Binary to decimal converter

Question

Binary to decimal converter

As far as I can see, we don't have a simple binary to decimal conversion challenge.

---

Write a program or function that takes a positive binary integer and outputs its decimal value.

You are not allowed to use any builtin base conversion functions. Integer-to-decimal functions (e.g., a function that turns 101010 into [1, 0, 1, 0, 1, 0] or "101010") are exempt from this rule and thus allowed.

Rules:

• The code must support binary numbers up to the highest numeric value your language supports (by default)
• You may choose to have leading zeros in the binary representation
• The decimal output may not have leading zeros.
• Input and output formats are optional, but there can't be any separators between digits. (1,0,1,0,1,0,1,0) is not a valid input format, but both 10101010 and (["10101010"]) are.
  • You must take the input in the "normal" direction. 1110 is 14 not 7.

Test cases:

1
1

10
2

101010
42

1101111111010101100101110111001110001000110100110011100000111
2016120520371234567

This challenge is related to a few other challenges, for instance this, this and this.

Answer 1

Clojure, 114 105 63 41 bytes

V4: 41 bytes

-22 bytes thanks to @cliffroot. Since digit is a character, it can be converted to it's code via int, then have 48 subtracted from it to get the actual number. The map was also factored out. I don't know why it seemed necessary.

#(reduce(fn[a d](+(* a 2)(-(int d)48)))%)

V3: 63 bytes

(fn[s](reduce #(+(* %1 2)%2)(map #(Integer/parseInt(str %))s)))

-42 bytes (!) by peeking at other answers. My "zipping" was evidently very naïve. Instead of raising 2 to the current place's power, then multiplying it by the current digit and adding the result to the accumulator, it just multiplies the accumulator by 2, adds on the current digit, then adds it to the accumulator. Also converted the reducing function to a macro to shave off a bit.

Thanks to @nimi, and @Adnan!

Ungolfed:

(defn to-dec [binary-str]
  (reduce (fn [acc digit]
            (+ (* acc 2) digit))
          (map #(Integer/parseInt (str %)) binary-str)))

V2: 105 bytes

#(reduce(fn[a[p d]](+ a(*(Integer/parseInt(str d))(long(Math/pow 2 p)))))0(map vector(range)(reverse %)))

-9 bytes by reversing the string so I don't need to create an awkward descending range.

V1: 114 bytes

Well, I'm certainly not winning! In my defense, this is the first program I've ever written that converts between bases, so I had to learn how to do it. It also doesn't help that Math/pow returns a double that requires converting from, and Integer/parseInt doesn't accept a character, so the digit needs to be wrapped prior to passing.

#(reduce(fn[a[p d]](+ a(*(Integer/parseInt(str d))(long(Math/pow 2 p)))))0(map vector(range(dec(count %))-1 -1)%))

Zips the string with a descending index representing the place number. Reduces over the resulting list.

Ungolfed:

(defn to-dec [binary-str]
  (reduce (fn [acc [place digit]]
            (let [parsed-digit (Integer/parseInt (str digit))
                  place-value (long (Math/pow 2 place))]
              (+ acc (* parsed-digit place-value))))
          0
          (map vector (range (dec (count binary-str)) -1 -1) binary-str)))

Answer 2

Perl, 21 19 16 + 4 = 20 bytes

-4 bytes thanks to @Dada

Run with -F -p (including the extra space after the F). Pipe values to the function using echo -n

$\+=$_+$\for@F}{

Run as echo -n "101010" | perl -F -pE '$\+=$_+$\for@F}{'

I feel this is sufficiently different from @Dada's answer that it merits its own entry.

Explanation:

-F                              #Splits the input character by character into the @F array
-p                              #Wraps the entire program in while(<>){ ... print} turning it into
while(<>){$\+=$_+$\for@F}{print}
                   for@F        #Loops through the @F array in order ($_ as alias), and...
          $\+=$_+$\             #...doubles $\, and then adds $_ to it (0 or 1)...
while(<>){              }       #...as long as there is input.
                         {print}#Prints the contents of $_ (empty outside of its scope), followed by the output record separator $\

This uses my personal algorithm of choice for binary-to-decimal conversion. Given a binary number, start your accumulator at 0, and go through its bits one by one. Double the accumulator each bit, then add the bit itself to your accumulator, and you end up with the decimal value. It works because each bit ends up being doubled the appropriate number of times for its position based on how many more bits are left in the original binary number.

Answer 3

Haskell, 31 bytes

f=foldl(\a b->2*a+(read$b:[]))0

Takes input in string format (e.g. "1111"). Produces output in integer format (e.g. 15).

:[] Converts from an element to an array -- in this chase from Char to [Char] (String). read Converts from string to whatever context it's in (in this case the context is addition, so converts to Num)

so (read$b:[]) converts b from Char to Num. a is the accumulator, so multiply that by two and add the Num version of b.

If input in the format [1,1,1,1] was allowed, the 18 byte

f=foldl((+).(2*))0

would work, but since it's not, it doesn't.

Answer 4

Dyalog APL, 12 bytes

(++⊢)/⌽⍎¨⍞

⍞ get string input

⍎¨ convert each character to number

⌽ reverse

(...)/ insert the following function between the numbers

 ++⊢ the sum of the arguments plus the right argument

---

ngn shaved 2 bytes.

Answer 5

7, 25 bytes (65 characters)

07717134200446170134271600446170001755630036117700136161177546635

Try it online!

Explanation

077                             Initialize counter to 0 (an empty section)
   17134200..77546635           Push the main loop onto the frame

Once the original code has finished executing, the frame is

||6||7134200447013427160044700017556300377700136177754635

The last section of this is the main loop, which will run repeatedly until it gets deleted. All of it except the last two commands is data and section separators, which push commands onto the frame to leave it as:

||6|| (main loop) |73426644|6734216644|6667553663||00137||54

3 outputs the last section (54) and discards the last two bars. On the first iteration, 5 is interpreted as a switch to output format 5 ("US-TTY"), which converts each pair of commands to a character. The 4 following it does not form a group, so it is ignored. On future iterations, we are already in output format 5, so 54 is interpreted as a group meaning "input character". To do this, the character from STDIN is converted into its character code (or, on EOF, -1), and 1 is added. Then, the last section (which is now the 00137) is repeated that many times. In summary:

• If the character is 0, 00137 is repeated 49 times.
• If the character is 1, 00137 is repeated 50 times.
• If there are no more characters, 00137 is repeated 0 times (i.e. deleted).
• If this was the first iteration, 00137 is left alone (i.e. repeated 1 time).

5 removes the last section (00137 repeated some number of times) from the frame and executes it, resulting in this many sections containing 6673.

The main loop has now finished executing, so the last section (which is 6673 unless we reached EOF) runs. 66 concatenates the last three sections into one (and has some other effects that don't affect the number of sections), which 73 deletes. Thus, the last three sections are deleted, and this repeats until the last section is no longer 6673.

• If we reached EOF, nothing happens.
• If this is the first iteration or the character was 0, after removing the last three sections 0 or 16 times, there is just one copy left on the frame. This copy deletes itself and the two sections before it, leaving everything up to the main loop and the section after it.
• If the character was 1, after removing the last three sections 17 times, all 50 copies and the third section after the main loop (the EOF handler) have been deleted.

Now, the last section left on the frame runs. This could, depending on the inputted character, be any of the three sections after the main loop.

If this is the first iteration or the character was 0, the section just after the main loop, 73426644, runs. 73 deletes this section, and 4 swaps the main loop with the counter behind it. This counter keeps track of the number we want to output, stored as the number of 7s and 1s minus the number of 6s and 0s. This metric has the property that it is not changed by pacification (which changes some 6s into 0s, some 7s into 1s, and sometimes inserts 7...6 around code). 2 duplicates the counter and 6 concatenates the two copies (after pacifying the second, which, as we saw, does not change the value it represents), so the value is doubled. If this was the first iteration, the counter was previously empty, and doubling it still results in an empty section. Then, 6 gets rid of the empty section inserted by 4 (and pacifies the counter again, leaving the value unchanged), and 44 swaps the counter back to its original position. This leaves two empty sections on the end of the frame, which are removed automatically, and the main loop runs again.

If the character was 1, the second section after the main loop (6734216644) runs. This is just like the section before it (explained in the last paragraph), except for two extra commands: a 6 at the start, which joins the section with the previous one before 73 deletes it, and a 1 in the middle, which adds a 7 to the counter (increasing its value by 1) after it gets doubled.

If we reached EOF, the third section after the main loop (6667553663) runs. 666 joins the last four sections (everything after the counter) into one section, to be deleted by 3. 7553 exits output format 5 by outputting 55 and deletes the section before it, leaving only ||6| (counter) on the frame. 66 pacifies the two sections and combines them, turning the 6 back into a 0 (and not changing the value of the counter). Finally, the last 3 outputs everything. The 0 at the start enters output format 0 ("Numerical output"), and the rest of the section (i.e. the counter) is converted to a number by taking the number of 7s and 1s minus the number of 6s and 0s. This value, which is the input converted from binary, is output in decimal and the program terminates.

Answer 6

Excel, 74 70 55

Trailing parens already discounted. Tested in Excel Online.

Formulae:

• A1: Input
• B1: =LEN(A1) (7)

Main Code (48):

A pretty simple "add all the powers of 2" formula:

=SUM(MID(A1,1+B1-SEQUENCE(B1),1)/2*2^SEQUENCE(B1))

Verify with:

=DECIMAL(A1,2)

Answer 7

R, 48 37 bytes

-11 bytes thanks to Dominic van Essen

sum(utf8ToInt(scan(,""))%%2*2^(31:0))

Takes input as a string, left-padded with 0s to 32 bits. Try it online!

I copied some good ideas from djhurio's much earlier R answer, so go give that an upvote too. As with the first solution there, this solution won't work for the last test case because it's too large to fit into R's default size of integer.

Explanation

To get a vector of the bits as integers 0 and 1, we use uft8ToInt to convert to a vector of character codes. This gives us a list of 48s and 49s, which we take mod 2 to get 0s and 1s.

Then, to get the appropriate powers of 2 in descending order, we construct a range from 31 down to 0. Then we convert each of those numbers to the corresponding power of two (2^).

Finally, we multiply the two vectors together and return their sum.

Answer 8

x86-16 machine code, 10 bytes

Binary:

00000000: 31d2 acd0 e811 d2e2 f9c3                 1.........

Listing:

31 D2       XOR  DX, DX         ; clear output value
        CHRLOOP:
AC          LODSB               ; load next char into AL
D0 E8       SHR  AL, 1          ; CF = LSb of ASCII char
11 D2       ADC  DX, DX         ; shift CF left into result
E2 F9       LOOP CHRLOOP        ; loop until end of string
C3          RET                 ; return to caller

Callable function, input string in [SI] length in CX. Result in DX.

Example test program I/O:

Answer 9

K (ngn/k), 11 10 bytes

(+/2 1*,)/

Try it online!

Answer 10

C, 48 bytes

i;f(char *s){for(i=0;*++s;i+=i+*s-48);return i;}

Heavily inspired by @edc45 's answer but I made some significant changes. I made 'i' a global variable to save having to typecast. I also used a for loop to save a few bytes. It does now require a single leading 0 in the binary representation.

More readable version

f(char *s){
  int i = 0; // Total decimal counter
  for(;;*++s;i+=i+*s-48); // Shift left and add 0 or 1 depending on charecter
  return i;
}

Answer 11

k, 8 bytes

Same method as the Haskell answer above.

{y+2*x}/

Example:

{y+2*x}/1101111111010101100101110111001110001000110100110011100000111b
2016120520371234567

Answer 12

JavaScript (ES7), 56 47 bytes

Reverses a binary string, then adds each digit's value to the sum.

n=>[...n].reverse().reduce((s,d,i)=>s+d*2**i,0)

Demo

f=n=>[...n].reverse().reduce((s,d,i)=>s+d*2**i,0)
document.write( f('101010') ) // 42

Answer 13

Java 7, 87 bytes

long c(String b){int a=b.length()-1;return a<0?0:b.charAt(a)-48+2*c(b.substring(0,a));}

For some reason I always go straight to recursion. Looks like an iterative solution works a bit nicer in this case...

Answer 14

JavaScript (ES6), 38

Simple is better

s=>eval("for(i=v=0;c=s[i++];)v+=+c+v")

Test

f=s=>eval("for(i=v=0;c=s[i++];)v+=+c+v")

console.log("Test 0 to 99999")
for(e=n=0;n<100000;n++)
{  
  b=n.toString(2)
  r=f(b)
  if(r!=n)console.log(++e,n,b,r)
}
console.log(e+" errors")

  

Answer 15

Turing Machine Code, 272 bytes

(Using, as usual, the morphett.info rule table syntax)

0 * * l B
B * * l C
C * 0 r D
D * * r E
E * * r A
A _ * l 1
A * * r *
1 0 1 l 1
1 1 0 l 2
1 _ * r Y
Y * * * X
X * _ r X
X _ _ * halt
2 * * l 2
2 _ _ l 3
3 * 1 r 4
3 1 2 r 4
3 2 3 r 4
3 3 4 r 4
3 4 5 r 4
3 5 6 r 4
3 6 7 r 4
3 7 8 r 4
3 8 9 r 4
3 9 0 l 3
4 * * r 4
4 _ _ r A

AKA "Yet another trivial modification of my earlier base converter programs."

Try it online, or you can also use test it using this java implementation.

Answer 16

JavaScript, 18 83 bytes

f=n=>parseInt(n,2)

f=n=>n.split('').reverse().reduce(function(x,y,i){return(+y)?x+Math.pow(2,i):x;},0)

Demo

f=n=>n.split('').reverse().reduce(function(x,y,i){return(+y)?x+Math.pow(2,i):x;},0)
document.write(f('1011')) // 11

Answer 17

MATL, 8, 7 bytes

"@oovsE

Try it online!

One byte saved thanks to @LuisMendo!

Alternate approach: (9 bytes)

ootn:PW*s

Answer 18

Befunge, 20 18 bytes

Input must be terminated with EOF rather than EOL (this lets us save a couple of bytes)

>+~:0`v
^*2\%2_$.@

Try it online!

Explanation

>             The stack is initially empty, the equivalent of all zeros.
 +            So the first pass add just leaves zero as the current total. 
  ~           Read a character from stdin to the top of the stack.
   :0`        Test if greater than 0 (i.e. not EOF)
      _       If true (i.e > 0) go left.
    %2        Modulo 2 is a shortcut for converting the character to a numeric value.
   \          Swap to bring the current total to the top of the stack.
 *2           Multiply the total by 2.
^             Return to the beginning of the loop,
 +            This time around add the new digit to the total.

                ...on EOF we go right...
       $      Drop the EOF character from the stack.
        .     Output the calculated total.
         @    Exit.

Answer 19

Ruby, 37 bytes

ruby -e 'o=0;gets.each_byte{|i|o+=o+i%2};p o/2'
         1234567890123456789012345678901234567

This depends on the terminating \n (ASCII decimal 10) being zero modulo 2 (and on ASCII 0 and 1 being 0 and 1 mod two, respectively, which thankfully they are).

Answer 20

아희(Aheui), 40 bytes

아빟뱐썩러숙
뎌반뗘희멍파퍄

Accepts a string composed of 1s and 0s.

To try online

Since the online Aheui interpreter does not allow arbitrary-length strings as inputs, this alternative code must be used (identical code with slight modifications):

Add the character 벟 at the end of the first line (after 우) length(n)-times.

어우
우어
뱐썩러숙
번댜펴퍼망희땨

If the input is 10110, the first line would be 어우벟벟벟벟벟.

When prompted for an input, do NOT type quotation marks. (i.e. type 10110, not "10110")

Try it here! (copy and paste the code)

Answer 21

ClojureScript, 36 bytes

(fn[x](reduce #(+(* 2 %)(int %2))x))

or

#(reduce(fn[a n](+(* 2 a)(int n)))%)

The straightforward reduction. Takes a string as input.

Answer 22

Minkolang v0.15, 23 19 bytes

n6ZrI[2%2i;*1R]$+N.

Try it online!

Explanation

n                             gets input in the form of a number
 6Z                           converts to string (so that it is split into an array)
   r                          reverses it
    I                         gets the stack length
     [        ]               for loop with the stack's length as the number of iterations
      2%                       gets the modulo of the ascii value
                               1 =(string conversion)> 49 =(after modulo)> 1
                               0 =(string conversion)> 48 =(after modulo)> 0
        2i;                    raises 2 to the power of the loop counter
           *                   multiplies it by the modulo
            1R                 rotates stack 1 time
              $+              sums everything
                N.            outputs as number and exit

Answer 23

Common Lisp, 99 88 72 bytes

Takes a string as input

(defun f(s)(reduce(lambda(a d)(+ d(* a 2)))(map'list #'digit-char-p s)))

Ungolfed:

(defun bin-to-dec (bin-str)
  (reduce (lambda (acc digit) (+ digit (* acc 2)))
          (map 'list #'digit-char-p bin-str)))

Answer 24

><> (Fish) 36 28 bytes

/i:1+?!v$2*$2%+!| !
/0| ;n~<

Edit 1: Forgot to put the output in the original. Added output and used MOD 2 instead of minus 48 to convert ascii to decimal to save the extra bytes lost. (no change in bytes)

Edit 2: Changed the algorithm completely. Each loop now does this; times current value by 2, then add the mod of the input. (saving of 8 bytes)

Online version

Try it Online! - This works with bigger numbers than the above link.

Answer 25

C, 44 bytes

d(s,v)char*s;{return*s?d(s,v+=v+*s++-48):v;}

Use as follows :

int main(){
  printf("%i\n", d("101010",0));
}

Remove two bytes and an unused parameter thanks to Steadybox

Answer 26

MATLAB, 49 bytes

@(a)dot(int2str(a)-'0',2.^(floor(log10(a)):-1:0))

Anonymous function that splits the input into an array with int2str(a)-'0', then does a dot product with powers of 2. Has rounding error for the last test case, will update the solution when I figure out a fix.

Answer 27

Forth (gforth 0.7.3), 47 bytes

: x 2 base ! bl parse s>number drop decimal . ;

: x - define new word with name 'x'
2 base ! - set base to binary
bl parse - read line until a space (bl) or EOL
s>number - try to convert the string to number
drop - we only want the converted number and not the success flag
decimal - set base to decimal
. - print value on top of stack
; - end of definition

Test cases:

x 1 1  ok
x 10 2  ok
x 101010 42  ok
x 1101111111010101100101110111001110001000110100110011100000111 2016120520371234567  ok

Answer 28

C, 41 37 bytes

i;b(char*s){i+=i+*s++%2;i=*s?b(s):i;}

Wandbox

Answer 29

SmileBASIC, 47 44 bytes

INPUT B$WHILE""<B$N=N*2+VAL(SHIFT(B$))WEND?N

Another program of the same size:

INPUT B$WHILE""<B$N=N*2OR"0"<SHIFT(B$)WEND?N

Answer 30

APL(NARS), 26 chars, 52 bytes

{+/a×⌽1,2x*⍳¯1+≢a←1-⍨⎕D⍳⍵}

test:

  f←{+/a×⌽1,2x*⍳¯1+≢a←1-⍨⎕D⍳⍵}
  ⎕fmt f"0"
0
~
  f"1"
1
  ⎕fmt f"101010"
42
~~
  f"10"
2 
  (≢,f)"11011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
152 4995366924470859583704000893073232977339613183 

152 bits... It should be limited from memory for store string and numbers.