Convert binary to unary

Question

The title says it all; Given a number in the binary (base-2) number system, output the same number expressed in unary (base-1).

You should take the binary number as a string (optionally with a separator), list, or equivalent structure of digits, using any number or digit you like for the digits. The unary output can use any aforementioned format, and it should use a single-digit throughout the number but does not have to be consistent for different input values. You may choose to assume that the binary number will have no more than 8 bits, and/or that it will always be left-padded with zero bits to a certain length.

One method you may use to achieve this task is documented at this esolangs.org article:

1. Replace all 1s with 0* (Using * as the unary digit; you may use any character.)
2. Replace all *0s with 0**.
3. Remove all 0s.

Here are some examples, using 0 and 1 for the binary digits and * for the unary digit:

• 1001 → *********
• 1010 → **********
• 00111111 → ***************************************************************
• 00000 → (empty output)

This is code-golf, so the shortest answer in each language wins.

Answer 1

sed 4.2.2, 28

:
s/1/0u/
s/u0/0uu/
t
s/0//g

Implementation algorithm as per the question.

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Answer 2

dc, 17

[q]sz2iA?^9/d0=zn

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Explanation

[ ]                # define a macro to...
 q                 #   quit (i.e. print the empty string)
   sz              # save the macro to register z
     2i            # set input radix to 2 (binary)
       A           # pop 10
        ?          # read input
         ^         # exponentiate 10 to the power of the input
          9/       # divide by 9.  This gives a number consisting of n 1s, or 0
            d      # duplicate top of stack
             0=z   # if equal to 0, call macro stored in z register to print the empty string
                n  # otherwise print the unary number

Answer 3

Julia 1.0, 24 bytes

~x="*"^parse(Int,"0b"*x)

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Generally, any base other than 10 needs to be specified: parse(Int,x,base=2). Two bytes are saved by appending a 0b integer prefix instead.

Answer 4

Brain-Flak, 42 bytes

([]){{}({}<>({}){})<>([])}<>{({}[(())])}{}

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The code works by accumulating the values of partial sums on the second stack, then it expands the second stack from the single value n, to n copies of 1.

([]) Push the stack height
{ While the stack is not empty
  {} pop the stack height
  (
    {} pop the top of the stack
    <> Jump to the second stack
    ({}){} Add the top element of the second stack twice, popping it
  ) Push the sum
  <> 
  ([]) Push new height
}
<> On the second stack:
{ While n>0
  ({} Pop n
    [(()) Push 1
    ] add -1
  ) Push n-1
}{} Pop the zero

Answer 5

Thue, 46 bytes

a::=:::
1::=0*
*0::=0**
*>::=!>
!::=~*
::=
a>

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Answer 6

Swift 5.9, 48 bytes

let f={{String.init}()("*",Int($0+"",radix:2)!)}

Answer 7

YASEPL, 65 61 49 47 45 41 39 bytes

=i=1'=l®1+`1--!a¥i,1=r$2^l*a;a,r~!+!l+[

1 is the unary digit, prompts you for input

YASEPL, 37 35 bytes

this one exits with an error but works the same.

=i=1'=l®1-i-!a¥i,1=r$2^l*a;a,r~!+?6

Answer 8

BrainChild, 73 bytes

include*;int j=0;gets()(function(int c)j=j*2+(c-48))while(j--)putchar(65)

Readable

include io.bc;
int j=0;
gets()(function(int c){
    j=j*2;
    j=j+(c-48)
})
while(j--)
    putchar('A')

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Answer 9

AWK, 48 bytes

{gsub(1,"0*");while(sub(/*0/,"0**"));gsub(0,X)}1

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Another version:

{for(;i++<=NF;)for(k=1;$i&&k++<=2^(NF-i);)printf "*"}

Answer 10

Lua, 33 bytes

print(('*'):rep(tonumber(...,2)))

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Answer 11

Scala 3, 25 bytes

"*"*Integer.parseInt(_,2)

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Answer 12

brainfuck, 78 bytes

+>>,[>+++++++[-<------->]<+<<[->+>>++<<<]<++++++[->++++++<]>>>[-<[-<.>]>]>>>,]

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Someone already did it in brainfuck and got a better score here, but I thought I'd try it out myself.

Answer 13

Japt, 2 bytes

Takes input as a binary string, outputs a string of spaces.

Íç

Try it (Footer replaces spaces with 1s)

Íç     :Implicit input of binary string
Í      :Convert to decimal
 ç     :Repeat a space that many times