1,11)  # numbers 1 to 10
      *             # splat to print
print(            ) # print the numbers out

Answer 3

Lua, 25 Bytes

for i=1,10 do print(i)end

Answer 4

Brachylog, 6 5 bytes

Saved one byte thanks to @mat

10yb.

Try it online!

Explanation

10y        Get the list [0:1:2:3:4:5:6:7:8:9:10]
   b.      Output is that list minus the first element

Answer 5

PHP, 20 bytes

1 2 3 4 5 6 7 8 9 10

Standard output is implicit outside of PHP tags when reading from a file or standard input.

I'm assuming that an empty separator does not meet the 'arbitrary separator' constraint.

Answer 6

Perl 5.10, 14 13 bytes

Almost the same as Perl 6:

say for 1..10

I could've done something like say"@{[1..10]}" which is the exact same thing as in Perl 6, but it is too long. :p

Each number is outputted on a separate line.

Thanks to manatwork for saving 1 byte!

Try it here!

Answer 7

Brainfuck, 66 Bytes

Sadly, Someone found a 58 Bytes solution, but as we both have different way of solving this, I'll post mine anyway.

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

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

Here's what the tape looks like at each step, the * shows where the pointer lies.

+++[>+++>++<<-] tape (0*:9:6)cell-2: counter for printing;cell-3: used for ascii-value of 1
>>+[>+++++++<-] tape (0:9:0*:49) 49 is the ascii value of 1
>[>+>+<<-<+>]   tape (0:9:49:0*:49:49) duplicate it 3 times
<--<            tape (0:9*:47:0:49:49) one will be used as a delimiter "/"
[>>>.+<<.<-]    tape (0:0*:47:0:57:49) outputs the numbers 1..9 separated by "/"
>>>>.-.         tape (0:0:47:0:49:48) outputs 10

Answer 8

Common Lisp, 29 bytes

(dotimes(n 10)(print(+ n 1)))

Answer 9

Convex, 3 bytes

A´`

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Answer 10

Swift 3 (Jun 20, 2016), 20 bytes

print(Array(1...10))

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Just doing print(1...10) (un)surprisingly prints 1...10 to STDOUT, so we need to unwrap the Range object into an Array. A Set would be shorter, but does not guarantee order (and in practice doesn't, either).

Answer 11

Befunge-93, 12 bytes

1::.9`#@_1+#

Explanation:

1             Start stack with the number 1
  :.          Print the number on top of the stack
 :  9`        Set (number on top of the stack) > 9
      #       Skip the next character, which would otherwise stop the program
        _     Is the number in the top of the stack true (i.e. not 0)?
       @      If yes, stop the program
         1+   If no, add 1 to the top of the stack
           #  Skip the next character. Since we're at the edge of the program,
              it would wrap to the beginning, so the character being skipped is
              the "1" at the left edge. That is, "don't put 1 on the stack again"

Answer 12

Julia 0.2, 11 bytes

print(1:10)

Pretty self explanatory. 1:10 is a range, and print expands it.

Answer 13

Cheddar, 7 bytes

->1|>10

Prett simple. |> is range operator. [0, 10] is:

->|>10

Answer 14

UGL, 20 bytes

cuuuuu$+$Rl$d:_lorO:

Try it online!

Answer 15

Hexagony, 37 Bytes

Minified:

........./..\..10$\("\.$...).>{<!@.._

Try It Online!

Layed Out:

   . . . .
  . . . . .
 / . . \ . .
1 0 $ \ ( " \
 . $ . . . )
  . > { < !
   @ . . _

Answer 16

Latex, 62 Bytes

\documentclass{book}\begin{document}12345678910\end{document}

Thanks @egreg

Answer 17

R, 9 bytes

cat(1:10)

cat prints, 1:10 creates the numbers 1 to 10.

Answer 18

Brain-Flak, 30 bytes

((()()()()()){}){(({})[()])}{}

Try it online!

Explanation

For a documentation of what each command does, see the GitHub page.

((()()()()()){}){(({})[()])}{}

((()()()()()){})                 push 10
                {(({})[()])}     while top != 0: pop top, push top, push top-1
                            {}   pop

Answer 19

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!

Answer 20

Python3 : 22 19 bytes

print(*range(1,11))

---

Output:

1 2 3 4 5 6 7 8 9 10

Answer 21

TrumpScript, 45 bytes

Say "1 2 3 4 5 6 7 8 9 10"!
America is great.

Don't even try using a loop... it's much much longer.

Edit: to clarify, TrumpScript uses only numbers greater than 1,000,000, so counting from 1 to 10 would require a lot of variables.

Answer 22

Dip, 2 bytes

Tl

^{Body must be at least 30 characters; you entered 16.}

Answer 23

Befunge-98, 9 bytes

1+:.:9`kq

Shortened from 11 thanks to a suggestion from ais523.

A step by step description:

1    push 1 to stack.
+    pop top two stack items and add them. push result.
     popping an empty stack returns 0, which is why
     this works the first loop around.
:    dup top.
.    print top of stack as integer.
     Funge-98 standard states that integer printing
     puts a space afterward automatically.
:    dup top.
9    push 9.
`    pop two items, first b then a.
     push 1 if a > b, else 0.
k    execute next char n times.
     n is top of stack.
     if n is 0, the next char is just skipped.
q    quit.

This program loops because Funge-98 lines wrap around.

Old Versions

11 bytes

1+:.:a\`!kq

Answer 24

QBIC, 9 bytes

[1|10:?a]

Explanation

[       ]  FOR <x> = ...NEXT; note that <x> becomes the symbol a-z, based on availability
 1|10      <x> goes from 1 to 10, an increment could be specified with ||
     :     End of FOR-loop specification
      ?a   Print the value of a

EDIT: A non-competing entry saves me one byte two bytes:

[q|z:?a

QBIC now supports variable params for FOR, and q through z are initialised in QBIC to represent 1-10, so saved a byte on the 10. Woohoo! Also, QBIC now detects any unclosed language constructs (IF/DO/FOR) at the end of a program and auto-closes 'm.

EDIT2: QBIC now has a variable number of parameters for its FOR loops. It by default assumes a loop from 1 to 10. Providing 1 parameter (y) makes it run from 1 to y, Providing 2 parameters (x and y) loops from x to y. Providing 3 parameters changes the interval: FOR a = x TO y STEP z. Using the defaults gives us this 4-byter:

[|?a

Answer 25

Python 3 (REPL): 17 Bytes

list(range(1,11))

Outputs a list of ints!

[1,2,3,4,5,6,7,8,9,10]

Try it online!

Answer 26

Dyvil, 20 bytes

for(x<-1..10)print x

21 bytes

for x<-1..10{print x}

Two ways to do the same thing: for-each loop over the range 1..10 with type-inferred loop variable x. The loop body calls the built-in print(int) function with the argument x.

---

22 bytes

print{for x<-1..10{x}}

This version calls the built-in print(AutoPrinter.()->void) method. An AutoPrinter instance is implicitly available to the {for ...} block, which is passed as an anonymous function. The AutoPrinter class has an applyStatement(any) method, which is called when an expression like x appears in a statement context. The implementation forwards to print(any), which prints the expression.

---

24 bytes

(1..10).forEach(print _)

Functional approach; creates an IntRange object and calls its forEach(int -> void) (higher order) method. The argument is an anonymous lambda function that passes the argument to the aforementioned print(int) method. In this case, print _ is syntactic sugar for the lambda expression x => print x. Again, the types are inferred by the compiler, so that it internally becomes (int x) => print x.

---

Output in all cases:

1
2
3
4
5
6
7
8
9
10

Answer 27

TI-Basic, 9 bytes

seq(I,I,1,10

Answer 28

PHP, 23 bytes

while($i<10)echo++$i,_;

prints a trailing delimiter, run with -nr.

or

<?=join(_,range(1,10));

(already posted by BusinessCat a year ago)

or

for(;$i++<10;)echo$i,_;

(by Ismael Miguel in a comment to BusinessCat´s answer)

---

21 bytes without the delimiters: while($i<10)echo++$i; or <?=join(range(1,10));

21 bytes with native array output: print_r(range(1,10)); (run with -r)

Answer 29

Brain-Flak, 30 bytes

((()()()()()){}){(({})[()])}{}

Try it online!

Answer 30

Commercial, 649 bytes

A now 1 dollar off!
B now 1 dollar off!
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B
A has been selling out worldwide!
A: made by the makers of B

Try it online!

I tried to do this in a loop, but Jumps in Commercial are weird and I couldn't get it to work properly