Print all integers

Question

Write a program or function which will provably print all integers exactly once given infinite time and memory.

Possible outputs could be:

0, 1, -1, 2, -2, 3, -3, 4, -4, …

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -1, -2, -3, -4, -5, -6, -7, -8, -9, 10, 11, …

This is not a valid output, as this would never enumerate negative numbers:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, …

• The output must be in decimal, unless your language does not support decimal integer (in that case use the natural representation of integers your language uses).

• Your program has to work up to the numbers with the biggest magnitude of the standard integer type of your language.

• Each integer must be separated from the next using any separator (a space, a comma, a linebreak, etc.) that is not a digit nor the negative sign of your language.

• The separator must not change at any point.

• The separator can consist of multiple characters, as long as none of them is a digit nor the negative sign (e.g. ,  is as valid as just ,).

• Any supported integer must eventually be printed after a finite amount of time.

Scoring

This is code-golf, so the shortest answer in bytes wins

Leaderboard

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Answer 1

><>, 19 15 bytes

1::1$-naonao1+!

This prints the following:

0
1
-1
2
-2
3
-3

... and so on. The separator is a newline.

Re-written after reading @xnor's answer to use a version of that algorithm. Starting at n=1, the program prints 1-n and n, each followed by a newline, before incrementing n. After overflowing the maximum value the program will end with an error of something smells fishy.... Exactly when this will happen depends on the interpreter implementation.

---

Previous version:

0:nao0$-:10{0(?$~+!

Starting at 0, the program loops indefinitely. On each loop, the current value is printed along with a newline. It is then negated, and incremented if positive.

Answer 2

Pyke, 7 2 bytes

~I

Try it here!

7 bytes

oDID_)r

Try it here!

If printing +-0 is ok, oD_r

Answer 3

Pyth, 6 bytes

0f!
_

Try it online!

In string representation, this is "0f!\n_\n".

We print 0 explicitly, then use f in "count up from 1 until result truthy" mode, where the function prints the input, negates it, prints that, and then boolean negates the result so f will never halt.

Answer 4

convey, 14 bytes

}"0
 v^0
1-"-}

Try it online!

How it works (so cool):

Answer 5

APOL, 19 bytes

w(T p(∈) p(-(0 ∈)))

Pretty self-explanatory.

Answer 6

Jelly, 5 bytes

Ṅ~ṄNß

Try it online!

How it works

Ṅ~ṄNß  Main link. Argument: n. Implict argument: 0

Ṅ      Print n and a linefeed.
 ~     Apply bitwise NOT, yielding -(n + 1).
  Ṅ    Print -(n + 1) and a linefeed.
   N   Negate, yielding n + 1.
    ß  Recursively call the main link with argument n + 1.

Answer 7

SQL, 84 76 Bytes

Saved 8 Bytes thanks to steenbergh :)

Golfed:

DECLARE @n INT SET @n= 1 WHILE 1=1 BEGIN PRINT 1-@n PRINT @n SET @n=@n+1 END

Ungolfed:

DECLARE @n INT 
SET @n= 1
WHILE 1=1 
BEGIN 
PRINT 1-@n 
PRINT @n 
SET @n=@n+1
END

Prints:

0
1
-1
2
-2
3
-3
4
-4
5
-5
...

Answer 8

Husk, 2 bytes

İZ

Try it online!

I know Husk is created after the post of the challenge but I think it is still worth mentioning.

Answer 9

Prolog (SWI), 33+3 = 36 bytes

A*B:-C is B-A,writeln(C),C*(1-B).

Try it online!

Called as 0*0.

Prints 0, 1, -1, 2, -2 ... separated by newlines.

Saved 5 bytes thanks to SQB
Saved 16 bytes thanks to user3744156

Answer 10

MarioLANG, 17 bytes

:<
+"
)(
-:
>!
=#

Try it online!

Vertical loops are normally shorter in MarioLANG. This outputs:

0 -1 1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 .......

Answer 11

Hexagony, 14 bytes

Compacted:

 !~{/")!;/;"~{/

Formatted:

   ! ~ {
  / " ) !
 ; / ; " ~
  { / . .
   . . .

Try it online!

Improves upon @Sunny Pun's answer by not using the middle -> edge branches, but instead uses a more linear control flow, while delimiting with null bytes (didn't see anything against this) which are the default state of memory in Hexagony.

Answer 12

Dyalog APL, 18 17 15 14 12 bytes

{∇-⎕←1-⎕←⍵}0

Try it online!

{       Begin function
  ∇     Call recursively with new argument and repeat: -⍵-1
  -     Negate sign to make it negative again: -⍵-1
  ⎕←    Print current value: ⍵+1
  1-    Subtract it from 1, flipping the sign and incrementing the absolute value: ⍵+1
  ⎕←    Print current value: -⍵
  -     Negate sign to make it negative: -⍵
  ⍵     Argument to function, starts at 0 and increments by one each time: ⍵
}       End function
0       Call function with 0

Thanks to @Razetime & @Adám for their help golfing my original 18 bytes to 12. It started as ⎕←0⋄{∇⊃1+|⎕←-⎕←⍵}1 and is far more elegant now, not to mention more competitive!

Answer 13

Bash, with no external utilities, and no integer size limits, 210 208 bytes

echo 0
while
j=$i@
do
while
k=${j/9@/@0}
[ $k != $j ]
do
j=$k
done
j=${j/0@/1}
j=${j/1@/2}
j=${j/2@/3}
j=${j/3@/4}
j=${j/4@/5}
j=${j/5@/6}
j=${j/6@/7}
j=${j/7@/8}
j=${j/8@/9}
i=${j/@/1}
echo $i
echo -$i
done

Try it online!

If you wish to see it run with large integers, initialize i to a large integer.

Answer 14

Regenerate -a, 22 bytes

0|($2[0-9]!-?[1-9]())+

Attempt This Online!

Thanks to thejonymyster for the idea to use this language to answer this question.

In pretty-printed format:

    0
|
    (
        $2       # only matches after $2 has been set
        [0-9]    # all subsequent digits
    !
        -?       # negative or positive
        [1-9]    # first digit
        ()       # capture group $2
    )+           # Repeat the above 1 to infinity times

The first main alternative is 0. Once that has been used, it moves on to the second main alternative, which is split into two sub-alternatives. They are split using ! instead of |, so whichever choice is made will be locked-in once it no longer results in any changes of state.

The first time through, the $2[0-9] branch cannot be taken, since $2 is unset and results in a non-match. So the branch -?[1-9]() will be taken – and it results in a change of state: the capture group $2 becomes set rather than unset.

Due to the change of state, the (...!...) set of alternatives is given another chance to be chosen, if it is repeated. Since $2 is now set, the first alternative will be taken, and this locks it in – from this point forward, it is the only alternative that will be taken.

...or at least, that's my current best understanding of how it works. I don't think it's correct, otherwise 0|($4($3[0-9]![1-9]())!-?()){2,} (32 bytes) should do the same thing as the 22 byte version, but it doesn't. The ! operand is undocumented, so it'll take analyzing Regenerate's source code or asking to author to figure out how exactly it works.

...but as it turns out, 0|($2[0-9]!$3[1-9]()!-?‍()){2,} (31 chars / 33 bytes), with a zero width joiner inserted after the -?, does work, so my explanation may be accurate. The problem may just be that alternatives don't work the same way when they make a zero-width match.

Regenerate -a (staging), 15 bytes

0|-?[1-9][0-9]*

Attempt This Online!

In the latest version of the language, the obvious solution now works. Before the fix, this would only print nonnegative integers beginning with the digit 1:

Attempt This Online! - presumably this link will eventually show the same behavior as the above one, when the new version of Regenerate graduates from staging

Regenerate -a, 12 bytes (10 characters)

0|(‍+)-?#1

Attempt This Online!

If outputting invisible whitespace is allowed, this can be used to "cheat":

    0
|
    (‍+)  # $1 = Repeat U+200D (Zero Width Joiner), 1 to infinity times
    -?   # Optionally match a negative sign
    #1   # Match the length of $1 as a number in decimal

Regenerate would theoretically keep printing all integers indefinitely, but in practice it stops due to hitting a recursion limit. It can't actually repeat anything more than 994 times on Attempt This Online, for example.

Answer 15

PowerShell v2+, 26 bytes

0;for(;$i-lt2gb){$i;-++$i}

Covers all 32-bit [int] values, as that's the default number type for PowerShell. Prints 0, then loops up to 2gb (which is a special operator, not a constant, yielding 2147483648). Must be called as a full program so that $i properly defaults back to $null.

Example

PS C:\Tools\Scripts> 0;for(;$i-lt2gb){$i;-++$i}
0
-1
1
-2
2
...
2147483646
-2147483647
2147483647
-2147483648

---

Truly infinite, 34 bytes

0;for([bigint]$i=1;;$i+=1){$i;-$i}

For a truly infinite variation that will (eventually) print every single integer in existence given infinite time and memory, try the above. We simply tack on the [bigint] cast and change slightly how the loop is calculated.

Answer 16

><>, 11 bytes

lnao0l-naol

Try it online! Uses the stack length as a counter.

ln             Output length of stack
  ao           Output newline
    0l-n       Output 0 - (length of stack + 1), +1 because of the additional 0
        ao     Output newline
          l    Push length of stack, increasing the stack length by 1
               (Implicit loop since ><> is toroidal)

Answer 17

Perl, 19 bytes

perl -E 'say-$}while say$}++'

In reality will get stuck when at some huge number $} switched to floating point and starts losing precision.

This 27 byte version will really print forever:

perl -E 's//0/;say"-".++$_ while say'

Answer 18

Hexagony, 22 18 bytes

!(~2016}Q2;'Oct4!~

Try it Online!

Excuse me for keep editing.. After typing the explanations I managed to squeeze.. not so squeeze this into a 3-hexagon and there are still plenty of space to put today's date in.

Old Answer

Try it Online! (old answer)

I have a love at first sight with this language...

!(._/;'<~.2/~/!}Q/>.$>

Expanded

   ! ( . _        When n<=0, print (-ve number) and n--
  / ; ' < ~       After printing , 
 . 2 / ~ / !      If n<=0, n=-n and print (the +ve number)
} Q / > . $ >     Else n=-n and go into auto-if
 . . . . . .
  . . . . . 
   . . . .        When n>0, do nothing as the number is printed at line 3

The basic algorithm is,

Loop: print, if(n<=0) n--, n=-n, then print ,

I come into this answer with the thought of using implicit if by going out of the corner:

   ! ( . .
  . . . . .
 . . . . . .
. ~ } Q 2 ; '
 . . . . . .
  . . . . .
   ! . . .

However it is real using a lot of bytes for no-ops in putting the ! there, and in the hope of getting the ! (print) back into the main middle loop, I found it hard to print the 0 since the if(n<=0)n-- is run before the main loop for printing. So keep drawing on a whiteboard (it is easier to overwrite a byte on a whiteboard than most of other tools) I came up with the above which puts one extra ~ (negation) after branching at line 2 but saves me from using the no-ops at the end.

Anyone who can guide me how to make beautiful Hexagony explanation images?

Answer 19

q KDB+, 25 bytes

n:-0W;while[n<0W;0N!n+:1]

Set n to negative infinity.

n:-0W

Increment and output n.

0N!n+:1

Loop until n < infinity.

n<0W

Possible alternate 17 bytes

(0N!1+)\[0W>;-0W]

Uses the \ (iterate) dyadic function to apply left hand side (plus 1 then output) from negative infinity (-0W) to infinity(0W).

Only problem is that it may throw a wsfull error eventually or at the very end after outputting all the integers.

Answer 20

Wumpus, 8 bytes

=nON
=)N

Try it online!

Prints the integers in the order -1, 0, -2, 1, -3, 2, ... using linefeed separation.

Explanation

Let's look at the actual grid first:

The instruction pointer starts in the top left corner moving east and will reflect off the edge whenever it reaches a boundary of the code. Hence, this program loops through the code indefinitely, but we reuse the single O (because it's executed both before and after entering the top right corner).

So the loop body looks like this:

=nONON)=

Let's go through this:

=   Duplicate the top of the stack. Initially, this is an implicit zero,
    but in general this will be the non-negative number of each pair we print.
n   Bitwise NOT. Turns the copy of n into -n-1.
O   Output -n-1.
N   Output a linefeed.
O   Output n.
N   Output a linefeed.
)   Increment n to n+1.
=   Duplicate it (because one copy of it will be printed in the next iteration).

Answer 21

C (gcc), 40 38 36 bytes

main(i){for(;printf("%d ",i),i++;);}

Try it online!

Explanation:

The first argument to main() is argc, which is 1 if the program is run without additional parameters. When we reach the max positive integer, i wraps around and becomes negative. The program stops when i==0, after printing it.

Thanks @ceilingcat for -3 bytes

Answer 22

Turing Machine Code, 460 bytes

This prints the list in Unary so I'm hoping this still counts since, in Turing Machine Code, the concept of a 'number base' isn't built in. I'm going to mull over a base ten solution and post it separately if it exists and I find it.

0 * 1 r 2
2 * , r 3
3 * - r 4
4 * 1 r 5
5 * * r 5
5 _ , r 6
6 * * r 6
6 _ 1 * 7
7 * * l 7
7 , , l 8
8 * * l 8
8 - - r 9
9 1 @ r a
8 , , r g
g 1 @ r a
g - - r a
a * * r a
a _ 1 l b
b * * l b
b @ 1 r c
c 1 @ r a
c , , r d
d * * l d
d - - r 6
d , , r e
e * * r e
e _ , r f
f _ - r h
h * * l h
h , , l i
i * * l i
i , , r j
j 1 @ r k
j , , r m
k * * r k
m * * r m
m _ , r n
k _ 1 * o
o * * l o
o , , l p
p * * l p 
p , , r q
p @ 1 r j
q * * r q
q _ , r n
n _ 1 r 7

Try it online!

Answer 23

V (vim), 18 bytes

i0<esc>qqYp<c-a>Yi-<esc>p@qq@q

Try it online!

a small improvement by shuffling around the loop and removing the 1 from James' answer.

Answer 24

Vyxal, 5 bytes

{N…ꜝ…

Try it Online!

{     # Forever...
 N    # Negate
  …   # Print
   ꜝ  # ~
    … # Print

Answer 25

iogii, 7 5 bytes

e;1<-

run it here!!

-2 inspired by Darren Smith!

e        Expand (iterate from nil (coerces to 0)):
  1<     Is the value less than 1?
 ;  -    Subtract the original value from that.

expand and iterate technically aren't actual iteration constructs, but they can be reasoned about as such.

Answer 26

Piet + ascii-piet, 20 bytes (10×2=20 codels)

tL RrN?FmBnDbAjBsCiq

Try Piet online!

The separator is ASCII 1. You can confirm the separator is ASCII 1 by copy pasting the output from the Piet code into the input of this Python script.

Answer 27

Octave, 30 bytes

x=0;while 1;disp([x++;-x]);end

Explanation:

x=1;       % Initialize x to 1
while 1;   % Infinite loop since 1 == true
[x++,-x]   % x, then post increment and show the negative version
disp(___)  % display it
end

Prints the following sequence, starting from 0:

   0
  -1
   1
  -2
   2
  -3
   3
   .
   .

Answer 28

MATLAB 64 bytes

As a start, with nothing clever:

fprintf('0');a=int64(1);while(1) fprintf(' %d',[a -a]);a=a+1;end

Enter at the console, generates 0 1 -1 2 -2...

There must be a better way than this! Come on, folks...

Answer 29

Perl 6, 22 bytes

loop {say $--;say ++$}

Explanation:

loop {
  say   (state $ = 0)--; # prints ｢0␤｣ first time around, then ｢-1␤｣ ｢-2␤｣ etc
  say ++(state $ = 0)    # prints ｢1␤｣ first time around, then ｢2␤｣ ｢3␤｣ etc
}

Answer 30

Befunge 93, 8 bytes

:.-:0`!-

Try it online!

Explanation

Befunge's stack is thankfully filled with an implicit infinite amount of zeros, and printing a number also prints a trailing space.

:.    Print the top of the stack.
-     Subtract it from the implicit zero underneath, effectively multiplying by -1.
:0`   Check whether its greater than 0.
!     Logical NOT. Gives 0 if the current value is positive and 1 otherwise.
-     Subtract from current value.

The source code is toroidal so this program repeats indefinitely.