Split a byte array into a bit array

Question

Write a function that when given a buffer b (1 - 104857600 bytes long) and a number of bits n (1 <= n <= 64), splits the buffer into chunks of n bits. Right-pad the last chunk with 0s up to n bits.

e.g.

Given the buffer b = "f0oBaR" or equivalently [102,48,111,66,97,82] and n = 5, return

[12, 24, 24, 6, 30, 16, 19, 1, 10, 8]

This is because the above buffer, when represented as binary looks like:

01100110 00110000 01101111 01000010 01100001 01010010

And when re-grouped into 5s looks like:

01100 11000 11000 00110 11110 10000 10011 00001 01010 010[00]

Which when converted back into decimal gives the answer.

Notes

• You may use whatever data type makes the most sense in your language to represent the buffer. In PHP you'd probably use a string, in Node you might want to use a Buffer
  • If you use a string to represent the buffer, assume it's ASCII for the char -> int conversion
  • You may use an array of ints (0-255) for input if you prefer
• Return value must be an array or list of ints

Test Cases

> b = "Hello World", n = 50
318401791769729, 412278856237056

> b = [1,2,3,4,5], n = 1
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1

> b = "codegolf", n = 32
1668244581, 1735355494

> b = "codegolf" n = 64
7165055918859578470

> b = "codegolf" n = 7
49, 91, 108, 70, 43, 29, 94, 108, 51, 0

> b = "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Pellentesque vel est eu velit lacinia iaculis. Nulla facilisi. Mauris vitae elit sapien. Nullam odio nulla, laoreet at lorem eu, elementum ultricies libero. Praesent orci elit, sodales consectetur magna eget, pulvinar eleifend mi. Ut euismod leo ut tortor ultrices blandit. Praesent dapibus tincidunt velit vitae viverra. Nam posuere dui quis ipsum iaculis, quis tristique nisl tincidunt. Aliquam ac ligula a diam congue tempus sit amet quis nisl. Nam lacinia ante vitae leo efficitur, eu tincidunt metus condimentum. Cras euismod quis quam vitae imperdiet. Ut at est turpis.", n = 16
19567, 29285, 27936, 26992, 29557, 27936, 25711, 27759, 29216, 29545, 29728, 24941, 25972, 11296, 25455, 28275, 25955, 29797, 29813, 29216, 24932, 26992, 26995, 25449, 28263, 8293, 27753, 29742, 8272, 25964, 27749, 28276, 25971, 29045, 25888, 30309, 27680, 25971, 29728, 25973, 8310, 25964, 26996, 8300, 24931, 26990, 26977, 8297, 24931, 30060, 26995, 11808, 20085, 27756, 24864, 26209, 25449, 27753, 29545, 11808, 19809, 30066, 26995, 8310, 26996, 24933, 8293, 27753, 29728, 29537, 28777, 25966, 11808, 20085, 27756, 24941, 8303, 25705, 28448, 28277, 27756, 24876, 8300, 24943, 29285, 25972, 8289, 29728, 27759, 29285, 27936, 25973, 11296, 25964, 25965, 25966, 29813, 27936, 30060, 29810, 26979, 26981, 29472, 27753, 25189, 29295, 11808, 20594, 24933, 29541, 28276, 8303, 29283, 26912, 25964, 26996, 11296, 29551, 25697, 27749, 29472, 25455, 28275, 25955, 29797, 29813, 29216, 28001, 26478, 24864, 25959, 25972, 11296, 28789, 27766, 26990, 24946, 8293, 27749, 26982, 25966, 25632, 28009, 11808, 21876, 8293, 30057, 29549, 28516, 8300, 25967, 8309, 29728, 29807, 29300, 28530, 8309, 27764, 29289, 25445, 29472, 25196, 24942, 25705, 29742, 8272, 29281, 25971, 25966, 29728, 25697, 28777, 25205, 29472, 29801, 28259, 26980, 30062, 29728, 30309, 27753, 29728, 30313, 29793, 25888, 30313, 30309, 29298, 24878, 8270, 24941, 8304, 28531, 30053, 29285, 8292, 30057, 8305, 30057, 29472, 26992, 29557, 27936, 26977, 25461, 27753, 29484, 8305, 30057, 29472, 29810, 26995, 29801, 29045, 25888, 28265, 29548, 8308, 26990, 25449, 25717, 28276, 11808, 16748, 26993, 30049, 27936, 24931, 8300, 26983, 30060, 24864, 24864, 25705, 24941, 8291, 28526, 26485, 25888, 29797, 28016, 30067, 8307, 26996, 8289, 28005, 29728, 29045, 26995, 8302, 26995, 27694, 8270, 24941, 8300, 24931, 26990, 26977, 8289, 28276, 25888, 30313, 29793, 25888, 27749, 28448, 25958, 26217, 25449, 29813, 29228, 8293, 29984, 29801, 28259, 26980, 30062, 29728, 28005, 29813, 29472, 25455, 28260, 26989, 25966, 29813, 27950, 8259, 29281, 29472, 25973, 26995, 28015, 25632, 29045, 26995, 8305, 30049, 27936, 30313, 29793, 25888, 26989, 28773, 29284, 26981, 29742, 8277, 29728, 24948, 8293, 29556, 8308, 30066, 28777, 29486

> b = [2,31,73,127,179,233], n = 8
2, 31, 73, 127, 179, 233

Answer 1

Pyth, 18 17 bytes

iR2c.[t.B+C1z\0QQ

Thanks to @lirtosiast for a byte!

            z      get input
         +C1       prepend a 0x01 to prevent leading zeroes from disappearing
       .B          convert to binary string
      t            remove the leading 1 from ^^
    .[       \0Q   pad right with zeroes to multiple of second input
   c            Q  get chunks/slices of length second input
iR2                map(x: int(x, 2))

Answer 2

Jelly, 13 bytes

1;ḅ256æ«BḊsḄṖ

This takes the input as a list of integers. Try it online!

How it works

1;ḅ256æ«BḊsḄṖ  Main link. Arguments: A (list), n (integer)

1;             Prepend 1 to A.
  ḅ256         Convert from base 256 to integer.
      æ«       Bitshift the result n units to the left.
        B      Convert to binary.
         Ḋ     Discard the first binary digit (corresponds to prepended 1).
          s    Split into chunks of length n.
           Ḅ   Convert each chunk from binary to integer.
            Ṗ  Discard the last integer (corresponds to bitshift/padding).

Answer 3

Julia, 117 bytes

f(x,n,b=join(map(i->bin(i,8),x)),d=endof,z=rpad(b,d(b)+d(b)%n,0))=map(i->parse(Int,i,2),[z[i:i+n-1]for i=1:n:d(z)-n])

This is a function that accepts an integer array and an integer and returns an integer array. It's an exercise in function argument abuse.

Ungolfed:

function f(x::Array{Int,1},                  # Input array
           n::Int,                           # Input integer
           b = join(map(i -> bin(i, 8), x)), # `x` joined as a binary string
           d = endof,                        # Store the `endof` function
           z = rpad(b, d(b) + d(b) % n, 0))  # `b` padded to a multiple of n

    # Parse out the integers in base 2
    map(i -> parse(Int, i, 2), [z[i:i+n-1] for i = 1:n:d(z)-n])
end

Answer 4

JavaScript (ES6), 120 bytes

f=(a,n,b=0,t=0,r=[])=>b<n?a.length?f(a.slice(1),n,b+8,t*256+a[0],r):b?[...r,t<<n-b]:r:f(a,n,b-=n,t&(1<<b)-1,[...r,t>>b])

Recursive bit twiddling on integer arrays. Ungolfed:

function bits(array, nbits) {
    var count = 0;
    var total = 0;
    var result = [];
    for (;;) {
        if (nbits <= count) {
            // We have enough bits to be able to add to the result
            count -= nbits;
            result.push(total >> count);
            total &= (1 << count) - 1;
        } else if (array.length) {
            // Grab the next 8 bits from the array element
            count += 8;
            total <<= 8;
            total += array.shift();
        } else {
            // Deal with any leftover bits
            if (count) result.push(total << nbits - count);
            return result;
        }
    }
}

Answer 5

Python 3, 102 bytes

j=''.join
lambda s,n:[int(j(k),2)for k in zip(*[iter(j([bin(i)[2:].zfill(8)for i in s+[0]]))]*n)][:-1]

use iter trick to group string

• s: the input string/buffer
• n: the number of bits in each splitted chunk

Results

>>> f([102,48,111,66,97,82],4)
[6, 6, 3, 0, 6, 15, 4, 2, 6, 1, 5, 2, 0]

>>> f([102,48,111,66,97,82],5)
[12, 24, 24, 6, 30, 16, 19, 1, 10, 8]

>>> f([102,48,111,66,97,82],6)
[25, 35, 1, 47, 16, 38, 5, 18]

>>> f([102,48,111,66,97,82],8)
[102, 48, 111, 66, 97, 82]

Answer 6

Ruby, 114 bytes

->s,n{a=s.bytes.map{|b|b.to_s(2).rjust 8,?0}.join.split""
r=[]
r<<a.shift(n).join.ljust(n,?0).to_i(2)while a[0]
r}

Slightly cleaner:

f = -> str, num {
    arr = str.bytes.map {|byte|
        byte.to_s(2).rjust(8, "0")
    }.join.split("")
    result = []
    while arr.size > 0
        result << arr.shift(num).join.ljust(num, "0").to_i(2)
    end
    result
}

puts f["f0oBaR", 5]

Answer 7

Perl 6, 93 68 bytes

{@^a».&{sprintf "%08b",$_}.join.comb($^b)».&{:2($_~0 x$b-.chars)}}

Answer 8

PHP, 262 217 189 bytes

function f($b,$n){$M='array_map';return$M('bindec',$M(function($x)use($n){return str_pad($x,$n,0);},str_split(implode('',$M(function($s){return str_pad($s,8,0,0);},$M('decbin',$b))),$n)));}

(updated with tips from Ismael Miguel)

Formatted for readability:

function f($b, $n) {
    $M = 'array_map';
    return $M('bindec', $M(function ($x) use ($n) {
        return str_pad($x, $n, 0);
    }, str_split(implode('', $M(function ($s) {
        return str_pad($s, 8, 0, 0);
    }, $M('decbin', $b))), $n)));
}

Example:

> implode(', ',f(array_map('ord',str_split('f0oBaR')),5));
"12, 24, 24, 6, 30, 16, 19, 1, 10, 8"

Answer 9

CJam, 30 bytes

{_@{2b8 0e[}%e_0a@*+/-1<{2b}%}

Try it online!

This is an unnamed block which expects the int buffer and the amount of chunks on the stack and leaves the result on the stack.

Decided to give CJam a try. Only took me 2 hours to get it done ^^ This is probably too long, suggestions are very welcome!

Explanation

_         e# duplicate the chunk count
@         e# rotate stack, array now on top and chunk counts on the bottom
{         e# start a new block
 2b       e# convert to binary
 8 0e[    e# add zeros on the left, so the binary is 8 bits
}         e# end previous block
%         e# apply this block to each array-element (map)
e_        e# flatten array
0a        e# push an array with a single zero to the stack
@         e# rotate stack, stack contain now n [array] [0] n
*         e# repeat the array [0] n times
+         e# concat the two array
/         e# split into chunks of length n, now the stacks only contains the array
-1<       e# discard the last chunk
{2b}%     e# convert every chunk back to decimal

Answer 10

JavaScript (ES6) 104

Iterative bit by bit fiddling,

Edit 5 bytes save thx @Neil

(s,g,c=g,t=0)=>(s.map(x=>{for(i=8;i--;--c||(s.push(t),c=g,t=0))t+=t+(x>>i)%2},s=[]),c-g&&s.push(t<<c),s)

Less golfed

( 
 // parameters
 s, // byte source array
 g, // output bit group size
 // default parameters used as locals 
 c = g, // output bit counter
 t = 0  // temp bit accumulator
) => (
  s.map(x => 
    { // for each byte in s
      for(i = 8; // loop for 8 bits
        i--; 
        )
        // loop body
        t += t + (x>>i) % 2, // shift t to left and add next bit
        --c // decrement c,if c==0 add bit group to output and reset count and accumulator
          ||(s.push(t), c=g, t=0)
    }, 
    s=[] // init output, reusing s to avoid wasting another global
  ),
  c-g && s.push(t<<c), // add remaining bits, if any
  s // return result
)

Test

f=(s,g,c=g,t=0)=>(s.map(x=>{for(i=8;i--;--c||(s.push(t),c=g,t=0))t+=t+(x>>i)%2},s=[]),c-g&&s.push(t<<c),s)

function test()
{
  var a = A.value.match(/\d+/g)||[]
  var g = +G.value
  var r = f(a,g)
  
  O.textContent = r
  K.innerHTML = a.map(x=>`<i>${(256- -x).toString(2).slice(-8)}</i>`).join``
  + '\n'+ r.map(x=>`<i>${(256*256*256*256+x).toString(2).slice(-g)}</i>`).join``
}  

test()

#A { width: 50% }
#G { width: 5% }
i:nth-child(even) { color: #00c }
i:nth-child(odd) { color: #c00 }

Input array <input id=A value="102,48,111,66,97,82">
Group by bits <input id=G value=5> (up to 32)<br>
Output <button onclick="test()">-></button>
<span id=O></span>
<pre id=K></pre>

Answer 11

Ruby, 66 bytes

->s,n{(s.unpack('B*')[0]+?0*~-n).scan(/.{#{n}}/).map{|x|x.to_i 2}}

Try it online!

Takes input buffer as a string, so that a few test strings were constructed directly in the footer to avoid unprintables.

Answer 12

J, 24 bytes

[:#.-@[>\;@(_8:{."1#:@])

This is an anonymous function, which takes n as its left argument and b as numbers as its right argument.

Test:

      5 ([:#.-@[>\;@(_8:{."1#:@])) 102 48 111 66 97 82
12 24 24 6 30 16 19 1 10 8

Explanation:

[:#.-@[>\;@(_8:{."1#:@])

                   #:@]   NB. Convert each number in `b` to bits
            _8:{."1       NB. Take the last 8 items for each
                          NB.    (padding with zeroes at the front)
         ;@               NB. Make a list of all the bits
    -@[                   NB. Negate `n` 
                          NB. (\ gives non-overlapping infixes if [<0)
       >\                 NB. Get non-overlapping n-sized infixes
 [:#.                     NB. Convert those back to decimal 

Answer 13

Haskell, 112 109 bytes

import Data.Digits
import Data.Lists
n#x=unDigits 2.take n.(++[0,0..])<$>chunksOf n(tail.digits 2.(+256)=<<x)

Usage example: 5 # [102,48,111,66,97,82] -> [12,24,24,6,30,16,19,1,10,8].

How it works

import Data.Digits                  -- needed for base 2 conversion
import Data.Lists                   -- needed for "chunksOf", i.e. splitting in
                                    -- sublists of length n

           (                  =<<x) -- map over the input list and combine the
                                    -- results into a single list:
            tail.digits 2.(+256)    -- convert to base two with exactly 8 digits    
         chunksOf n                 -- split into chunks of length n    
       <$>                          -- convert every chunk (<$> is map)
    take n.(++[0,0..])              -- pad with 0s
unDigits 2                          -- convert from base 2   

Answer 14

Java, 313 306 322 bytes

I hope this beats PHP... And nope. Stupid long function names.

-7 thanks to @quartata for getting rid of public +16 to fix an error when the split was exact, thanks to @TheCoder for catching it

int[] f(String b,int s){int i=0,o[]=new int[(int)Math.ceil(b.length()*8.0/s)],a=0;String x="",t;for(char c:b.toCharArray()){t=Integer.toString(c,2);while(t.length()<8)t="0"+t;x+=t;a+=8;while(a>=s){o[i++]=Integer.parseInt(x.substring(0,s),2);x=x.substring(s,a);a-=s;}}while(a++<s)x+="0";o[i]=Integer.parseInt(x,2);return o;}

Answer 15

MATL, 9 bytes

8&B!we!XB

Try it online!

Takes input b as a string delimited by '' or as an array of comma-separated values like [102, 48, 111], then n.

8           # push 8
&B          # implicitly take input b, and use 2-element convert to binary
            # to push a binary matrix of 8 bits
!           # transpose, so each column represents an input
w           # implicitly take input n and swap it with binary matrix to top of stack
e           # reshape into n rows, padding with zeros at end
            # this matrix will have each column as an n-bit integer
!           # transpose, so each row is now the n-bit integer
XB          # convert each row to decimal
            # implicit output

Answer 16

Perl 5 -nl -MData::Dump=pp , 96 bytes

$}=$_;pp map{$_=sprintf"%-$}s",$_;y/ /0/;oct"0b$_"}(join'',map{sprintf"%08b",$_}<>)=~m/.{1,$_}/g

Try it online!

Requires the Data::Dump module.

Takes n on the first line of input and the numbers on each line after that.

Outputs to STDERR (the Debug field on TIO).

Deparsed and tidied:

BEGIN { $/ = "\n"; $\ = "\n"; }
use Data::Dump ( split( /,/, 'pp', 0 ) );
LINE: while ( defined( $_ = readline ARGV ) ) {
    chomp $_;
    $} = $_;
    pp(
        map( {
                $_ = sprintf( "%-$}s", $_ );
                tr/ /0/;
                oct "0b$_";
            } join( '', map( { sprintf '%08b', $_; } readline ARGV ) ) =~
              /.{1,$_}/g )
    );
}

Answer 17

Powershell 146 bytes

param([int[]][char[]]$b,$n)-join($b|%{[convert]::ToString($_,2).PadLeft(8,"0")})-split"(.{$n})"|?{$_}|%{[convert]::ToInt32($_.PadRight($n,"0"),2)}

---

Take in the buffer and convert it to a char array and then an integer array. For each of those convert to binary, pad the entries with 0's where needed, and join as one large string. Split that string on n characters and drop the blanks that are created. Each element from the split is padded (only the last element really would need it) and converted back into an integer. Output is an array

Answer 18

Python 3.5 - 312 292 bytes:

def d(a, b):
    o=[];o+=([str(bin(g)).lstrip('0b')if str(type(g))=="<class 'int'>"else str(bin(ord(g))).lstrip('0b')for g in a]);n=[''.join(o)[i:i+b]for i in range(0,len(''.join(o)),b)];v=[]
    for t in n:
        if len(t)!=b:n[n.index(t)]=str(t)+'0'*(b-len(t))
    v+=([int(str(f),2)for f in n])
    return v

Although this may be long, this is, in my knowledge, is the shortest way to accept both functions and arrays without errors, and still being able to retain some accuracy in Python 3.5.

Answer 19

Java, 253 247 bytes

Golfed

int i,l,a[];Integer I;String f="";int[]c(String s,int n){for(char c:s.toCharArray())f+=f.format("%08d",I.parseInt(I.toString(c, 2)));while(((l=f.length())%n)>0)f+="0";for(a=new int[l=l/n];i<l;)a[i]=I.parseInt(f.substring(i*n,i++*n+n),2);return a;}

UnGolfed

int i,l,a[];
Integer I;
String f="";
int[]c(String s,int n) {
    for(char c:s.toCharArray())
        f+=f.format("%08d",I.parseInt(I.toString(c,2)));
    while(((l=f.length())%n)>0)
        f+="0";
    for(a=new int[l=l/n];i<l;)
        a[i]=I.parseInt(f.substring(i*n,i++*n+n),2);
    return a;
}

Answer 20

Jelly, 13 bytes

+256BḊ€Ẏsz0ZḄ

Try it online!

Different from Dennis's answer.

Note: The input is actually a list of non-negative integers, but the TIO link eases the pain for you, and accepts either such a list or a string.

Answer 21

Stax, 12 bytes

è■àåk┘K¥xk└╣

Run and debug it

This isn't a function as specified in the challenge, but a program, since stax doesn't support functions. It supports input of strings or array literals.

Answer 22

Python 2, 101 bytes

lambda a,n:[int(''.join(bin(x+256)[3:]for x in a+[0]*n)[n*i:][:n],2)for i in range((len(a)*8-1)/n+1)]

Try it online!

Answer 23

Dyalog APL, 36 bytes

{2∘⊥¨↓A⍴(×/A←⍺,⍨⌈⍺÷⍨8×≢⍵)↑∊↓⍉⍵⊤⍨8/2}

Try it online!

This probably could be golfed down more.

Answer 24

Elixir, 63 60 bytes

&(s=&2-1)&&for<<x::size(&2)<-<<"#{&1}",0::size(s)>> >>,do: x

Try it online!

Takes input as Elixir binary, outputs a list of integers.

This code makes use of Elixir bitstring generator comprehension to chunk input binary &1 into bit blocks of size provided as argument &2. To account for any leftover bits at the end, we pad the binary with &2 - 1 zero bits. Here is also the place where some unwanted verbosity kicks in: Elixir complains if we don't explicitly declare &1 as bitstring, and it also doesn't support expressions in size(...), hence the need for an extra variable assignment.

Easter egg: in the footer, replace IO.inspect with IO.puts, and our function magically "translates" Lorem ipsum from Latin to Chinese - Try it online!

Answer 25

Japt, 16 bytes

After a long day of meetings, it feels like I've forgotten how to golf! Will play around with this on the train home later, see if I can improve on it.

c_¤ùT8ÃòV úTV mÍ

Try it

Answer 26

PHP, 135 129 124 bytes

function($b,$n){foreach($b as$c)$a.=sprintf('%08b',$c);foreach(str_split($a,$n)as$s)$d[]=bindec(str_pad($s,$n,0));return$d;}

Try it online!

Implemented as a function, input buffer is an array of ints and returns an array of ints.

Output

> b = "f0oBaR", n = 5
[12,24,24,6,30,16,19,1,10,8]

> b = "Hello World", n = 50
[318401791769729,412278856237056]

> b = [1,2,3,4,5], n = 1
[0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1]

Verify all test cases

Answer 27

APL(NARS), 471 chars, 942 bytes

TH←{v←↑⍴⍴⍵⋄v>2:64⋄v=2:32⋄(v=1)∧''≡0↑⍵:20⋄''≡0↑⍵:4⋄v=1:16⋄⍵≢+⍵:8⋄⍵=⌈⍵:2⋄1}
TV←{x←TH¨⍵⋄k←↑x⋄t←↑⍴⍵⋄t=+/x=2:2⋄t=+/x≤2:1⋄(k≤8)∧⍬≡x∼k:k⋄0}
T←{v←↑⍴⍴⍵⋄v>2:64+TV⍵⋄v=2:32+TV⍵⋄(v=1)∧''≡0↑⍵:20⋄''≡0↑⍵:4⋄v=1:16+TV⍵⋄⍵≢+⍵:8⋄⍵=⌈⍵:2⋄1}
RI←{t←T⍵⋄(t≠1)∧(t≠2)∧(t≠17)∧(t≠18):0⋄∧/((1⊃⍺)≤⍵)∧⍵≤(2⊃⍺)}
B←{(8⍴2)⊤⍵}⋄C←{¯1+⎕AV⍳⍵}⋄f←{t←T⍵⋄(0 255 RI⍵)∧18=t:∊B¨⍵⋄(0 255 RI x←C¨⍵)∧20=t:∊B¨x⋄,¯1}⋄W←{((↑⍴⍵)⍴2)⊥⍵}
q←{(∼1 64 RI,⍺)∨2≠T⍺:,¯1⋄x←f⍵⋄¯1=↑x:,¯1⋄t←↑⍴x⋄k←(⍺-m)×0≠m←⍺∣t⋄W⍉((t+k)÷⍺)⍺⍴(((t⍴1),k⍴0)\x)}

commented code and test:

  ⍝TH⍵ return type its argument
  TH←{v←↑⍴⍴⍵⋄v>2:64⋄v=2:32⋄(v=1)∧''≡0↑⍵:20⋄''≡0↑⍵:4⋄v=1:16⋄⍵≢+⍵:8⋄⍵=⌈⍵:2⋄1}
  ⍝ TV⍵ check if type each element of array ⍵ is the same and basic 
  ⍝ (float(int and float too),int,char,complex) and return its number (or 0 if it is not basic)
  TV←{x←TH¨⍵⋄k←↑x⋄t←↑⍴⍵⋄t=+/x=2:2⋄t=+/x≤2:1⋄(k≤8)∧⍬≡x∼k:k⋄0}
  ⍝ T⍵ return the type of ⍵ [it would be ok if ⍵ is not a function]
  ⍝|1 Float|2 Int|4 Char|8 Complex,Quaternion or Oction|16 List|32 Matrix|64 Tensor
  ⍝|17 List Float|18 List Int|20 List Char=string|etc
  T←{v←↑⍴⍴⍵⋄v>2:64+TV⍵⋄v=2:32+TV⍵⋄(v=1)∧''≡0↑⍵:20⋄''≡0↑⍵:4⋄v=1:16+TV⍵⋄⍵≢+⍵:8⋄⍵=⌈⍵:2⋄1}
  ⍝ ⍺RI⍵ check if the numeric array ⍵ has elements in [1⊃⍺ 2⊃⍺]; if type is not ok return 0(false)
  RI←{t←T⍵⋄(t≠1)∧(t≠2)∧(t≠17)∧(t≠18):0⋄∧/((1⊃⍺)≤⍵)∧⍵≤(2⊃⍺)}

  B←{(8⍴2)⊤⍵}   ⍝ from decimal to binary of element 0..255
  C←{¯1+⎕AV⍳⍵}   ⍝ return the number of char that is in array AV seems the ascii number
  ⍝ f⍵ with ⍵ List int element in 0..255 or String with numeric element 0..255 
  ⍝ return the corrispondence disclosed binary array 
  f←{t←T⍵⋄(0 255 RI⍵)∧18=t:∊B¨⍵⋄(0 255 RI x←C¨⍵)∧20=t:∊B¨x⋄,¯1}
  W←{((↑⍴⍵)⍴2)⊥⍵} ⍝ from list of binary digit to decimal
  ⍝ the function for the exercise
  q←{(∼1 64 RI,⍺)∨2≠T⍺:,¯1⋄x←f⍵⋄¯1=↑x:,¯1⋄t←↑⍴x⋄k←(⍺-m)×0≠m←⍺∣t⋄W⍉((t+k)÷⍺)⍺⍴(((t⍴1),k⍴0)\x)}


  5 q    'f0oBaR'
12 24 24 6 30 16 19 1 10 8 
  50 q "Hello World"
318401791769729 412278856237056 
  1  q 1 2 3 4 5
0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 
  32 q "codegolf"
1668244581 1735355494 
  7 q "codegolf"
49 91 108 70 43 29 94 108 51 0 
  8 q 2 31 73 127 179 233
2 31 73 127 179 233 
  64 q 2 31 73 127 179 233
1.529217252E17 
  65 q 2 31 73 127 179 233
¯1 
  0 q 2 31 73 127 179 233
¯1 
  23 q '123'
1612057 4194304 
  23 q '123∞'
¯1 
  23 q '1' 2 3
¯1 
  23 q 2 3.3
¯1 
  23 q 2 
¯1 
  23 q '1'
¯1 
  23 q ,2 
65536 
  23 q ,'1'
1605632