Decimal floating point number to binary

Decimal floating point number to binary is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Task

Create a program that takes a decimal floating point number and displays its binary representation and vice versa: takes a floating point binary number and outputs its decimal representation.

The output might be something like this:

       23.34375  => 10111.01011
     1011.11101  =>    11.90625

dc

Works with: dc version 1.3.95 (GNU bc 1.06.95)

Interactively: <lang bash>$ dc 2o 23.34375 p 10111.01011000000000000 q $ dc 2i 1011.11101 p 11.90625 q $</lang>

Directly on the command line: <lang bash>$ dc -e '2o 23.34375 p' 10111.01011000000000000 $ dc -e '2i 1011.11101 p' 11.90625 $ echo '2o 23.34375 p' | dc 10111.01011000000000000 $ echo '2i 1011.11101 p' | dc 11.90625 $</lang>

From the manpage: "To enter a negative number, begin the number with '_'. '-' cannot be used for this, as it is a binary operator for subtraction instead." <lang bash>$ dc -e '2o _23.34375 p' -10111.01011000000000000 $ dc -e '2i _1011.11101 p' -11.90625 $</lang>

D

Translation of: Python

<lang d>import std.stdio, std.conv, std.array, std.string, std.range, std.algorithm, std.typecons;

immutable string[string] hex2bin, bin2hex;

static this() pure @safe {

   hex2bin = 16.iota.map!(x => tuple("%x".format(x), "%04b".format(x))).assocArray;
   bin2hex = 16.iota.map!(x => tuple("%b".format(x), "%x".format(x))).assocArray;

}

string dec2bin(real d) pure @safe {

   immutable neg = d < 0;
   if (neg)
       d = -d;
   immutable hx = "%a".format(d);
   immutable p = hx.countUntil('p');
   immutable bn = hx[2 .. p].split("").map!(ch => hex2bin.get(ch, ch)).join;
   // Currently Phobos lacks a strip(string, string) function.
   // return (neg ? "-" : "") ~ bn.strip("0")
   return (neg ? "-" : "") ~ bn.tr("0", " ").strip.tr(" ", "0")
          ~ hx[p .. p + 2] ~ "%b".format(hx[p + 2 .. $].to!int);

}

real bin2dec(string bn) pure /*@safe*/ in {

   assert(!bn.empty);

} body {

   immutable neg = bn[0] == '-';
   if (neg)
       bn = bn[1 .. $];
   immutable dp1 = bn.countUntil('.');
   immutable extra0a = "0".replicate(4 - dp1 % 4);
   immutable bn2 = extra0a ~ bn;
   immutable dp2 = bn2.countUntil('.');
   immutable p = bn2.countUntil('p');
   auto hx = iota(0, dp2 + 1, 4)
             .map!(i => bin2hex.get(bn2[i..min(i + 4, p)]
                                    .tr("0", " ")
                                    .stripLeft
                                    .tr(" ", "0"),
                                    bn2[i .. i + 1]))
             .join;
   immutable bn3 = bn2[dp2 + 1 .. p];
   immutable extra0b = "0".replicate(4 - bn3.length % 4);
   immutable bn4 = bn3 ~ extra0b;
   hx ~= iota(0, bn4.length, 4)
         .map!(i => bin2hex[bn4[i .. i + 4].tr("0", " ").stripLeft.tr(" ", "0")])
         .join;
   hx = (neg ? "-" : "") ~ "0x" ~ hx ~ bn2[p .. p+2] ~ bn2[p + 2 .. $].to!int(2).text;
   return hx.to!real;

}

void main() /*@safe*/ {

   immutable x = 23.34375;
   immutable y1 = x.dec2bin;
   y1.writeln;
   writefln("%.6f", y1.bin2dec);
   immutable y2 = dec2bin(-x);
   y2.writeln;
   y2.bin2dec.writeln;
   writefln("%.6f", "1011.11101p+0".bin2dec);

}</lang>

Output:

1.011101011p+100
23.343750
-1.011101011p+100
-23.3438
11.906250

Elixir

<lang elixir>defmodule RC do

 def dec2bin(dec, precision\\16) do
   [int, df] = case String.trim(dec) |> String.split(".") do
     [int] -> [int, nil]
     [int, df] -> [int, df]
   end
   {sign, int} = if String.first(int)=="-", do: String.split_at(int, 1), else: {"", int}
   bin = sign <> (String.to_integer(int) |> Integer.to_string(2)) <> "."
   if df && String.to_integer(df)>0 do
     String.to_float("0."<>df) |> dec2bin(precision, bin)
   else
     bin <> "0"
   end
 end
 
 defp dec2bin(fp, digit, bin) when fp==0.0 or digit<=0, do: bin
 defp dec2bin(fp, digit, bin) do
   fp = fp * 2
   n = trunc(fp)
   dec2bin(fp-n, digit-1, bin<>Integer.to_string(n))
 end
 
 def bin2dec(bin) do
   [int, df] = case String.trim(bin) |> String.split(".") do
     [int] -> [int, nil]
     [int, df] -> [int, df]
   end
   {sign, int} = if String.first(int)=="-", do: String.split_at(int, 1), else: {"", int}
   dec = sign <> (String.to_integer(int, 2) |> Integer.to_string)
   dec <> if df && String.to_integer(df,2)>0 do
            1..String.length(df)
            |> Enum.reduce(String.to_integer(df, 2), fn _,acc -> acc / 2 end)
            |> to_string
            |> String.slice(1..-1)
          else
            ".0"
          end
 end

end

data = ~w[23.34375 11.90625 -23.34375 -11.90625] Enum.each(data, fn dec ->

 bin  = RC.dec2bin(dec)
 dec2 = RC.bin2dec(bin)
 :io.format "~10s => ~12s =>~10s~n", [dec, bin, dec2]

end)

data = ~w[13 0.1 -5 -0.25] Enum.each(data, fn dec ->

 bin  = RC.dec2bin(dec)
 dec2 = RC.bin2dec(bin)
 :io.format "~10s => ~18s =>~12s~n", [dec, bin, dec2]

end)</lang>

Output:

  23.34375 =>  10111.01011 =>  23.34375
  11.90625 =>   1011.11101 =>  11.90625
 -23.34375 => -10111.01011 => -23.34375
 -11.90625 =>  -1011.11101 => -11.90625
        13 =>             1101.0 =>        13.0
       0.1 => 0.0001100110011001 =>0.0999908447
        -5 =>             -101.0 =>        -5.0
     -0.25 =>              -0.01 =>       -0.25

Factor

<lang factor>USING: interpolate io kernel math.parser sequences ;

bin>dec ( x -- y )

   number>string "0b${}p0" interpolate>string string>number ;

23.34375 dup >bin 1011.11101 dup bin>dec [ [I ${} => ${}I] nl ] 2bi@</lang>

Output:

23.34375 => 1.011101011p4
1011.11101 => 11.90625

Fortran

This is a cut-back version of a free-format routine EATREAL that worked in a more general context. The text was to be found in an external variable ACARD(1:LC) with external fingers L1 and L2; L1 marked the start point and L2 advanced through the number. If a problem arose then an error message could denounce the offending text ACARD(L1:L2) as well as just say "Invalid input" or similar. The routine worked in base ten only, but it is a trivial matter to replace *10 by *BASE. Here however a possible base may extend beyond just the decimal digits, so it is no longer possible to rely on zero to nine only and their associated character codes, thus the "digit" is now identified by indexing into an array of digits, thereby enabling "A" to follow "9" without character code testing. As the INDEX function works with CHARACTER variables that are indexed from one, to get zero for the first character in DIGIT, one must be subtracted. For handling the rescaling needed for fractional digits, a table of powers of ten up to sixteen was defined, but now the base may not be ten so BASE**DD is computed on the fly. The original routine was intended for usages in the hundreds of millions of calls, so this version would be unsuitable! Further, the exponent addendum (as in 35E+16) can no longer be recognised because "E" is now a possible digit. In handling this, the exponent part was added to DD and who knows, the result may produce a zero DD as in "123.456E-3" but otherwise a MOD(DD,16) would select from the table of powers of ten, and beyond that would be handled by successive squaring. Few numbers are presented with more than sixteen fractional digits, but I have been supplied data supposedly on electric power consumption via the national grid with values such as 1.21282E-31 kilowatt-hours, and other values with twenty-eight digits of ... precision?

Because on a binary computer most decimal fractions are recurring sequences of binary digits, it is better to divide by ten than to multiply by 0·1. Thus, although the successive fractional digits could be incorporated by something like P = P*BASE; X = X + D/P if the computer's arithmetic was conducted in a base that is not compatible with BASE (for example, two and ten) each step would introduce another calculation error. It is better to risk one only, at the end.

An alternative method is to present the text to the I/O system as with READ (ACARD,*) X, except that there is no facility for specifying any base other than ten. In a more general situation the text would first have to be scanned to span the number part, thus incurring double handling. The codes for hexadecimal, octal and binary formats do not read or write numbers in those bases, they show the bit pattern of the numerical storage format instead, and for floating-point numbers this is very different. Thus, Pi comes out as 100000000001001001000011111101101010100010001000010110100011000 in B64 format, not 11·0010010000111111011010101... Note the omitted high-order bit in the normalised binary floating-point format - a further complication.

The source is F77 style, except for the MODULE usage simply for some slight convenience in sharing DIGIT and not having to re-declare the type of EATNUM. <lang Fortran> MODULE REBASE !Play with some conversions between bases.

      CHARACTER*36 DIGIT	!A set of acceptable digit characters.
      PARAMETER (DIGIT = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ")	!Not only including hexadecimal.
      CONTAINS		!No great complication.
       LOGICAL FUNCTION EATNUM(ACARD,BASE,V)	!Reads a text number using the specified base.

Chews into the likes of 666, -666.666, .666 with their variations. Completes with the value in V, success as the result. Could check that no digit exceeds the specified BASE usage, but that would mean an error message... Concocted by R.N.McLean (whom God preserve) May XXMM. Clunky usage of ICHAR encourages the compaq compiler to employ single-character-at-a-time usage.

        CHARACTER*(*) ACARD	!The text.
        INTEGER BASE		!The base may not be ten.
        DOUBLE PRECISION V	!The object of the exercise.
        DOUBLE PRECISION X	!Maximum precision for all this fun. No sign of REAL*10.
        INTEGER D,DD		!A single digit. and a digit count.
        INTEGER L2,LC		!Finger and limit.
        INTEGER*1 C		!Since ICHAR is in use.
        LOGICAL ADIGIT,XNEG	!Things noticed along the way.
         ADIGIT = .FALSE.	!No digits seen.
         XNEG = .FALSE.	!No negative number.
         DD = 0		!No decimal digits.
         X = 0			!No value.
         L2 = 1		!The starting point.
         LC = LEN(ACARD)	!The ending point
         IF (L2.GT.LC) GO TO 20	!Urk! Off the end even before I start.

Chew into the number. Admit a possible leading sign, then digits.

         C = ICHAR(ACARD(L2:L2))	!Grab, since there are two comparisons.
         IF (C.EQ.ICHAR("+")) GO TO 1	!First consider
         IF (C.NE.ICHAR("-")) GO TO 2	!Possible signs.
         XNEG = .TRUE.		!To be acted upon later.
   1     L2 = L2 + 1		!Advance one.
         IF (L2.GT.LC) GO TO 20	!Off the end?
   2     D = INDEX(DIGIT,ACARD(L2:L2)) - 1	!No. Taste the candidate digit.
         IF (D .LT. 0) GO TO 10	!Is it to my taste?
         X = D				!Yes. Yum.
         ADIGIT = .TRUE.		!A digit has been seen. One is enough to be seen.
   3     L2 = L2 + 1			!More may follow.
         IF (L2.GT.LC) GO TO 20	!Perhaps not.
         D = ICHAR(ACARD(L2:L2)) - ICHAR("0")	!Taste the candidate digit.
         IF (D .LT. 0) GO TO 10	!Digitish?
         X = X*BASE + D		!Yes. Assimilate.
         GO TO 3			!Even more might well follow.

Consider any decimal digits, introduced by a decimal point.

  10     IF (ICHAR(ACARD(L2:L2)).EQ.ICHAR(".")) GO TO 11	!A full stop as a decimal point?
         IF (ICHAR(ACARD(L2:L2)).NE.ICHAR("·")) GO TO 20	!So, is there a decimal point?
  11     L2 = L2 + 1			!Advance one.
         IF (L2.GT.LC) GO TO 20	!Sudden end?
         D = INDEX(DIGIT,ACARD(L2:L2)) - 1	!No. Taste the digit candidate.
         IF (D .LT. 0) GO TO 20	!Suitable?
         X = X*BASE + D		!Yes. Continue augmenting the number.
         DD = 1			!This is the first decimal digit.
         ADIGIT = .TRUE.		!There may have been none before the decimal point.
  12     L2 = L2 + 1			!If once one digit is seen, is not the jungle full of digits?
         IF (L2.GT.LC) GO TO 20	!Perhaps not.
         D = ICHAR(ACARD(L2:L2)) - ICHAR("0")	!Taste the digit candidate.
         IF (D < 0 .OR. 9 < D) GO TO 20	!Suitable?
         X = X*BASE + D			!Yes. Accept as before.
         DD = DD + 1			!Now, no need to set ADIGIT to true again.
         GO TO 12			!Carry on.

Can't consider any exponent part, started by an "E" or "D", as these may be possible digit symbols.

  20     IF (DD .GT. 0) X = X/BASE**DD	!Rescale for the fractional digits.
         IF (XNEG) X = -X		!Fix the sign.
         V = X				!Place the result.
         EATNUM = ADIGIT		!Report success.
       END FUNCTION EATNUM	!And awayt.

       SUBROUTINE FP8DIGITS(X,BASE,TEXT,L)	!Full expansion of the value of X in BASE.

Converts a number X to a specified BASE. For integers, successive division by BASE, for fractions, successive multiplication.

        REAL*8 X,T		!The value, and an associate.
        INTEGER BASE		!As desired.
        CHARACTER*(*) TEXT	!Scratchpad for results.
        INTEGER L		!The length of the result.
        INTEGER N,ND		!Counters.
        INTEGER D		!The digit of the moment.
        LOGICAL NEG		!Annoyance with signs.
         IF (BASE.LE.1 .OR. BASE.GT.LEN(DIGIT)) BASE = 10	!Preclude oddities.
         WRITE (TEXT,1) BASE	!Scrub the TEXT with an announcement.
   1     FORMAT ("Base",I3)	!A limited range is expected..
         T = X			!Grab the value.
         N = T			!Its integer part, with truncation.
         T = ABS(T - N)	!Thus obtain the fractional part.
         NEG = N .LT. 0	!Negative numbers are a nuisance.
         IF (NEG) N = -N	!So simplify for what follows.
         L = LEN(TEXT)		!Limit of the scratchpad.
         ND = 0		!No digits have been rolled.

Crunch the integer part. Use the tail end of TEXT as a scratchpad, as the size of N is unassessed.

  10     D = MOD(N,BASE)		!Extract the low-order digit in BASE.
         TEXT(L:L) = DIGIT(D+1:D+1)	!Place it as text.
         ND = ND + 1			!Count another digit rolled.
         N = N/BASE			!Drop down a power.
         L = L - 1			!Move back correspondingly.
         IF (L.LE.0) THEN		!Run out of space?
           TEXT = "Overflow!"		!Then, this will have to do!
           L = MIN(9,LEN(TEXT))	!TEXT might be far too short.
          RETURN			!Give up.
         END IF			!But, space is expected.
         IF (N.GT.0) GO TO 10		!Are we there yet?
         IF (NEG) THEN			!Yes! Is a negative sign needed?
           TEXT(L:L) = "-"		!Yes. Place it.
           L = L - 1			!And retreat another place.
         END IF			!No + sign for positive numbers.
         N = LEN(TEXT) - L		!So, how much scratchpad was used?
         TEXT(9:9 + N - 1) = TEXT(L + 1:)	!Append to the initial TEXT(1:8) from the start.
         L = 9 + N			!Finger what follows the units position.
         TEXT(L:L) = "."		!Laziness leads to a full stop for a decimal point.

Crunch through the fractional part until nothing remains.

         DO WHILE(T.GT.0)	!Eventually, this will be zero.
           IF (L.GE.LEN(TEXT)) THEN	!Provided I have enough space!
             L = LEN(TEXT)		!If not, use the whole supply.
             TEXT(L:L) = "~"		!Place a marker suggesting that more should follow.
            RETURN		!And give up.
           END IF		!Otherwise, a digit is to be found.
           T = T*BASE		!Shift up a power.
           N = T		!The integer part is the digit.
           T = T - N		!Remove that integer part from T.
           L = L + 1		!Advance the finger.
           TEXT(L:L) = DIGIT(N+1:N+1)	!Place the digit.
           ND = ND + 1		!Count it also.
         END DO		!And see if anything remains.

Cast forth an addendum, to save the reader from mumbling while counting long strings of digits.

         IF (LEN(TEXT) - L .GT. 11) THEN	!Err, is there space for an addendum?
           WRITE (TEXT(L + 2:),11) ND		!Yes! Reveal the number of digits.
  11       FORMAT ("Digits:",I3)		!I expect no ore than two-digit digit counts.
           L = L + 1 + 10			!So this should do.
         END IF				!So much for the addendum.
       END SUBROUTINE FP8DIGITS	!Bases play best with related bases, such as 4 and 8. Less so with (say) 3 and 7...
     END MODULE REBASE	!Enough for inspection.

     PROGRAM TESTSOME

Check some conversions from one base to another.

     USE REBASE
     INTEGER N		!Some number of tests.
     PARAMETER (N = 5)		!This number.
     CHARACTER*12 TEXT(N)	!Sufficient size texts.
     DATA TEXT/"23.34375","10111.01011","1011.1101","11.90625","-666"/	!Also demonstrate a negative.
     DOUBLE PRECISION V	!The value in the computer's own representation.
     INTEGER I,L,BASE		!Assistants.
     CHARACTER*88 BACK		!A scratchpad.

     WRITE (6,1)	!A heading would be nice.
   1 FORMAT ("Test text in base",3X,"Value in base 10")

Chug through the tests.

     DO BASE = 10,2,-8	!Odd loop generates BASE = 10 then BASE = 2.
       DO I = 1,N		!Step through the test texts.
         WRITE (6,11) TEXT(I),BASE	!Start the line with the input.
  11     FORMAT (A,I5,$)		!The $, obviously, means no new line.
         IF (.NOT.EATNUM(TEXT(I),BASE,V)) THEN	!There shouldn't be any trouble.
           WRITE (6,*) "Not a good number!"		!But...
          ELSE				!So then,
           WRITE (BACK,*) V			!Reveal the resulting value.
           WRITE (6,12) BACK(1:20)		!Sufficient space, I hope.
  12       FORMAT (A,$)			!All to produce tabular output.
           CALL FP8DIGITS(V,2,BACK,L)		!Convert back to a text string.
           WRITE (6,13) BACK(1:L)		!And reveal.
  13       FORMAT (A)				!Thus end the line.
           CALL FP8DIGITS(V,10,BACK,L)		!And in another base,
           WRITE (6,14) BACK(1:L)		!The same value.
  14       FORMAT (37X,A)			!Nicely aligned.
         END IF			!So much for that test.
       END DO			!On to the next test.
     END DO		!And another base.
     END	!Enough of that. </lang>

Rather than mess about with invocations, the test interprets the texts firstly as base ten sequences, then base two. It makes no complaint over encountering the likes of "666" when commanded to absorb according to base two. The placewise notation is straightforward: 666 = 6x2² + 6x2¹ + 6x2⁰

Test text in base   Value in base 10
23.34375       10   23.3437500000000 Base  2 10111.01011 Digits: 10
                                     Base 10 23.34375 Digits:  7
10111.01011    10   10111.0101100000 Base  2 10011101111111.00000010100101101001000110100111010111 Digits: 52
                                     Base 10 10111.01010999999925843439996242523193359375 Digits: 43
1011.1101      10   1011.11010000000 Base  2 1111110011.0001110000101111100000110111101101001010001 Digits: 53
                                     Base 10 1011.1100999999999885403667576611042022705078125 Digits: 47
11.90625       10   11.9062500000000 Base  2 1011.11101 Digits:  9
                                     Base 10 11.90625 Digits:  7
-666           10  -666.000000000000 Base  2 -1010011010. Digits: 10
                                     Base 10 -666. Digits:  3
23.34375        2   10.4687500000000 Base  2 1010.01111 Digits:  9
                                     Base 10 10.46875 Digits:  7
10111.01011     2   23.3437500000000 Base  2 10111.01011 Digits: 10
                                     Base 10 23.34375 Digits:  7
1011.1101       2   11.8125000000000 Base  2 1011.1101 Digits:  8
                                     Base 10 11.8125 Digits:  6
11.90625        2   8.53125000000000 Base  2 1000.10001 Digits:  9
                                     Base 10 8.53125 Digits:  6
-666            2  -42.0000000000000 Base  2 -101010. Digits:  6
                                     Base 10 -42. Digits:  2

Note again that a decimal value in binary is almost always a recurring sequence and that the exact decimal value of the actual binary sequence in the computer (of finite length) is not the same as the original decimal value. 23·34375 happens to be an exact decimal representation of a binary value whose digit count is less than that available to a double-precision floating-point variable. But although 1011·1101 has few digits, in decimal it converts to a recurring sequence in binary just as does 0·1.

FreeBASIC

<lang freebasic>' FB 1.05.0 Win64

' Expresses (or rounds) the fractional part to the same number of places in binary as the decimal to be converted. ' This is accurate for the numbers used in this exercise but in general would not be. Function decToBin(d As Double) As String

 Dim neg As String = ""
 If d < 0.0 Then
   d = -d
   neg = "-"
 End If
 Dim i  As Integer = Fix(d)
 Dim f  As Double  = Frac(d)
 If f = 0 Then
   Return neg + Bin(i) 
 End If
 Dim le As Integer = Len(Str(d)) - Len(Str(i)) - 1
 Return neg + Bin(i) + "." + Bin(Fix((2.0 ^ le) * f + 0.5), le)    

End Function

Function binToDec(s As String) As Double

 If s = "" Then Return 0.0
 Dim neg As Integer = 1
 If Left(s, 1) = "-" Then
    s = Mid(s, 2)
    neg = -1
 End If
 Dim index As Integer = Instr(s, ".")   
 If index = 0 Then 
   Return ValLng("&B" + s) * neg
 Else
   Dim a  As Integer  = ValLng("&B" + Left(s, index - 1))
   Dim b  As Integer  = ValLng("&B" + Mid(s, index + 1))
   Dim le As Integer  = Len(Mid(s, index + 1))
   Return (a + b / (2.0 ^ le)) * neg
 End If

End Function

Print "23.34375 => "; decToBin(23.34375) Print "1011.11101 => " ; binToDec("1011.11101") Print "-23.34375 => " ; decToBin(-23.34375) Print "-1011.11101 => " ; binToDec("-1011.11101") Print "64 => "; decToBin(64) Print "-100001 => " ; binToDec("-100001") Print Print "Press any key to quit" Sleep</lang>

Output:

23.34375     =>  10111.01011
1011.11101   =>  11.90625
-23.34375    => -10111.01011
-1011.11101  => -11.90625
64           =>  1000000
-100001      => -33

Go

Translation of: Kotlin

<lang go>package main

import (

   "fmt"
   "math"
   "strconv"
   "strings"

)

func decToBin(d float64) string {

   whole := int64(math.Floor(d))
   binary := strconv.FormatInt(whole, 2) + "."
   dd := d - float64(whole)
   for dd > 0.0 {
       r := dd * 2.0
       if r >= 1.0 {
           binary += "1"
           dd = r - 1.0
       } else {
           binary += "0"
           dd = r
       }
   }
   return binary

}

func binToDec(s string) float64 {

   ss := strings.Replace(s, ".", "", 1)
   num, _ := strconv.ParseInt(ss, 2, 64)
   ss = strings.Split(s, ".")[1]
   ss = strings.Replace(ss, "1", "0", -1)
   den, _ := strconv.ParseInt("1" + ss, 2, 64)
   return float64(num) / float64(den)

}

func main() {

   f := 23.34375
   fmt.Printf("%v\t => %s\n", f, decToBin(f))
   s := "1011.11101"
   fmt.Printf("%s\t => %v\n", s, binToDec(s))

}</lang>

Output:

23.34375	 => 10111.01011
1011.11101	 => 11.90625

Haskell

float to binary part only: <lang haskell>import Data.Char (intToDigit) import Numeric (floatToDigits, showIntAtBase)

dec2bin :: RealFloat a => a -> String dec2bin f = "0." ++ map intToDigit digits ++ "p+" ++ showIntAtBase 2 intToDigit ex ""

 where (digits, ex) = floatToDigits 2 f

main :: IO () main = putStrLn $ dec2bin 23.34375</lang>

Output:

0.1011101011p+101

J

In this draft, the task does not give any guidelines for handling precision. So we will use 99 places after the decimal point and trim any trailing zeros (and the decimal point, for integer case).

Also, since J does not have a "Decimal floating point number" data type, we will use a list of characters to represent a decimal or binary number (this corresponds roughly with the relevant feature set of REXX which seems to have had a strong influence on this draft of this task), and use internal (mantissa,exponent) representations during the conversion.

Implementation:

<lang J>b2b=:2 :0

 NB. string to rational number
 exp=. (1x+y i.'.')-#y
 mant=. n#.x:0"."0 y-.'.'
 number=. mant*n^exp*'.' e. y
 NB. rational number to string
 exp=. _99
 mant=. <.1r2+number*m^x:-exp
 s=. exp&(}.,'.',{.) (":m#.inv mant)-.' '
 ((exp-1)>.-+/*/\|.s e.'.0') }. s

)</lang>

Example use:

<lang J> 2 b2b 10 '23.34375' 10111.01011

  10 b2b 2 '1011.11101'

11.90625</lang>

Kotlin

<lang scala>// version 1.1.0

fun decToBin(d: Double): String {

   val whole  = Math.floor(d).toLong()
   var binary = whole.toString(2) + "." 
   var dd = d - whole
   while (dd > 0.0) { 
       val r = dd * 2.0
       if (r >= 1.0) {  
           binary += "1"
           dd = r - 1
       }
       else {    
           binary += "0"
           dd = r
       }
   } 
   return binary

}

fun binToDec(s: String): Double {

   val num = s.replace(".", "").toLong(2)
   val den = ("1" + s.split('.')[1].replace("1", "0")).toLong(2)
   return num.toDouble() / den

}

fun main(args: Array<String>) {

   val d = 23.34375
   println("$d\t => ${decToBin(d)}")
   val s = "1011.11101"
   println("$s\t => ${binToDec(s)}")

}</lang>

Output:

23.34375         => 10111.01011
1011.11101       => 11.90625

LiveCode

LiveCode's baseConvert only works on integers. Only the first part of this task is complete. <lang LiveCode>function float2bin n

   put 15 into limit
   put 0 into i
   split n using "."
   put  baseConvert(n[1],10,2) into significand
   put "0." before n[2]
   repeat until (n[2] = 0 or i > limit)
       put n[2] * 2 into tmp
       split tmp with "."
       put tmp[1] after insignificand
       put "0." & tmp[2] into n[2]
       add 1 to i
   end repeat
   return significand & "." & insignificand

end float2bin

put float2bin(23.34375) // 10111.01011 put float2bin(11.90625) //1011.11101</lang>

M2000 Interpreter

<lang M2000 Interpreter> Module Checkit {

     Conv2dec=lambda (n$, frombase=10, dp$=".") -> {
          neg=left$(n$,1)="-": if neg then n$=mid$(n$,2)
          if instr(n$, dp$)>0 then {
                n2$=Piece$(n$,dp$,2)
                n$=Piece$(n$, dp$,1)
          } else n2$=""
          n0=0
          l1=len(n$)+1
          For i=len(n$) to 1
                dig$=Mid$(n$,l1-i,1)
                dig=asc(dig$)-48
                if dig$>"9" then dig-=7
                if dig>=frombase then error "not in base:"+frombase
                n0+=dig*frombase^(i-1)
          next i
          if n2$<>"" then {
             For i=1 to len(n2$)
                dig$=Mid$(n2$,i,1)
                dig=asc(dig$)-48
                if dig$>"9" then dig-=7
                if dig>=frombase then error "not in base:"+frombase
                n0+=dig/frombase^i
             Next i 
          }
          if neg then n0-!
          =n0
     }
     Conv2Any$=Lambda$ (dec, tobase=10, dp$=".", prec=16) -> {
          a$=""
          neg=false
          if dec<0 then neg=true
          dec=abs(dec)
          n2=frac(dec)
          if dec=0 then {
                a$="0" 
          } else {
                do {
                       n=dec mod tobase
                       if n>=10 then n+=7
                       a$=chr$(n+48)+a$
                       dec=dec div tobase
                 } until dec==0
           }
           if n2<>0 then {
                a$+=dp$
                prec--
                do {
                     prec--
                     dec=n2*tobase           
                     n2=frac(dec)
                     dec-=n2
                     n2=round(n2)
                     if dec>=10 then dec+=7
                     a$+=chr$(dec+48)
                } until n2=0  or prec<0      
           }
           if neg then {="-"+a$} else =a$
     }
     Rem : Locale 1033 ' use . for all print out for decimal point
     Print Conv2dec("10111.01011",2); " => ";Conv2Any$(23.34375,2)
     Print Conv2Any$(11.90625, 2); " => "; Conv2dec("1011.11101",2)
     \\ using , for decimal point
     Print Conv2Any$(Conv2dec("1011,11101",2, ","), 10, ",") 
     Print 12312321.1212
     Print Conv2Any$(12312321.1212, 2)
     \\ using . for 1033 locale
     Print Str$(Conv2Dec(Conv2Any$(12312321.1212, 2), 2), 1033)="12312321.1211853"
     Print Str$(Conv2Dec(Conv2Any$(12312321.1212, 2,,52), 2), 1033) ="12312321.1212"

} Checkit </lang>

Output:

23.34375 => 10111.01011
1011.11101 => 11.90625
11,90625
12312321.1212
101110111101111100000001.0001111100000110
     True
     True

Maple

<lang Maple> convert(23.34375,binary,decimal);

convert(1011.11101,decimal,binary); </lang> Output:

                          10111.01011

                          11.90625000

Mathematica

<lang Mathematica>dec2bin[x_] := Block[{digits, pos},

 {digits,  pos} = (RealDigits[ToExpression@x,  2] /. {a_, b_} :> {ToString /@ a, b});
 StringJoin @@ 
  Join@{Take[digits, pos], {"."}, Quiet@NestWhile[Most, Drop[digits, pos], Last@# === "0" &]}]

bin2dec[x_] := FromDigits[RealDigits@ToExpression@x, 2] // N

Print[NumberForm[#, {9, 5}], " => ", dec2bin@#] &@23.34375; Print[NumberForm[#, {9, 5}], " => ", NumberForm[bin2dec@#, {9, 5}]] &@1011.11101;</lang>

Output:

23.34375 => 10111.01011
1011.11101 => 11.90625

OCaml

Well, I am no expert in OCaml, and my code may seem a bit messy, but I actually took a rather naive aproach... Anyway, the program seems to work, but the algorithm(s) can probably be improved. After reading the discussion, I took into account the suggestion that the program should perform conversions from any base to any other base.

Works with: OCaml version 4.03+

<lang ocaml>

1. load "str.cma"

(* Using the interpteter or the compiler:

*
* Interpreter:
*     $ ocaml convert.ml
*
* Compiler:
*     First of all, delete the line '#load "str.cma"'.
*     Then, using the native compiler:
*         $ ocmalopt -o convert str.cmxa convert.ml
*         $ ./convert
*     Or using ocamlbuild:
*         $ ocamlbuild -pkg str convert.native
*         $ ./convert.native
*)

(* This program converts from any numerical base to any numerical base. *) (* The numbers are wrapped within a type named 'value'.

* Ex: Float 23.7
*     String "1011.110101"
*
* Conversion is performed by the function 'convert'.
* Ex: Convert.convert ~to_base:2 (Convert.Float 23.7)
*     Convert.convert ~from_base:2 to_base:10 (Convert.String "1011.110101")
*
* (Parameter 'from_base' is optional and defaults to 10)
*)

(* This signature should be located in a separate file 'convert.mli'... *) module Convert : sig

 type value = | Float of float
              | String of string
 val convert : ?from_base:int -> to_base:int -> value -> value

end = struct

type value = | Float of float

            | String of string

(* =========================== *) (* === Auxiliary functions === *) (* =========================== *) (* Digits available: 0 to 9 plus A to Z.

* A base less than 2 does not make sense... Does it?
* A base greater than 36 would need more letters... *)

let min_base = 2 let max_base = 10 + int_of_char 'Z' - int_of_char 'A' + 1

(* A maximum number of decimal positions just to avoid infinite loops while

* computing them... I think 30 is enough... *)

let max_decs = 30

(* Convert an integer into the corresponding digit:

* 0->'0'; ...; 9->'9'; 10->'A'; 11->'B'; etc. *)

let dig_of_int n =

 if n >= 0 && n <= 9
 then char_of_int (int_of_char '0' + n)
 else if n >= 10 && n <= max_base
      then char_of_int (n + int_of_char 'A' - 10)
      else failwith (Printf.sprintf "Incorrect digit: %c" (char_of_int n))

(* Convert a digit into the corresponding integer:

* '0'->0; ...; '9'->9; 'A'->10; 'B'->11; etc. *)

let int_of_dig c =

 match c with
 | '0' .. '9' -> int_of_char c - int_of_char '0'
 | 'A' .. 'Z' -> int_of_char c - int_of_char 'A' + 10
 | _ -> failwith (Printf.sprintf "Incorrect character: %c" c)

(* A numerical base must be within some limits. *) let check_base b =

 if b >= min_base && b <= max_base
 then true
 else invalid_arg ("Invalid base " ^ string_of_int b)

(* A number must have its digits within the range [0..b-1] for any base b. *) let check_number b n =

 let max_digit = dig_of_int (b - 1)
 and str = match n with
           | Float f -> string_of_float f
           | String s -> s
        |> Str.replace_first (Str.regexp "-") ""  (* strip off one sign *)
        |> Str.replace_first (Str.regexp "\\.") "" (* strip off one dot *)
 in
 let rec scan i =
   if i >= String.length str then true
   else
     if str.[i] <= max_digit
     then scan (i + 1)
     else invalid_arg (Printf.sprintf "Invalid digit %c for base %d" str.[i] b)
 in
 scan 0

(* =============================== *) (* === End Auxiliary functions === *) (* =============================== *)

(* ============================ *) (* === Conversion functions === *) (* ============================ *) (* Convert a floating point number, which is always base 10, to any base. *) let conv_float to_base fl =

 let d = abs_float fl in
 let int_part = truncate d in
 let dec_part = d -. float int_part
 in
 let rec ft_int n ls = 
   (* Conversion of the integer part. *)
   let quot = n / to_base
   and rest = n mod to_base in
   if quot = 0 then rest :: ls
               else ft_int quot (rest :: ls)
 in
 let rec ft_dec nd x ls =
   (* Conversion of the decimal part. nd is the maximum number of decinals to
    * be computed. *)
   if x = 0. || nd = 0 then List.rev ls
   else let prod = x *. float to_base in
        let intpart = truncate prod in
        ft_dec (nd - 1) (prod -. float intpart) (intpart :: ls)
 in
 let join_digs ls =
   (* Convert integers into digits and join them into a string. *)
   List.map (fun n -> dig_of_int n |> Char.escaped) ls |> String.concat ""
 in
 let sign = if fl < 0. then "-" else "" in 
 let left = sign ^ join_digs (ft_int int_part [])
 and right = join_digs (ft_dec max_decs dec_part []) in
 (* Decimal point only if a decimal part exists. *)
 let str = left ^ (if String.length right = 0 then "" else  "." ^ right) in
 if to_base = 10 then Float (float_of_string str)
                 else String str

(* Convert a value from one base to another.

* Using base 10 as an intermediate conversion. *)

let conv_string from_base to_base st =

 let digs = Str.replace_first (Str.regexp "-") "" st in
 let splitted = Str.split (Str.regexp "\\.") digs in
 let max_weight = String.length (List.hd splitted) - 1
 in
 let intdigs = List.map (Str.split (Str.regexp "")) splitted
            |> List.flatten
            |> List.map (fun s -> int_of_dig s.[0])
 in
 let conv10 w ns =
   List.mapi (fun i n -> float n *. float from_base ** float (w - i)) ns
   |> List.fold_left (+.) 0.
 in
 let sign = if st.[0] = '-' then -1. else 1. in
 let num_b10 = sign *. conv10 max_weight intdigs in
 if to_base = 10 then Float num_b10
                 else conv_float to_base num_b10

(* ================================ *) (* === End Conversion functions === *) (* ================================ *)

(* ============================== *) (* === Conversion starts here === *) (* ============================== *) let convert ?(from_base = 10) ~to_base number =

 let _ = check_base from_base && check_base to_base
 and _ = check_number from_base number
 in
 match number, from_base, to_base with
 | (Float fl, fb, tb) when fb = 10 && tb = 10 ->
     number
 | (Float fl, fb, _) when fb = 10 ->
     conv_float to_base fl
 | (Float _, _, _) ->
     invalid_arg "With a float, base of origin is always 10"
 | (String st, _, _) ->
     conv_string from_base to_base st

end (* of module Convert *)

(* ======================= *) (* === Some testing... === *) (* ======================= *) open Convert let () =

 let values_both = [
   (10, 2, Float 23.34375, String "10111.01011");
   (10, 2, Float (-23.34375), String "-10111.01011");
   (10, 2, Float 11.90625, String "1011.11101");
   (10, 2, Float (-11.90625), String "-1011.11101");
   (10, 2, Float 0., String "0");
   (2, 16, String "1111", String "F");
   (2, 16, String "-1111", String "-F")
 ]
 and values = [
   (10, 10, String "23.7", Float 23.7);
   (* conversion of Float to base 10 results in the same Float, not a String *)
   (10, 10, Float 23.7, Float 23.7)
 ] in
 let get_float = function
   | Float f -> f
   | _ -> failwith "Incorrect Float..."
 in
 let get_string = function
   | String s -> s
   | _ -> failwith "Incorrect String..."
 in
 let result pred =
   if pred then "PASS" else "FAIL"
 in
 let pretty_print v1 v2 calc =
   match v1, v2 with
   | Float f, String s ->
      Printf.sprintf "%f => %s; expected %s [%s]\n"
                     (get_float v1) (get_string calc) (get_string v2)
                     (result (calc = v2))
   | String s, Float f -> 
      Printf.sprintf "%s => %f; expected %f [%s]\n"
                     (get_string v1) (get_float calc) (get_float v2)
                     (result (calc = v2))
   | String s1, String s2 -> 
      Printf.sprintf "%s => %s; expected %s [%s]\n"
                     (get_string v1) (get_string calc) (get_string v2)
                     (result (calc = v2))
   | Float f1, Float f2 -> 
      Printf.sprintf "%f => %f; expected %f [%s]\n"
                     (get_float v1) (get_float calc) (get_float v2)
                     (result (calc = v2))
 in
 let testit (base1, base2, num1, num2) =
   let calc1 = convert ~from_base:base1 ~to_base:base2 num1 in
   pretty_print num1 num2 calc1
 in
 let testit_both (base1, base2, num1, num2) =
   testit (base1, base2, num1, num2) ^ testit (base2, base1, num2, num1)
 in
 let _ = List.iter (fun tpl -> print_endline (testit_both tpl)) values_both
 and _ = List.iter (fun tpl -> print_endline (testit tpl)) values
 in ()

</lang>

Output:

23.343750 => 10111.01011; expected 10111.01011 [PASS]
10111.01011 => 23.343750; expected 23.343750 [PASS]

-23.343750 => -10111.01011; expected -10111.01011 [PASS]
-10111.01011 => -23.343750; expected -23.343750 [PASS]

11.906250 => 1011.11101; expected 1011.11101 [PASS]
1011.11101 => 11.906250; expected 11.906250 [PASS]

-11.906250 => -1011.11101; expected -1011.11101 [PASS]
-1011.11101 => -11.906250; expected -11.906250 [PASS]

0.000000 => 0; expected 0 [PASS]
0 => 0.000000; expected 0.000000 [PASS]

1111 => F; expected F [PASS]
F => 1111; expected 1111 [PASS]

-1111 => -F; expected -F [PASS]
-F => -1111; expected -1111 [PASS]

23.7 => 23.700000; expected 23.700000 [PASS]

23.700000 => 23.700000; expected 23.700000 [PASS]

Perl 6

<lang perl6>given "23.34375" { say "$_ => ", :10($_).base(2) } given "1011.11101" { say "$_ => ", :2($_).base(10) }</lang>

Output:

23.34375 => 10111.01011
1011.11101 => 11.90625

Python

Python has float.hex() and float.fromhex() that can be used to form our own binary format. <lang python>hex2bin = dict('{:x} {:04b}'.format(x,x).split() for x in range(16)) bin2hex = dict('{:b} {:x}'.format(x,x).split() for x in range(16))

def float_dec2bin(d):

   neg = False
   if d < 0:
       d = -d
       neg = True
   hx = float(d).hex()
   p = hx.index('p')
   bn = .join(hex2bin.get(char, char) for char in hx[2:p])
   return (('-' if neg else ) + bn.strip('0') + hx[p:p+2]
           + bin(int(hx[p+2:]))[2:])

def float_bin2dec(bn):

   neg = False
   if bn[0] == '-':
       bn = bn[1:]
       neg = True
   dp = bn.index('.')
   extra0 = '0' * (4 - (dp % 4))
   bn2 = extra0 + bn
   dp = bn2.index('.')
   p = bn2.index('p')
   hx = .join(bin2hex.get(bn2[i:min(i+4, p)].lstrip('0'), bn2[i])
                for i in range(0, dp+1, 4))
   bn3 = bn2[dp+1:p]
   extra0 = '0' * (4 - (len(bn3) % 4))
   bn4 = bn3 + extra0
   hx += .join(bin2hex.get(bn4[i:i+4].lstrip('0'))
                 for i in range(0, len(bn4), 4))
   hx = (('-' if neg else ) + '0x' + hx + bn2[p:p+2]
         + str(int('0b' + bn2[p+2:], 2)))
   return float.fromhex(hx)</lang>

Output:

Run the above in idle then you can do the following interactively:

>>> x = 23.34375
>>> y = float_dec2bin(x)
>>> y
'1.011101011p+100'
>>> float_bin2dec(y)
23.34375
>>> y = float_dec2bin(-x)
>>> y
'-1.011101011p+100'
>>> float_bin2dec(y)
-23.34375
>>> float_bin2dec('1011.11101p+0')
11.90625
>>> 

Racket

The binary to number conversion is easy because it's supported by Racket. We can use string->number, wrap it in a dedicated function or use the read extension. <lang Racket>#lang racket

(define (string->number/binary x)

 (string->number x 2))

(string->number/binary "10.0101") (newline) (string->number/binary "0.01")

1. b0.01

(string->number "0.01" 2) (newline)</lang>

Output:

2.3125

0.25
0.25
0.25

Racket only supports the number to binary conversion for integer numbers, so we multiply the original number by a power of two, to get all the binary digits, and then we manipulate the string to place the point in the correct place. <lang Racket>(define (number->string/binary x)

 (define decimals-places 10)
 (define digits-all (~r (inexact->exact (round (* x (expt 2 decimals-places))))
                        #:base 2
                        #:pad-string "0"
                        #:min-width (add1 decimals-places)))
 (define digits-length (string-length digits-all))
 (define integer-part (substring digits-all 0 (- digits-length decimals-places)))
 (define decimal-part* (string-trim (substring digits-all (- digits-length decimals-places))
                                   "0" 
                                   #:left? #f
                                   #:repeat? #t))
 (define decimal-part (if (string=? decimal-part* "") "0"  decimal-part*))
 (string-append integer-part "." decimal-part))
 

(number->string/binary 9.01) (number->string/binary 9) (number->string/binary 0.01) (newline)</lang>

Output:

"1001.000000101"
"1001.000000000"
"0.000000101"

Some additional interesting examples <lang Racket>(number->string/binary (string->number/binary "010110.0011010")) (string->number/binary (number->string/binary .1)) (newline)

(number->string/binary (string->number/binary "0.11111111111")) (string->number/binary "0.11111111111")</lang>

Output:

"10110.001101000"
0.099609375

"1.000000000"
0.99951171875

REXX

version 1

This REXX version will handle any number of digits, with bases up to 242 (using extended ASCII characters).

Bases up to 62 will just use decimal digits along with upper and lowercase (Latin) letters.

This REXX program is a modified version of the original program which can handle any base (no limit),   and the original program did more extensive error checking.

This program handles numbers with leading signs (-, +).

Bases   that are   negative   are also supported   (which won't be explained here).

This REXX version used a single function   (base)   to perform the base conversions.

Quite a bit of additional code was added to verify that the number   (and other arguments)   are valid   (has too many signs,   too many decimal points,   or has invalid digits for the base specified, etc).

The number of digits (numerals) in the result may be specified   (for continued fractions).   The default is 60.

The REXX program can handle any sized number   (as per the number of digits)   that can be entered at the command line. <lang rexx>/*REXX pgm converts any number in a base to another base including fractions; bases≤242.*/ parse arg number toBase inBase digits . /*obtain optional arguments from the CL*/ if toBase== | toBase=="," then toBase= 10 /*Not specified? Then use the default.*/ if inBase== | inBase=="," then inBase= 10 /* " " " " " " */ if digits== | digits=="," then digits= 60 /* " " " " " " */ if number== | number=="," then call err "no number specified." if \datatype(toBase, 'W') then call err "invalid toBase: " toBase if \datatype(inBase, 'W') then call err "invalid inBase: " inBase if \datatype(digits, 'W') then call err "invalid digits: " digits numeric digits max(digits, length(number)) + 5 /*use a bigger numeric decimal digits. */ $= base(number, toBase, inBase) /*convert the # from a base to a base. */ numeric digits digits /*use a smaller numeric digs*/ if toBase==10 then if pos(., $)\==0 then $= format($) /*maybe use the FORMAT BIF.*/ say number ' (in base' inBase") = " $ ' (in base' toBase")." exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ base: procedure; parse arg x 1 s 2 1 ox,tt,ii,,oS

     @#= 0123456789;   @abc= 'abcdefghijklmnopqrstuvwxyz';    @abcu= @abc;    upper @abcu
     dontUse= @#'.+-'@abc || @abcu"0708090a0b0c0d"x;          OK= @# || @abcu || @abc
     $= OK || space( translate( xrange('1'x, "fe"x), , dontUse), 0)   /*max base string*/
     m= length($) - 1                                     /*M:  is the maximum base.   */
     if tt==           then tt= 10                      /*assume base 10  if omitted.*/
     if ii==           then ii= 10                      /*assume base 10  if omitted.*/
     i= abs(ii);              t= abs(tt)                  /*use absolute values for I&T*/
     if t==999 | t=="*"  then t= m                        /*T=999 or *?   Then use max.*/
     if t>m  then call err  'invalid range for ToBase:'     t";  the range is: "   2   m
     if i>m  then call err  'invalid range for InBase:'     i";  the range is: "   2   m
     != substr($, 1   +   10 * (tt<0), t)                 /*character string for base. */
     if tt<0   then != 0 || !                             /*Net base?   Prefix a zero. */
     if x==  then return left(!, t)                     /*Is X null?  Show base chars*/
     @=substr($, 1   +   10 * (ii<0), i)                  /*@:  legal chars for base X.*/
     if s='-' | s="+"  then do;  x= substr(x, 2);   oS= s /*elide the sign character,  */
                            end                           /* ··· and save the original·*/
     if (ii>10 & ii<37) | (ii<0 & ii>-27)  then upper x   /*should  X  be uppercased ? */
     parse var x  w  '.'  g                               /*sep whole from fraction.   */
     if pos('-', x)>0 | pos("+", x)>0 | x==. | x== | ,  /*too many signs,  bare .  or*/
        pos(., g)>0  then call err 'illegal number: '  ox /*too many decimal points ?  */
     __=w || g;                     _= verify(__, @'.')   /*# have any unusual digits? */
     if _\==0  then call err 'illegal char in number:'   ox   'char='    substr(__, _, 1)
     if i\==10 then do    /*────────────────────────────────convert # base I─►base ten.*/
                    _=0;  p=0;      do j=length(w)  to 1  by -1  while  w\==
                                    _= _ + (( pos( substr(w, j, 1), @)-1) * i**p);  p=p+1
                                    end   /*j*/           /*increase power of base  [↑]*/
                    w=_;  p=1;  _=0                       /*[↓] convert fractional part*/
                                    do j=1  for length(g)
                                    _= _ + (( pos( substr(g, j, 1), @)-1)/i**p);    p=p+1
                                    end   /*j*/           /*increase power of base  [↑]*/
                    g= _
                    end
               else if g\==  then g= "."g               /*reinsert period if needed. */
     if t\==10 then do    /*────────────────────────────────convert # base ten─►base T.*/
                    if w\==  then do                    /*convert whole number part. */
                                        do j=1  until t**j>w;  end   /*j*/
                                    n=
                                        do k=j-1  to 1  by -1;         _= t**k
                                        n=n || substr(!, 1 + w%_, 1);          w= w // _
                                        end     /*k*/                /*remainder ≡  // */
                                    w= n || substr(!, 1+w, 1)
                                    end
                    if g\==  then do;  n=                /*convert fractional part.  */
                                            do digits()+1  while g\==0;  p= g*t;  g= p//1
                                            n= n || substr(!, 1 + p%1, 1)
                                            end   /*digits ···*/
                                    if n==0    then n=
                                    if n\==  then n= '.'n          /*only a fraction?*/
                                    g= n
                                    end
                    end
     return oS || word( strip( space(w), 'L', 0)strip( strip(g, , 0),  "T", .)  0,  1)

/*──────────────────────────────────────────────────────────────────────────────────────*/ err: say; say '***error***: ' arg(1); say; exit 13</lang>

output   when using the input of:     23.34375   2

23.34375  (in base 10)    =    10111.01011  (in base 2).

output   when using the input of:     1011.11101   10   2

1011.11101  (in base 2)    =    11.90625  (in base 10).

output   when using the input of:     3.14159265358979323846264338327950288419716939937510582097494   62

3.14159265358979323846264338327950288419716939937510582097494  (in base 10)    =    3.8mHUcirZ3g3aaX5Bn156eBkfOx43HPGx7xT3yBX1Aoh3TAAEolLiHWo8Z4XVLWesfA6  (in base 62).

version 2

<lang rexx>/* REXX ---------------------------------------------------------------

• 08.02.2014 Walter Pachl
• --------------------------------------------------------------------*/

Call df2bf 23.34375,10111.01011 Call bf2df 1011.11101,11.90625 Call df2bf -23.34375,-10111.01011 Call bf2df -1011.11101,-11.90625 Exit

bf2df: Procedure

 Parse Arg x,soll
 If left(x,1)='-' Then Do
   sign='-'
   x=substr(x,2)
   End
 Else sign=
 Parse Var x int '.' fract
 int=reverse(int)
 vb=1
 res=0
 Do while int<>
   Parse Var int d +1 int
   res=res+d*vb
   vb=vb*2
   End
 vb=1
 Do while fract<>
   vb=vb/2
   Parse Var fract d +1 fract
   res=res+d*vb
   End
 res=sign||res
 Say sign||x '->' res
 If res<>soll Then
   Say 'soll='soll
 Return

df2bf: Procedure

 Parse Arg x,soll
 If left(x,1)='-' Then Do
   sign='-'
   x=substr(x,2)
   End
 Else sign=
 res=
 Parse Var x int '.' +0 fract
 Do While int>0
   dig=int//2
   int=int%2
   res=dig||res
   End
 If res= Then res='0'
 vb=1
 bf=
 Do i=1 To 30 while fract>0
   vb=vb/2
   If fract>=vb Then Do
     bf=bf'1'
     fract=fract-vb
     End
   Else
     bf=bf'0'
   End
 res=sign||res'.'bf
 Say sign||x '->' res
 If res<>soll Then
   Say 'soll='soll
 Return</lang>

Output:

23.34375 -> 10111.01011
1011.11101 -> 11.90625
-23.34375 -> -10111.01011
-1011.11101 -> -11.90625

Ruby

<lang ruby>def dec2bin(dec, precision=16) # String => String

 int, df = dec.split(".")
 minus = int.delete!("-")
 bin = (minus ? "-" : "") + int.to_i.to_s(2) + "."
 if df and df.to_i>0
   fp = ("."+df).to_f
   digit = 1
   until fp.zero? or digit>precision
     fp *= 2
     n = fp.to_i
     bin << n.to_s
     fp -= n
     digit += 1
   end
 else
   bin << "0"
 end
 bin

end

def bin2dec(bin) # String => String

 int, df = bin.split(".")
 minus = int.delete!("-")
 dec = (minus ? "-" : "") + int.to_i(2).to_s
 if df
   dec << (df.to_i(2) / 2.0**(df.size)).to_s[1..-1]
 else
   dec << ".0"
 end

end

data = %w[23.34375 11.90625 -23.34375 -11.90625] data.each do |dec|

 bin  = dec2bin(dec)
 dec2 = bin2dec(bin)
 puts "%10s => %12s =>%10s" % [dec, bin, dec2]

end</lang>

Output:

  23.34375 =>  10111.01011 =>  23.34375
  11.90625 =>   1011.11101 =>  11.90625
 -23.34375 => -10111.01011 => -23.34375
 -11.90625 =>  -1011.11101 => -11.90625

Scala

Idiomatic (FP with tailrec)

<lang Scala>import java.lang.Long

import scala.annotation.tailrec

object DecimalFloatPoint2Binary extends App {

 def doubleToBin(d: Double): String = {
   require(d >= 0.0, "Only positive values are allowed.")

   def fraction2BinaryFractionString(s: String, frac: Double) = {
     @tailrec
     def loop(acc: String, mid: Double): String = {
       if (mid > 0.0) {
         val r = mid * 2.0
         if (r >= 1.0) loop(acc + "1", r - 1) else loop(acc + "0", r)
       } else acc
     }

     loop(s + ".", frac)
   }

   val whole = math.floor(d).toLong

   fraction2BinaryFractionString(Long.toString(whole, 2), d - whole)
 }

 def binToDec(s: String) = {
   def num = Long.parseLong(s.replace(".", ""), 2)

   def den = Long.parseLong("1" + s.split('.')(1).replace("1", "0"), 2)

   num.toDouble / den
 }

 { // main
   println( { def d = 23.34375;     s"$d\t => ${doubleToBin(d)}" } )
   println( { def s = "1011.11101"; s"$s\t => ${binToDec(s)}"    } )
 }

}</lang>

Output:

Experience running it in your browser by Scastie (remote JVM).

Sidef

<lang ruby>func dec2bin(String n) {

   Num(Num(n, 10).base(2), 10)

}

func bin2dec(String n) {

   Num(Num(n, 10).base(10), 2)

}

with("23.34375") { |s| say (" #{s} => ", dec2bin(s)) } with("1011.11101") { |s| say ( "#{s} => ", bin2dec(s)) }</lang>

Output:

  23.34375 => 10111.01011
1011.11101 => 11.90625

Tcl

By far the easiest way to do this is to use Tcl's built-in handling of IEEE arithmetic, converting the IEEE representation into the string representation we want (and vice versa) by simple string manipulations.

Works with: Tcl version 8.6

<lang tcl>package require Tcl 8.6

proc dec2bin x {

   binary scan [binary format R $x] B* x
   regexp {(.)(.{8})(.{23})} $x -> s e m
   binary scan [binary format B* $e] cu e
   if {$e == 0 && ![string match *1* $m]} {

# Special case for zero set m 0.0

   } else {

incr e -127

set m 1$m if {$e < 0} { set m [string repeat "0" [expr {-$e}]]$m set m [string trimright [regsub {^.} $m "&."] "0"] } else { set m [string trimright [regsub ^.[string repeat . $e] $m "&."] "0"] } if {[string match *. $m]} { append m 0 }

   }
   if {$s} {

return -$m

   } else {

return $m

   }

} proc bin2dec x {

   if {[regexp {^-} $x]} {

set s 1 set x [string trimleft $x -0]

   } else {

set s 0 set x [string trimleft $x +0]

   }
   lassign [split [string trimright $x 0] .] fore aft
   if {[string length $fore]} {

set e [expr {[string length $fore] - 1}] set digits [string range $fore$aft 1 end]

   } elseif {[string length $aft]} {

set digits [string range [string trimleft $aft 0] 1 end] set e [expr {[string length $digits] - [string length $aft]}]

   } else {

set e -127 set digits {}

   }
   incr e 127
   binary scan [binary format B* [format %b%08b%-023s $s $e $digits]] R x
   return $x

}

foreach case {77 0.25 0.15625 0.1 -33.8 0 1 2 3 23.34375 11.90625} {

   set b [dec2bin $case]
   set d [bin2dec $b]
   puts "$case => $b => $d"

}</lang>

Output:

77 => 1001101.0 => 77.0
0.25 => 0.01 => 0.25
0.15625 => 0.00101 => 0.15625
0.1 => 0.000110011001100110011001101 => 0.10000000149011612
-33.8 => -100001.110011001100110011 => -33.79999923706055
0 => 0.0 => 0.0
1 => 1.0 => 1.0
2 => 10.0 => 2.0
3 => 11.0 => 3.0
23.34375 => 10111.01011 => 23.34375
11.90625 => 1011.11101 => 11.90625

Adapting the code to work with IEEE double-precision floats is left as an exercise for the reader, as is dealing with the trickier special cases of the infinities and NaN.

zkl

Rather limited. <lang zkl>fcn bin2float(bstr){ // no exponents

  bstr=bstr.strip();
  m:=bstr[0,1]=="-"; if(m)bstr=bstr[1,*]; m=m and -1 or 1;
  a,b:=bstr.split(".").apply(fcn(s){ s and s or 0 }).append(0,0);
  (a.toInt(2).toFloat() + b.toInt(2).toFloat()/(2).pow(b.len()))*m

}</lang> <lang zkl>foreach bstr in (T("1011.11101","0.01","0.11111111111","-.1","","1")){

  println(bstr," --> ",bin2float(bstr).toString(20))

}</lang>

Output:

1011.11101 --> 11.90625
0.01 --> 0.25
0.11111111111 --> 0.99951171875
-.1 --> -0.5
 --> 0
1 --> 1

<lang zkl>fcn float2bin(x,digitsOfPrecision=20){

  m,zeros:="","0"*digitsOfPrecision;
  if(x<0){ m="-"; x=-x }
  a,b:=x.modf();  // int and fractional parts
  b=(b*(2).pow(digitsOfPrecision)).round().toInt().toString(2);
  b=zeros[0,digitsOfPrecision-b.len()] + b;   // don't drop leading zeros
  if(z:=b.reverse().prefix(zeros)) b=b[0,-z]; // remove trailing zeros
  String(m,a.toString(2),".",b);   

}</lang> <lang zkl>foreach x in (T(23.34375,(0.0).pi,-33.8,0.1,0.15625)){

  println(x," --> ",s:=float2bin(x)," --> ",bin2float(s).toString(20));

}</lang>

Output:

23.3438 --> 10111.01011 --> 23.34375
3.14159 --> 11.00100100001111110111 --> 3.1415929794311523438
-33.8 --> -100001.11001100110011001101 --> -33.800000190734863281
0.1 --> 0.0001100110011001101 --> 0.1000003814697265625
0.15625 --> 0.00101 --> 0.15625