# Print numbers from 1 to 10

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!

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• Related (duplicate?) – Luis Mendo Jul 21 '16 at 9:07
• 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. – Peter Taylor Jul 21 '16 at 9:54
• @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.) – Martin Ender Jul 21 '16 at 10:01
• @xnor Quite frankly, I'd rather close the other challenge as a duplicate of this one. The requirement pretty much ruins it. – Dennis Jul 21 '16 at 14:09
• I can't believe every single of the (currently) 71 answers assumes the base should be decimal… – Skippy le Grand Gourou Jul 22 '16 at 15:05

# 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


## 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.

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"}

• 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 – CocoaBean Jul 23 '16 at 11:22
• @CocoaBean This IS kolmogorov-complexity – SuperJedi224 Jan 26 '17 at 18:54

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

• 1 byte shorter using k shorthand 1+(!)10 – streetster Jun 15 '17 at 22:26

# Python 3: 24 bytes

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


Simply print a list of the range.

Outputs:

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

# Microscript, 7 bytes

10c1p]h


# Microscript II, 10 bytes

0s{1+Ps}s10*


# Perl, 19 bytes

@x=1..10;print"@x";


The ".." operator can print an ascending list of numbers or letters.

So @x = A..Z; print "@x"; will print capital letters A through Z.

I saved quite a few bytes by removing all spaces.

# J-uby, 4 bytes (non-competing)

10.+


In J-uby, n.+ is the same as [*1..n].

• So, what is J-uby? Is it like Ruby's analog of Pyth, combined with J or something like that? – Zacharý Jun 19 '17 at 20:41
• @ZacharyT it's not like pyth. J-uby is an extension of ruby that gives it concise fictional programming features in the style of J. I'm now adding other methods as aliases to make it more suitable for code golfing. The most important principle is that any valid Ruby code is valid J-uby code. – Cyoce Jun 19 '17 at 20:45
• (It's not fictional, it's functional). So it's just somewhat like the Babel ES6 transpiler, in the fact that it converts J-uby code to Ruby? – Zacharý Jun 19 '17 at 20:47
• @ZacharyT auto correct is annoying. And since Ruby is awesome, you can just require 'juby' and then it is a j-uby program through metaprogramming. – Cyoce Jun 19 '17 at 20:52
• Yeah, I love how you used Ruby's wacky language features to enhance it. I wonder what someone could do with Perl this way, considering you can define new operators? – Zacharý Jun 19 '17 at 21:02

# Carrot, 20 bytes

1 2 3 4 5 6 7 8 9 10


Or if we don't need a separator for 11 bytes:

12345678910


Anything before a ^ in Carrot is placed onto the stack and then that is implicitly output at the end of the program.

With the current version of Carrot this seems to be the shortest way of doing it although I would like to post a none hard coded version if I find one.

# Charcoal, 5 bytes

Ｉ…·¹χ


Try it online!

Explanation:

Ｉ        Implicitly print the elements (casted as string) of
…·      the inclusive range
¹χ    from 1 to 10 (default value of the χ variable)


# Cubically, 16 bytes

(*1+1/1%@LUL'B)8


Try it online!

(*1+1/1%@LUL'B)8
(                 do..while loop
@          output notepad as ASCII (delimiters)
LUL'B     algorithm with period 10
)8  loop while cube unsolved


Output (? denotes an unprintable):

1?2?3?4?5?6?7?8?9   10


# Pyth, 4 2 bytes

Well, I guess I think more about legitimate functionality/usability, but this is Code Golf, where usability comes after functionality.

-2 by not formatting my output.

ST


Explanation:

S        1-indexed range from 1 - ...
T       Ten


Try It Online!

# Pyth, 4 bytes

VSTN


Explanation:

V        For...
ST      in the 1-indexed range from 1 - 10 (T)
N     Print the current item


Try it online!

Or alternatively...

VThN


Explanation:

V        For...
T       In range from 0 - 9
hN     Print (the current item + 1)


Try it online!

Pretty simple.

• Your output format can be whatever your language supports. This includes arbitrary separators (commas, semicolons, newlines, combinations of those, etc., but no digits), prefixes and postfixes (like [...]). – Tornado547 Dec 29 '17 at 7:17
• This means that ST is valid – Tornado547 Dec 29 '17 at 7:18
• @Tornado547 wow, that is odd. many thanks – Stan Strum Dec 30 '17 at 3:49

# Kotlin: 30?/31/49 bytes

### Variable of function type: 30 bytes

val a={(1..10).map(::println)}


This might be considered cheating.

Invokable via a() like the later function. Functions are first-class citizens in Kotlin, so they can be assigned to variables aswell. The type of ´val a´ is technically () -> List<Unit>, a function that takes nothing and returns a "list of nothing", but we wouldn't be on code golf using a language that doesn't have implicit types, would we? 😊

### Function: 31 bytes

fun a(){(1..10).map(::println)}

### Executable: 49 bytes

fun main(a:Array<String>){(1..10).map(::println)}


# Julia 0.6, 11 bytes

show.(1:10)


Try it online!

# Pip, 3+1 = 4 bytes

\,t


Runs with the -n flag to separate the output with newlines.

Try it online!

Explanation:

\,         Inclusive range of 1 to
t        10

• @DLosc Thanks, that shortens it considerably. But how does it owrk though? Does \, create a range-object with the values 1 to t==10 inclusive, which it auto-prints (as that range-object is the last variable touched by Pip?) – steenbergh Feb 3 '18 at 11:54
• Yep! Exactly. (It's possible this is "using features newer than the question," because I can't recall when I changed the behavior so non-infinite ranges would output like lists, but that's allowable anyway. \, is not an issue--it was added last January.) – DLosc Feb 3 '18 at 17:48

# Forte, 64 bytes

12PRINT42-41:LET42=42+1:LET11=11+3
13LET99=11
99LET12=12+3
40END


Try it online!

### How?

In Forte, you do things by redefining numbers. For example, LET 4=5 is a statement that defines 4 to be 5. From now on, anytime 4 occurs in the program, it is replaced by 5--even as the result of expressions like 2+2. There are no looping constructs in the language; control flow is accomplished by redefining line numbers.

With that introduction, let's look at what this program does.

The first line executed is line 12. This prints 42-41, which (currently) results in 1. Next, it redefines 42 to be 42+1--that is, 43. From now on, anytime 42 occurs in the program, we'll actually use the value 43. Finally, it redefines 11 to be 11+3--that is, 14. This is setup for the loop we're about to enter.

• Redefinitions: 11->14; 42->43

Line 13 is executed next. It redefines 99 to be 11--except 11 has previously been redefined, so 99 actually becomes 14.

• Redefinitions: 11,99->14; 42->43

Now, the previous instruction has an effect on the control flow, because 99 happens to be a line number. Line 99 is now considered to be line 14, and thus we execute it next (rather than line 40, which would otherwise have come next). This line redefines 12 to be 12+3--that is, 15.

• Redefinitions: 11,99->14; 12->15; 42->43

And now the next instruction is the newly christened line 15, which prints 42-41 again... except that 42 is now 43, so it actually outputs 2. It then redefines 42 43 as 44 and 11 14 as 17.

• Redefinitions: 11,14,99->17; 12->15; 42,43->44

The next instruction is line 17 (the original line 99), which redefines 15 to be 18. And so lines 12/15/18 and 99/11/14/17 keep alternating back and forth, incrementing each other and printing ever-increasing values of 42-41, until we reach line 39. This prints 10 (42 now having reached a value of 51), redefines 42 and 11 again (but that won't matter), and then we move on to the next statement, line 40--which ends the program.

Simple, right?

# 17, 48 bytes

17 was made after the challenge was released, so it is not competing, though it still scored worse than most other answers because it isn't a golfing language, or very usable at all.

Outputs 1 to 10 seperates by \x11 (ascii value 17)

0{#
1 +
:
10 @
9 >
0 @
# 
$}777{0 10 @ 0 0 @}  Block 777(first block run): Initialises value 17 to 0. Runs block 0 Block 0: Loads value 17(relies on returning of 17 from stack when empty stack), adds 1, duplicates, stores at 17, if value > 9 pushes 1, else 0, stores value at 0, loads from 17, prints number, print ascii charater 17(relies on returning of 17 from stack when empty stack again) ## Perl 6, 11 bytes Two solutions with the same count: print ^10+1  and print 1..10  Output: 1 2 3 4 5 6 7 8 9 10 Try both of them on Try It Online! # Sd, 50 Bytes ++++++++++@**********.-/.-/.-/.-/.-/.-/.-/.-/.-/.!  ### How it works: ++++++++++ | Set variable 1 to 10 @ | Mark read location ********** | Set variable 2 to 10 . | Replace the instruction in position given by variable 2 with the ascii character given by variable 1 -/. | Subtract one from each variable and repeat -/.-/.-/.-/.-/.-/.-/.-/. | Repeat some more ! | Print all values before the @ (as ascii)  # Pyt, 3 bytes 1ᴇř  1 push 1 ᴇ 10^1 ř range from 1 to 10 Implicit output  Try it online! # Javascript (potentially non-competing) As a function (6 bytes): _=>110  Full program (10 bytes): alert110  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... # Stax, 2 bytes Am  Run online Added for completeness. A Push 10 m Map over range, implicit output  ## Canvas, 4 2 bytes ＡＲ  Try it here! # Jotlin, 15 12 bytes p((0..9)+10)  Gets the numbers 0..10, makes a list out of them, and then prints the result # QUARK, Non competing 7 bytes 5.625 bytes encoded. 1 10⋯  This is incredibly simple. Just pushes 1 and 10 to the stack, and creates a inclusive range out of them. The interpreter prints out the contents of the stack when the program ends, so no print command is needed. (Plus the print command doesn't work with arrays yet, and prints out a garbled mess, so :P) # (non-competing) rig, 3 bytes rig is a work-in-progress esoteric stack-based language. I'm just having fun with looking for challenges it can already solve, with its few commands. τr+  is a valid code that prints "1,2,3,4,5,6,7,8,9,10" Try it here Why? τ - push 10 to the stack [10] r - push range(10) to the stack [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] + - increment every element [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] - implicit print  • Why is this non-competing? – pppery Jun 11 at 23:52 # KoopaScript, 78 (28, 20) bytes def i a if \%va is \%v set a 0;setath a \%va + 1;if \%va smaller 11 print \%va  This prints the numbers 1 to 10, as well as a lot of information about the currently running code (because reasons). KS doesn't actually have a way to break loops yet, so this keeps running until the interpreter is reloaded. It stops printing at 10, though. def i a - Define a function with the name i and call it 1000/NaN (i.e. many) times a second if \%va is \%v - if the variable with the name a is undefined, set a 0; - set it to 0 setath a \%va + 1; - set the variable a to <the variable with the name a> + 1 if \%va smaller 11 - if the variable with the name a is smaller than 11, print \%va - trace the variable with the name a  To avoid printing debug info, either load this in an init script, or use this slightly longer (90) version that disables logging: def i a set verbose;if \%va is \%v set a 0;setath a \%va + 1;if \%va smaller 11 print \%va set verbose; - set the variable called 'verbose' (which handles whether to trace debugging stuff) to ""  Alternatively, print "1 2 3 4 5 6 7 8 9 10"  This is only 28 bytes and technically does what the specification entails. Other commands, e.g. if, display all the numbers, but, as it's caused by an error in the program, it doesn't display if logging is disabled. Technically, 1 2 3 4 5 6 7 8 9 10  prints all the numbers with commas between them, in the debug log. That brings it down to 20. I guess it's up to the reader (stacker? golfer?) to decide which one is the most correct # Stack Cats, 28 + 4 (-nm) = 32 bytes -(:!_:!_:-_-_]>{<:_-!:]}]<)-  Try it online! Stack Cats is a reversible esoteric language by Martin Ender. All of its commands can be reversed by applying the command's mirror image, and the whole code itself must be the mirror image of itself. Since half of the code is always redundant, it has -l and -m flags which allow to omit half of the program. -m appends the mirror image to the right (except the center character), so the full program is -(:!_:!_:-_-_]>{<:_-!:]}]<)-(>[{[:!-_:>}<[_-_-:_!:_!:)-. Stack Cats operates on an infinite tape of stacks, all of which are initially empty with infinite amount of implicit zeroes at the bottom. Here is a summary for the relevant commands: • - Negate the top. • ! Apply bitwise NOT to the top. • : Exchange top two elements of the current stack. • _ Pop a, Pop b, Push b, Push b-a. • >, < Move the cursor right or left. • ], [ Move the cursor right or left, along with the top. (Also called "Push") • (, ) Enter or exit a loop if the top is strictly positive. • { Enter a loop and remember the top. • } Exit a loop if the current top is equal to the remembered top. Also, due to the limitations of the language, the input/output occurs only at the start/end of the program. At the start of the program, each input is pushed to the initial stack, which has a -1 at the bottom. At the end of the program, all values on the stack are printed from top to bottom, with optional -1 at the bottom ignored. So this program starts with a single -1 on the stack, and ends with 1..10 on the current stack. How the code works: Code Stack Description [-1] -( [1] Negate the top; Enter the loop :!_ [1 0]->[1 -1]->[1 2] Swap top two (with implicit zero); Bit-NOT; Subtract :!_ [2 1]->[2 -2]->[2 4] Same :-_ [4 2]->[4 -2]->[4 6] Swap; Unary minus; Subtract -_ [4 -6]->[4 10] Generate 10 ]> [4] [10] [*] Push 10 to a new stack; Move right again { Remember 0 <: [4] [n 0*] [...] Swap with implicit zero _-! [4] [n n-1*] [...] Generate n-1 :] [4] [n-1] [... n*] Push n to the right } Loop while n is not zero ]< Remove last zero ) Top is 1; Exit the loop -(...)- Skip the rest  There is a standard construct <(...)*(...)> which allows to ignore half of the program, in order to program more effectively by human. I used a similar one -(...)-(...)- which utilizes the initial -1, but it only works with no-input programs and the final top should be strictly positive. Ignoring half of the program is quite a waste of space, so there is an open bounty for more efficient golfing with both halves. I'm aiming for this in the long run, though it's not yet clear how to even start tackling the problem. # cQuents, 6 4 bytes #t&$


Try it online!

## Explanation

#t       Set default input (n) to 10
&      Mode: Sequence 2 (print first n items in sequence, 1-indexed)
$Each item in the sequence is the current index  # Deadfish~, 4 bytes {io}  Try it online! Increment, then output; repeat ten times. ## (K+R)eg, 6 bytes (SBCS) 6 bytes: \ ï_(.  11 bytes: 9(9|:"1-)1$


Try it online!

One byte longer

## How it works

9(9|:"1-) # A countup program from 1 to 9 and then (astonishingly) adds a 0; I have no idea how it works.
1         # Push 1
\$         # Swap top 2 items
# The stack is implicitly outputted.

• 0ėɧ 3 bytes in the new interpreter – Lyxal Sep 16 '19 at 4:24
• ėø once I add the ø operator – Lyxal Sep 16 '19 at 4:26