Print 0 to 100 without 1-9 characters

Question

Print 0 to 100 without using characters 1,2,3,4,5,6,7,8,9 in your code.

Seperator of numbers can be comma, whitespace or newline.

Shortest code wins.

Answer 1

Vim, 20 bytes

i0<esc><C-A>s<C-R>=range(<C-R>"0<C-R>")

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Trying to improve on Razetime's answer, I stumbled upon the range function, which works wonders for this task. i<C-R>=range(101)\n would print the numbers we want, we just need to be a little creative to do it without 1.

Explanation

i0<esc><C-A>s<C-R>=range(<C-R>"0<C-R>")
i0<esc>                                    Insert a single 0
       <C-A>                               Increase it to a 1
            s                              Cut the 1 and go back to insert mode
             <C-R>=                        Write the result of the following function
                   range(             )    A range of numbers from 0 to N-1
                         <C-R>"            The last text that was deleted (1)
                               0           0
                                <C-R>"     1 again

Answer 2

v³, 96 93 bytes

^+++(###....+###....+++<..#+...-....###+.@#+...$)+).>+++.$#+...^##=.+###.-#+....+)<++(-+##++>

Unwrapped:

        ^ + + +
        ( # # #
        . . . .
        + # # #
. . . . + + + < . . # + . . . -
. . . . # # # + . @ # + . . . $
) + ) . > + + + . $ # + . . . ^
# # = . + # # # . - # + . . . .
        + ) < +
        + ( - +
        # # + +
        > . . .

I'm not able to provide a direct link, but here you should be able to fork the project and replace the script.txt with either of the above scripts.

Answer 3

Bash, 15

seq `dc<<<A0Kf`

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• 1 byte saved, thanks to @manatwork.

---

Previous answer:

Pure Bash (no external utilities), 23

• 9 bytes saved thanks to @ArcticKona.

eval echo {0..$[++x]00}

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Answer 4

Rust, 41 40 39 35 bytes

^{Thanks to Redwolf for -1 byte, Unrelated String for -4}

||for i in 0..b'e'{print!("{} ",i)}

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---

39 bytes

||(0..b'e').all(|i|print!("{} ",i)==())

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---

40 bytes

||(0..b'e').for_each(|i|print!("{} ",i))

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Answer 5

PICO-8, 50 45 bytes

i=0-#"0"repeat i+=#"0"?i
until#tostr(i)>#"00"

-5 bytes by replacing print with its shorthand, ?.

Demo (50 byte version; 45 byte version has same output):

Answer 6

APL (Dyalog), 19 bytes

⎕←0,⍳(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬

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≢⍬⍬⍬⍬ ⍝ This evaluates to 4
≢⍬⍬ ⍝ This evaluates to 2

⍳≢⍬⍬⍬⍬ ⍝ Evaluates to 1 2 3 4
(+/⍳≢⍬⍬⍬⍬) ⍝ Sums up previous list, 1+2+3+4 = 10
(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬ ⍝ Exponentiates previous result by 2
⍳(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬ ⍝ Generates 1 2 ... 100
0,⍳(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬ ⍝ Appends to 0 to front

Answer 7

Javascript, 48 bytes

for(a of Array("e".charCodeAt()).keys())alert(a)

Can be shorter, if you allow in reverse (36 chars)

for(i="e".charCodeAt();--i;)alert(i)

Answer 8

jq, 20 characters

range("e"|explode[])

Sample run:

bash-5.0$ jq -n 'range("e"|explode[])' | head
0
1
2
3
4
5
6
7
8
9

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Answer 9

APL(Dyalog Unicode), 8 bytes ^SBCS

⍳⎕UCS'e'

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A tradfn submission which prints with space separator.

Answer 10

Julia 1.0, 19 bytes

println.(0:0xA*0xA)

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Answer 11

bc, 17 bytes

for(;i<=A*A;i++)i

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Answer 12

K (ngn/k), 6 bytes

!0+"e"

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Uses 0+ to convert "e" to an integer, then takes the range from 0 up to, but not including, that value (101).

Answer 13

Rattle, 14 bytes

i+R`c0c0$[+i]~

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Explanation

i                    prints the value at the top of the stack (0) 
 +                   adds 1 to the value on top of the stack (which was 0, is now 1.0)
  R`                 reformats the top of the stack with the arg ` (the value at the top of the stack).
                        since ` = 1, it reformats the top of the stack as the integer 1
    c0c0             concatenates the value in storage at the current pointer (=0) to the top of the 
                        stack twice, resulting in "100"
        $            swaps the value on top of the stack (100) with the value in storage at the 
                        current pointer (0)
         [ .... ]~   loop structure: loops ~ times, where ~ = value_in_storage_at_pointer = 100
            +        adds one to the value on top of the stack
             i       prints the top of the stack as an integer

Note: the above code is based on version 1.0.* of Rattle. With the newest update (1.1.0), the code could be shortened to the following snippet (12 bytes) because the addition operator will now keep the top of the stack the same type (in this case, an integer) if possible.

i+c0c0$[+i]~

Answer 14

Whitespace, 58 bytes

Another in the theme of "this language doesn't even care about digit characters".

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The program with comments:

[Push  0 
][Label
  
][Dup 
 ][PrintNum	
 	][Push  10 	 	 
][PrintChar	
  ][Push  1 	
][Add	   ][Dup 
 ][Push  101 		  	 	
][Subtract	  	][JmpNeg
		
]

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Answer 15

MY BASIC, 29 bytes; BBC BASIC 27 bytes

for i=0 to 0xa*0xa
print i
next

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Edit: Thanks for upvoting! I have found there is a BASIC dialect with even shorter syntax for hexadecimals (it's in BBC BASIC):

for i=0 to &a*&a
print i
next

that's minus 2 bytes :)

Answer 16

Bash, 26 bytes

eval echo {$[x++]..${x}00}

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(Previously)

Bash, 28 bytes

eval echo {$((++x))..${x}00}

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Answer 17

Vim, 21 bytes

qqYP<C-x>qi0<esc><C-a>a00@q<esc>Yxx@0

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Explanation

qq       q                             Record macro q:
  Y                                     Yank the current line
   P                                    Paste a copy of it on the line above
    <C-x>                               Decrement the number under the cursor
          i0<esc>                      Insert a 0
                 <C-a>                 Increment it to 1
                      a00@q<esc>       Append 00@q
                                Y      Yank this line (100@q)
                                 xx    Delete the @q part
                                   @0  Execute the yanked text as commands
                                       (100@q executes the q macro 100 times)

Answer 18

Perl 5 (ppencode-compatible), 64 bytes

You didn't clarify that I must separate each with exactly one character, so here it is mine.

print length uc xor s qq q xor print while length ne ord qw q eq

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Explained

   # print(length) did not work for zero as $_ is not defined at then
   print length uc xor
   s qq q xor
   # delimiter
   print
while
   # equals to: length ne 101
   length ne ord qw q eq

Answer 19

Befunge-93, 16 13 bytes

Thanks to @Cinaski for saving 3 bytes with \! instead of "ba"-.

:.\!+:"d"`#@_

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Uses \! to NOT the 0 at the bottom of the stack and uses that to increment the loop, then tests if the counter is greater than d to end. Certainly not the shortest answer, but this is my first golf challenge, and I wanted to practice Befunge, which I decided to pick up yesterday. This is also my first time trying a stack-based language, and I'm having a lot of fun with it.

Answer 20

Zig, 63 66 47 72 bytes

fn a()void{for(" "**'e')|_,i|{@import("std").debug.print("{d} ",.{i});}}

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I've excluded the @import() boilerplate as it seems analogous to C's #include, which is excluded from other answers. If deemed necessary, I will add it back in.

Explanation

fn a() void {
    for (" " ** 'e') |_, i| {
        @import("std").debug.print("{d} ",.{i});
    }
}

• fn a() void Declare a function which takes no parameters and returns nothing
• for () |_, i| For every item in the array inside of (), iterate and capture the entree as _ (a throwaway variable) and the index as i
• " " ** 'e' Take the string (strings are slices, or pointer-arrays which know their length) and repeat it 'e' (101) times
• ** Requires a little bit more more explanation I think: In Zig, there is the concept of "comptime" (compile time) and runtime. ** is an operator which repeats any array literal or slice literal at comptime, because the resulting length is still known to the compiler.
• @import("std").debug.print("",.{}); Print to STDERR (I believe that's valid for this question, right?), the first argument is the formatting string, and the second is an "anonymous sctruct"/tuple with a variable number of arguments in it (Zig doesn't have var-args).
• "{d} " The format string. Zig denotes {} as the formatting characters, with d meaning a digit in this case.

Answer 21

Rust, 46 bytes

I did not see a rust solution, so here's my attempt:

||print!("{:?}",(0..b'e').collect::<Vec<_>>())

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Thanks to @ovs for pointing out the closure variant.

The range (0..b'e') is collected into a vector (using the placeholder _, letting the compiler figure out the type) and printed using the debug formatter {:?}, which "dumps" the entire vector.

The range upper bound is exclusive, and is represented using the byte literal b'e', which is equivalent to an u8 integer number literal; in this case 101 (e's ASCII value).

Answer 22

C, 53 bytes

int main(a,b){for(;a^'f';a++){printf("%d\n",a-!!a);}}

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43 42 bytes

(Thanks to Jo King♦)

k;main(){for(;k^'e';)printf("%d ",k++);}

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Answer 23

unsure, 99 bytes

ummmmm uhhhh errrrr uhhh um errrrr um um yeah err heh but um yeah err then uh okay um err then wait

It's not the shortest.

Explanation:

push 101                  ummmmm uhhhh errrrr uhhh um errrrr
push 0 to other stack     um um yeah err heh
loop                      but ... wait
  decrement                 um yeah err
  print + increment other   then uh okay um err then

Answer 24

C - 37 Bytes

i;f(){while(i<'e')printf("%d",i++);}

Ungolfed

i;
f()
{
    while(i < 'e')
        printf("%d", i++);
}

Explanation

Function to print numbers from 0 to 100 without digits. A global variable of type integer is created (so that it is automatically initialized to 0), the variable is incremented and printed 100 times through a loop which is executed while the variable is less than 'e' or 101 in ASCII.

Answer 25

Zsh, 12 bytes

seq 0 $[##d]

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• seq: count
  • from 0
  • $[##d]: to the character value of d

Alternative:

Zsh, 12 bytes

!
seq 0 $?00

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! does nothing, but fails with exit code 1; $? then retrieves the exit code.

Answer 26

Powershell, 18 bytes

0..[byte][char]'d'

Answer 27

Husk, 4 bytes

ŀc'e

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Surprised Husk doesn't have a builtin for 100.

How it works

ŀc'e - Main function, no arguments
  'e - Character literal "e"
 c   - Convert to charcode; 101
ŀ    - Lowered range; [0, 1, ..., 100]

Answer 28

MATL, 6 5 6 bytes

O'd'&:

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Explanation

O     % Letter O which is a pre-defined literal for zero
'd'   % String literal, 'd' (ASCII 100)
&:    % Create an array from 0...100
      % Implicitly print the result

Answer 29

Befunge-98 (FBBI), 13 11 bytes

.0!+::'e%j@

-1 byte thanks to @ovs

-1 byte thanks to @PizgenalFilegav

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Answer 30

Python 3, 96 bytes

import sys
n=0xb%0xa;m=0xa*0xa;p=lambda x:m if x>m else sys.stdout.write(str(x)+",")&p(x+n);p(n)

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