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

Excel, 23 22 23 bytes

=SEQUENCE(CODE("e"),,0)

Spreadsheet

Invalid Change

=SEQUENCE(CODE("e"))-1

Answer 2

Vim, 16 bytes

i0<Esc><C-a>s<C-r>=r    <C-r>"0<C-r>")

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Explanation:

i0<Esc>                                 # Insert `0`
       <C-a>                            # Increment
            s                           # Delete into register " and Insert
             <C-r>=                     # Start expression
                   r                    # Tab-autocomplete `range(`
                        <C-r>"          # 1
                              0         # 0
                               <C-r>"   # 1
                                     )  # Full expression is `range(101)`
                                        # Insert the range [0..101)

Alternatively (and more interesting, imo):

Vim, 18 bytes

iYp<C-v><C-a>0<Esc>d^@=!0
00@-

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Explanation:

iYp<C-v><C-a>0<Esc>        # Insert `Yp<C-a>0`
                   d^      # Delete `Yp<C-a>`
  @-                       # Execute `Yp<C-a>`... times
                     @=!0  #                  1
00                         #                   00

Answer 3

Java 106 bytes

class p {public static void main(String[] args){int i='A'/'A';while(i<=(int)'d')System.out.println(i++);}}}}

TIO

Answer 4

Oracle SQL, 56 bytes

select level-cos(0)from dual connect by level<ascii('f')

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Answer 5

JavaScript (V8), 30 bytes

for(i=0;+!print(i)+'00'-i++;);

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Jo king saves more bytes that I can count :)

Answer 6

!@#$%^&*()_+, 13 bytes

e($!#^$_^_
@)

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Explaination:

e($!#^$_^_
@)

e            Pushes 101 onto the stack
 (           While
  $          Swap
   !#        Output number without popping and newline
     ^       Top of stack = Top of stack + 1
      $      Swap
       _^_   Top of stack = Top of stack - 1
<newline>    Push 10
@            Output top of stack as ASCII and pop
 )           Close while

Answer 7

Labyrinth, 18 bytes

0) @
0 \(
)#":
 (!

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Push 101 () is increment and 0 is "append zero" command), and run "print stack height - 1, dup, decrement" until the top becomes zero.

Labyrinth, 19 bytes

0)
0
)}:!
 " \
@({)

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Keep track of two values a=0 and b=101, print a and increment and decrement b until b becomes zero.

Answer 8

Pure Bash, 34 + 1 = 35 bytes

Filename must be x; this is for extra one byte.

echo $[x++]
a=x
((x>${#a}00))||. x

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Answer 9

Lua (29 bytes)

for i=0,0xA..0 do print(i)end

Answer 10

Piet + ascii-piet, 39 bytes (7×6=42 codels)

Slightly cheating to use a language which isn't even text, but even the ascii-piet encoding of it contains no digits.

tlrtmE rraaD ? aaAdd? aAd ?aAk   Aletrq

This ascii-piet compiles into this piet program:

Try Piet online!

Answer 11

VBScript, 33 bytes

for i=0 to asc("d")
msgbox i
Next

Competitive answer in VBScript!

Answer 12

Javascript (Browser), 53 46 37 33 bytes

for(n=0;++n<+atob`MTAx`;)alert(n)

-15 bytes thanks to @EliteDaMyth

Answer 13

J, 12 bytes

a.i.@i.'e'"_

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Answer 14

Clojure, 24 bytes

(apply pr(range(int\e)))

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If it is acceptable that output is wrapped in parentheses, then we can remove apply for -6 bytes.

Answer 15

C# (.NET Core), 56 bytes

for(int i=0;i<=(0xb0e-0xaaa);i++){Console.WriteLine(i);}

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Answer 16

Nim, 26 bytes

for i in 0..'d'.ord:echo i

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Answer 17

Charcoal, 7 5 bytes

ＩＥ℅eι

Try it online! Link is to verbose version of code. Explanation:

   e    Literal string `e`
  ℅     ASCII code i.e. 101
 Ｅ      Map over implicit range
    ι   Current value
Ｉ       Cast to string
        Implicitly print

Answer 18

Alice, 17 bytes

aa*r\
 @Q
&d\
 O

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Explanation:

a       Push 10
 a      Push 10
  *     Pop x. Pop y. Push x * y
   r    Pop n. Push all integers from 0 to n, inclusive
    \   Switch to Ordinal mode. Redirect command flow to the southeast
        Command flow hits the boundary of the grid and is reflected to the southwest
   Q    Reverse the order of the stack
  \     Switch to Cardinal mode. Redirect command flow to the west
 d      Push the number of elements currently on the stack
&       Pop n. Add n to the iterator queue
        Command flow hits the boundary of the grid and wraps
  \     Switch to Ordinal mode. Redirect command flow to the southwest
 O      Pop s. Print s as a string followed by a newline 
          (Gets executed the number of times stored on the top of the iterator queue)
        Command flow hits the boundary of the grid and reflects to the northwest
&       Pop s. Add s to the iterator queue
          (Everything between here and the end of the program is just the command flow 
          bouncing around until it reaches the @)
        Command flow hits the boundary of the grid and reflects to the northeast
  *     Pop b. Pop a. Push the concatenation of a and b
          (Executes 0 times because & added an empty string to the iterator queue)
        Command flow hits the boundary of the grid and reflects to the southeast
   Q    Reverse the order of the stack
        Command flow hits the boundary of the grid and reflects to the southwest
        Command flow hits the boundary of the grid and reflects to the northwest
  \     Switch to Cardinal mode. Redirect command flow to the south
        Command flow hits the boundary of the grid and wraps
  *     Pop x. Pop y. Push x * y
  @     Terminate the program

Answer 19

Tcl, 46 bytes

set i 0
while \$i<[incr u]0$u {puts $i
incr i}

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Answer 20

SimpleTemplate 0.84, 45 bytes

This was fun, but quite difficult.

The code outputs all numbers from 0 to 100, with a trailing newline:

{@setC 0}{@for_ from"   "to"m"}{@echolC}{@incC}

Due to bugs in the compiler, the tab character (inside {@for_ from" "to"m"}) MUST be a real tab.

---

Ungolfed

This version should be easier to read, despite being functionally the same:

{@set counter 0}
{@for i from "  " to "m"}
    {@echo counter, EOL}
    {@inc counter}
{@/}

Closing the {@for [...]} is optional, but left here for the cleanest code possible.

---

You can try this on https://ideone.com/tLsDFn

Answer 21

Stax, 5 bytes

AJ^rJ

Run and debug it

AJ^   10 squared plus 1 (101)
   r  range from 0..n-1
    J join with spaces

Answer 22

Elixir, 42 28 bytes

(?b-?a)..?d|>Enum.join("\n")

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Answer 23

Japt, 2 bytes

Lò

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This outputs a list of numbers from 0 to 100 separated by commas.

How it works

Lò
L   -Number 100 
 ò  -Creates an inclusive range from 0 to L, and return it in the output

Answer 24

C++, 77 Bytes

#import<iostream>
main(){for(int i=0;i<=int('d');++i)  std::cout<<i<<" ";}

Here, I've used the ASCII value and ran the loop and printed the value. Simple!

Answer 25

TeX, 57 bytes

\newcount~\loop\advance~`^^A\the~ \ifnum~<`^^%\repeat\bye

Makes uses of these two tricks:

• \$\rm\TeX\$'s preprocessor runs through the code and replaces any instance of two consecutive superscript (category code 7) characters followed by a character token, and adds/subtracts 64 from its ascii code, hence ^^A becomes NUL.
• \$\rm\TeX\$ has a `backtick notation' of inputting numbers that reads the following character's ascii code instead.

Answer 26

GolfScript, 8/11 bytes

'e'{}/,`

Try it online! - 8 Bytes

Makes an array of values of e (101) elements, starting at 0, then formats with spaces. The format also has brackets at either end of the output, so it may not be valid. If not, they can be removed with 3 more bytes:

'e'{}/,' '*

Try it online! - 11 Bytes

Answer 27

V (vim), 30 bytes

i00<esc><C-a>hxpi0<esc>0"aDi0<esc>qqYp<C-a>q@a@qdd

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can definitely be improved.

Answer 28

K (oK), 4 bytes

Solution:

!"e"

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Explanation:

Scrolling through the other solutions tells me I wasn't as novel as I hoped when I came up with this.

!"e" / the solution
 "e" / ASCII 101
!    / til (i.e. range 0..n-1

Answer 29

Barrel, 9 bytes

#d(n+¶)n

Explanation:

#        // as many times as...
 d       // ...the ASCII value of 'd' (100)...
  (   )  // Create a single of instruction for the loop
   n     // print the accumulator of a number
    +    // increase the accumulator
     ¶   // print a newline
       n // print the final number

The final n is necessary because the loop only prints the numbers 0 to 99.

I could've shaved off 2 bytes by doing #e(n+¶, which would have used the ASCII value of 'e' (101) and also utilized the self-closing properties of the () instruction, but I had already assigned e to be the value of the mathematical constant \$e\$.

Answer 30

Pxem, 21 20 bytes (filename) + 0 bytes (content) = 23 21 20 bytes, requires unprintable character.

• Filename (escaped unprintable): \001.r.-.z.c.n,.o\001.+.ce.a
• Content: empty.

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With comments

XX.z
# push 1; push int(rand()*pop)
## NOTE rand() outputs 0<=n<1
## NOTE assuming NUL cannot be used for filename
.a\001.rXX.z
# while size<2 || pop!=pop; do
.a.zXX.z
  # dup; print pop; push comma; putc pop
  .a.c.n,.oXX.z
  # push 1; push pop+pop; dup; push 101
  .a\001.+.ce
# done
.a.a

---

Pxem, 3 bytes (filename) + 20 bytes (content) = 23 bytes, requires unprintable character.

• Filename: e.e
• Content (some unprintables are escaped): .c.w\001.-.e.+.n .o.d.a

With comment

e.eXX.z # push 101; call content
.a

XX.z
# dup; while pop!=0; do
.a.c.wXX.z
  # push 1; push abs(pop-pop); call content (* result of final stack will be pushed to original *)
  .a\001.-.eXX.z
  # push pop+pop; print pop; push space; print pop; return
  .a.+.n .o.dXX.z
# done; (* implicit return *)
.a.a

Try it online! (with pxem.posixism)