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

Keg, -pn, 15 9 bytes

0.1{:④±.⑨

Try it online!

Answer History

15 bytes

0. ,1{④± ,④ ,±⑨

Try it online!

Really messy I know, but it works. Integers are space seperated

Answer 2

cQuents, 7 bytes

=0:k,-k

Try it online!

Do you really need an explanation?

Answer 3

Red, 37 bytes

n: 1 forever[print[n 0 - n]n: n + 1]0

Answer 4

Symja, 38 37 bytes

i=0;While(0<1,Print(i);i++;Print(-i))

Try It Online!

To see that this will work for all integers, view this demonstration for numbers up to 100.

Answer 5

C, 55 Bytes

i=1;f(){putc('0');while(i>0){printf(",%d,%d",-i,i++);}}

Explanation

/*int*/ i = 1;
int f()
{
    putc('0');/*Print out 0 because -0 isn't a number*/
    while(i>0) /*i>0 handles overflow, ensuring all valid *
                 * integers are printed once and only once  */ {
        printf(",%d,%d",-i,i++);
        /* Unpacking the above line:
        the 1st %d prints out -i
        the 2nd %d prints out i
        i++ adds 1 to i(after it's passed to the printf)*/
    }
}

I put the comma at the beginning of the printf statement so that I didn't have to put it at the end of the puts and at the end of the printf.

Output: 0,-1,1,-2,2...

Edit: full C program version

@cleblanc made an answer that is a full C program, rather than a function, so I decided to do that too(Our answers are otherwise completely separate).

C, 58 Bytes

i=1;main(){putc('0');while(i>0){printf(",%d,%d",-i,i++);}}

Explanation

More or less the same as the original answer, but changing the function's name to main makes it a standalone program, rather than a function.

Answer 6

Julia, 24 bytes

f(n=0)=f(~n+(0>@show n))

Try it online!

Answer 7

Vyxal 5, 8 bytes

∞›(n,n⌐,

Explanation:

∞›(n,n⌐, # main program
∞›       # creates an infinite generator starting at 1
  (      # iterates over it
   n,    # prints the numbers [1, ∞)
     n⌐, # prints the numbers (-∞, 0]

Mapping would be better, but currently mapping generator output is not optimal.

Try it Online!

Answer 8

BRASCA, 20 18 bytes

-2 bytes thanks to RezNesX

nEo1[:nEo:0$-nEo}]

Try it online!

Explanation (old)

<implicit>           - The stack has infinite zeroes at the bottom
nEo                  - Print "0" and a space
   1[a            A] - Infinite loop:
      }:nEo          -   Print next positive integer and a space
           $         -   Swap to the negative number on the stack
            {:nEo    -   Print next negative integer and a space
                 $   -   Swap back to the positive number on the stack

Answer 9

AWK, 33 32 bytes

This code prints all integers in one line:

BEGIN{for(ORS=FS;;)print -i++,i}

An alternative 33-byte answer, which prints one integer per line:

BEGIN{for(;;)print n+=i++*(-1)^i}

Try it online!

Answer 10

GolfScript, 12 bytes

0{.p.~p).}do

0            # push 0 to the stack
 {           # this block will execute at least once, even though the TOS is 0
  .p         # print the current number
    .~p      # print the bitwise NOT of the current number
       )     # increments the current number
        .}do # creates an infinite loop

Because the bitwise NOT of x = (-x - 1), as the number increments, the inverse decrements. The bitwise NOT of 0 is -1, therefore it can be trivially proven that all integers are counted exactly once.

I know there's already a (longer) GolfScript solution here, I apologise if this is breaking any rules.

Answer 11

Kotlin, 49 48 bytes

(Int.MIN_VALUE..Int.MAX_VALUE).onEach(::println)

Saved one byte by using another iterator function

Try it online!

49 bytes variant

(Int.MIN_VALUE..Int.MAX_VALUE).forEach(::println)

Try it online!

Expected output

-2147483648
-2147483647
-2147483646
...
-1
0
1
2
...
2147483645
2147483646
2147483647

The program will take a while since this is only a single threaded printing and integer values reach -2147483648 to 2147483647.

Answer 12

Commodore C64 BASIC, 29 BASIC Bytes (does not end gracefully), 24 PETSCII characters with keyword abbreviations

0?i%,nO(i%),;:i%=i%+1:gO

By declaring a variable with a % identifier, you are telling Commodore BASIC to produce only integer representations of any floating point number, and this is within the range of -32768 to +32767 inclusive. The nO command, or NOT is a bitwise operator which may only be used with the same range previously mentioned (PRINT NOT (32768) for instance produces an ?ILLEGAL QUANTITY ERROR. Further reading about this is detailed here.

Because we are using line 0, we don't need to specify a line number with the GOTO command (at least not in BASIC 2.0 on the C64 and VIC-20; this may not be the same for all versions of Commodore BASIC).

The output is produced like:

0 -1 1 -2

and will continue until the maximum range is met, and produce an error.

With a graceful ending, 39 BASIC Bytes, 35 PETSCII characters with keyword abbreviations

0?i%,nO(i%),;:i%=i%+1:ifi%<32768then0

The same as above, but an IF condition is added. We can just use THEN0 rather than THEN GOTO 0 to save some typing.

Answer 13

Fortran (GFortran), 34 bytes

i=0
1 i=i+1;print*,1-i,i;goto1;end

Boringly had to waste 4 bytes setting i=0 otherwise Fortran just initialises the variable to anything

Try it online!

Answer 14

Commodore Basic, 18 bytes

1N=N+1:?N,1-N,:G┌1

PETSCII substitution: ┌ = SHIFT+O

Prints the sequence "1, 0, 2, -1, 3, -2...", with characters separated by tabs.

• Work up to the numbers with the biggest magnitude of the standard integer type? Yes. Commodore Basic integers have a range of -32768 to +32767; this program uses 40-bit floats with a 32-bit mantissa, and can print any number between -999999999 and 999999999.
• Output in decimal? Yes. Integers stored in a floating-point number are printed as integers as long as they are nine digits or less.
• Each integer separated from the next using any separator? Yes. There is always a tab character between two integers.
• The separator must not change? Yes. The separator is always exactly one tab character.
• The separator can consist of multiple characters, as long as none of them is a digit nor the negative sign? Yes. A tab character is neither a digit nor a negative sign.
• Any supported integer must eventually be printed after a finite amount of time? Yes. After about 25 minutes, it'll print the sequence 32767 -32766 32768 -32767 32769 -32768, finishing out the set of Commodore Basic-supported integers.

Answer 15

C, 64 58 Bytes

Thanks to @12431234123412341234123

f(n){printf(",%d,-%d",n,n);f(++n);}main(){f(printf("0"));}

Try it online!

A full program that terminates after printing all the integers it can represent.

Output

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

Answer 16

Uiua, 18 bytes

⍥(+⟜⊃±∵&s)∞¯1_1&s0

Explanation:

  ⍥(+⟜⊃±∵&s)∞¯1_1&s0             
# ^ ^ ^^^   ^^   ^
# | | |||   ||   └- print 0
# | | |||   |└----- [-1 1]
# | | |||   └------ do infinitely
# | | ||└---------- print each
# | | |└----------- sign of elements
# | | └------------ fork both functions
# | └-------------- increment by sign
# └---------------- repeat until

Answer 17

Swift 5.9, 32 bytes

(0...).map{print($0)
print(~$0)}

I tried desperately to merge the two calls to print(_:separator:terminator:) into one, but the fact that terminator is a newline by default seriously complicates things.

Answer 18

Oracle SQL 11.2, 77 73 72 68 bytes

SELECT CEIL((LEVEL-1)/2)*(MOD(LEVEL,2)*2-1)FROM DUAL CONNECT BY 1=1;

Answer 19

Non-wrapping Brain****, 9 bytes

Thanks to Scepheo for saving one byte!

+[.+<.->] 

Answer 20

Bash, 33 55 bytes

for((;;)){ echo -n $[-i++],$i,;((i==LLONG_MAX))&&exit;}

Unfortunately bash won't give an integer overflow error when $i exceeds the maximum size of the type used to store it. Therefore, I updated the code to include a termination criteria, that requires $LLONG_MAX to be defined in the environment or passed on the command line.

Run example: on my system bash uses signed 64-bit integers

LLONG_MAX=$[2**63-1] ./all_integers.sh

The following code is a 32 byte version (1 less) of my original post, that assumed undefined printing behavior after the integer overflow.

for((;;)){ echo -n $[-i++],$i,;}

Answer 21

C++ with Boost, 143 bytes

#include <boost/multiprecision/cpp_int.hpp>
#include <iostream>
boost::multiprecision::cpp_int i;main(){while(++i)std::cout<<1-i<<','<<i<<',';}

Not very short, but it does use an arbitrary-precision type. To test that, I tried replacing ++i with i=2*i+1, and very soon there were numbers several hundred digits long being printed correctly. This should work until the number takes up all your stack memory.

Answer 22

Swift 3 (34 bytes)

(Int.min...Int.max).map{print($0)}

Really, we should be able to get this down to 30 bytes like so:

(Int.min...Int.max).map(print)

But the compiler doesn't allow this yet.

Answer 23

k (20 bytes)

The function between curly braces is repeatedly iterated on it's result, starting at zero, until the result converges at 0W (k infinity).

{-1@'$?-1 1*x;x+1}/0

The function prints the argument and its negation and then returns incremented argument.

Answer 24

C# function 53 bytes

int j;void X(){for(;;)Console.Write($"{j++},-{j},");}

Explanation:

Create a variable "j" in class scope. This will default to have a value of zero. Write and increment value and then write the negative of that value. This will output "0,-1," on the first loop and "1,-2," on the 2nd loop.

Answer 25

Python, 40 bytes

p=print
p(0)
i=1
while 1:p(i);p(-i);i+=1

Answer 26

QBIC, 14 bytes

{?1-q ?q q=q+1

qis 1 by default in QBIC, so this prints (1-1=) 0, 1, -1, 2, -2... separated by line breaks.

Answer 27

Straw, 26 bytes

~(~$:>(,-)>:>(,)>#0+~:&):&

Try it online!

~(~$:>(,-)>:>(,)>#0+~:&):&
~ ~                 ~      Swap between the two stacks
 (                     )   Push a string literal
   $                       Unary -> decimal
    :      :         :     Duplicate
     >    > >   >          Output
      (,-)   (,)           String literal
                 #         Decimal -> unary
                  0        Push the character '0'
                   +       Concatenate
                      &  & Evaluate

It prints -0 too, not sure if it's allowed though

Answer 28

Pushy, 7 bytes

0[~#|h#

Try it online! (will eventually cut the output)

Explanation:

0     \ Push 0 as the starting counter
[     \ Infinitely:
 ~#   \   Negate counter (and print)
 |h   \   Push abs(counter)+1
 #    \   Print counter

This prints in the pattern 0 1 -1 2 -2 3 -3..., separated by newlines.

Pushy's reference implementation is in Python, which uses arbitrarily sized integers, and so there won't be any overflow if you run this for a long time. However, the integer would eventually get too large and cause a MemoryError.

Answer 29

CJam, 10 bytes

0{_n~_nz}h

Try it online!

Answer 30

Triangular, 21 bytes

()ip.%@,|<.>A@\/A|%p<

(Does not work on the TIO version yet.) Formats into this triangle:

     (
    ) i
   p . %
  @ , | <
 . > A @ \
/ A | % p <

This got really messy once I realized there had to be delimiters between the numbers.

The directionals (<,>\/) poke the IP around to get this code executed:

(i%|A@p%|A@p)

How it works:

• ( opens a loop.
• i increments the top of stack.
• % prints the top of stack as an integer.
• | negates the top of stack.
• A pushes 10 to the stack.
• @ prints the top of stack as ASCII \n.
• p pops the 10. Now the top of stack is the negated value.
• % prints the top of stack (negated value).
• | un-negates the top of stack.
• A pushes 10 to the stack again.
• @ prints as ASCII \n.
• p pops the top of stack.
• ) unconditionally jumps back to the loop.