Variable-length quantity

From Rosetta Code
Revision as of 08:52, 8 February 2013 by rosettacode>Georg Peter (Use radix operator instead of str function)
Task
Variable-length quantity
You are encouraged to solve this task according to the task description, using any language you may know.

Implement some operations on variable-length quantities, at least including conversions from a normal number in the language to the binary representation of the variable-length quantity for that number, and vice versa. Any variants are acceptable.

Task : With above operations,

  • convert these two numbers 0x200000 (2097152 in decimal) and 0x1fffff (2097151 in decimal) into sequences of octets (an eight-bit byte);
  • display these sequences of octets;
  • convert these sequences of octets back to numbers, and check that they are equal to original numbers.

Ada

<lang Ada>with Ada.Containers.Vectors; with Ada.Text_IO; with Ada.Unchecked_Conversion;

procedure VLQ is

  package Nat_IO is new Ada.Text_IO.Integer_IO (Natural);
  type Byte is mod 2**8;
  package Byte_IO is new Ada.Text_IO.Modular_IO (Byte);
  type Int7 is mod 2**7;
  package Int7_IO is new Ada.Text_IO.Modular_IO (Int7);
  type VLQ_Octet is record
     Value : Int7 := 0;
     Next  : Boolean := True;
  end record;
  pragma Pack (VLQ_Octet);
  for VLQ_Octet'Size use 8;
  function VLQ_To_Byte is new Ada.Unchecked_Conversion (VLQ_Octet, Byte);
  function Byte_To_VLQ is new Ada.Unchecked_Conversion (Byte, VLQ_Octet);
  package VLQ_Vectors is new Ada.Containers.Vectors (Natural, VLQ_Octet);
  procedure Hex_Print (Position : in VLQ_Vectors.Cursor) is
     Value : Byte := VLQ_To_Byte (VLQ_Vectors.Element (Position));
  begin
     Ada.Text_IO.Put (':');
     Byte_IO.Put (Item => Value, Width => 6, Base => 16);
  end Hex_Print;
  procedure Print (X : VLQ_Vectors.Vector) is
  begin
     X.Iterate (Hex_Print'Access);
     Ada.Text_IO.New_Line;
  end Print;
  function To_VLQ (From : Natural) return VLQ_Vectors.Vector is
     Result : VLQ_Vectors.Vector;
     Current : Natural := From;
     Element : VLQ_Octet;
  begin
     loop
        Element.Value := Int7 (Current mod 2**7);
        Result.Prepend (Element);
        Current := Current / 2**7;
        exit when Current = 0;
     end loop;
     Element := Result.Last_Element;
     Element.Next := False;
     VLQ_Vectors.Replace_Element (Result, Result.Last, Element);
     return Result;
  end To_VLQ;
  function To_Int (From : VLQ_Vectors.Vector) return Natural is
     use type VLQ_Vectors.Cursor;
     Result : Natural := 0;
     Iterator : VLQ_Vectors.Cursor := From.First;
  begin
     while Iterator /= VLQ_Vectors.No_Element loop
        Result := Result * 2**7;
        Result := Result + Natural(VLQ_Vectors.Element (Iterator).Value);
        VLQ_Vectors.Next (Iterator);
     end loop;
     return Result;
  end To_Int;
  Test : VLQ_Vectors.Vector;

begin

  Test := To_VLQ (16#7f#);
  Nat_IO.Put (To_Int (Test), 10, 16); Ada.Text_IO.Put (" = ");
  Print (Test);
  Test := To_VLQ (16#4000#);
  Nat_IO.Put (To_Int (Test), 10, 16); Ada.Text_IO.Put (" = ");
  Print (Test);
  Test := To_VLQ (16#0#);
  Nat_IO.Put (To_Int (Test), 10, 16); Ada.Text_IO.Put (" = ");
  Print (Test);
  Test := To_VLQ (16#3FFFFE#);
  Nat_IO.Put (To_Int (Test), 10, 16); Ada.Text_IO.Put (" = ");
  Print (Test);
  Test := To_VLQ (16#1FFFFF#);
  Nat_IO.Put (To_Int (Test), 10, 16); Ada.Text_IO.Put (" = ");
  Print (Test);
  Test := To_VLQ (16#200000#);
  Nat_IO.Put (To_Int (Test), 10, 16); Ada.Text_IO.Put (" = ");
  Print (Test);

end VLQ;</lang>

Output:

    16#7F# = :16#7F#
  16#4000# = :16#81#:16#80#: 16#0#
     16#0# = : 16#0#
16#3FFFFE# = :16#81#:16#FF#:16#FF#:16#7E#
16#1FFFFF# = :16#FF#:16#FF#:16#7F#
16#200000# = :16#81#:16#80#:16#80#: 16#0#

C

<lang c>#include <stdio.h>

  1. include <stdint.h>

void to_seq(uint64_t x, uint8_t *out) { int i, j; for (i = 9; i > 0; i--) { if (x & 127ULL << i * 7) break; } for (j = 0; j <= i; j++) out[j] = ((x >> ((i - j) * 7)) & 127) | 128;

out[i] ^= 128; }

uint64_t from_seq(uint8_t *in) { uint64_t r = 0;

do { r = (r << 7) | (uint64_t)(*in & 127); } while (*in++ & 128);

return r; }

int main() { uint8_t s[10]; uint64_t x[] = { 0x7f, 0x4000, 0, 0x3ffffe, 0x1fffff, 0x200000, 0x3311a1234df31413ULL};

int i, j; for (j = 0; j < sizeof(x)/8; j++) { to_seq(x[j], s); printf("seq from %llx: [ ", x[j]);

i = 0; do { printf("%02x ", s[i]); } while ((s[i++] & 128)); printf("] back: %llx\n", from_seq(s)); }

return 0; }</lang>output<lang>seq from 7f: [ 7f ] back: 7f seq from 4000: [ 81 80 00 ] back: 4000 seq from 0: [ 00 ] back: 0 seq from 3ffffe: [ 81 ff ff 7e ] back: 3ffffe seq from 1fffff: [ ff ff 7f ] back: 1fffff seq from 200000: [ 81 80 80 00 ] back: 200000 seq from 3311a1234df31413: [ b3 88 e8 a4 b4 ef cc a8 13 ] back: 3311a1234df31413</lang>

D

This implements a Variable-length Quantity struct for an ulong integer. <lang d>import std.stdio, std.string, std.file, std.algorithm;

/// Variable length quantity (unsigned long, max 63-bit). struct VLQ {

   ulong value;
   // This allows VLQ to work like an ulong.
   alias this = value;
   uint extract(in ubyte[] v) pure
   in {
       assert(v.length > 0);
   } body {
       immutable limit = min(v.length - 1, 8);
       ulong t = 0;
       size_t idx = 0;
       while ((idx < limit) && ((v[idx] & 0x80) > 0))
           t = (t << 7) | (0x7f & v[idx++]);
       if (idx > limit)
           throw new Exception(
               "Too large for ulong or invalid format.");
       else
           value = (t << 7) | v[idx];
       return idx + 1;
   }
   VLQ from(in ubyte[] v) pure {
       extract(v);
       return this;
   }
   @property ubyte[] toVLQ() const pure {
       ubyte[] v = [0x7f & value];
       for (ulong k = value >>> 7; k > 0; k >>>= 7)
           v ~= (k & 0x7f) | 0x80;
       if (v.length > 9)
           throw new Exception("Too large value.");
       v.reverse();
       return v;
   }
   static ulong[] split(in ubyte[] b) pure {
       ulong[] res;
       VLQ v;
       for (size_t i = 0; i < b.length; ) {
           i += v.extract(b[i .. $]);
           res ~= v.value;
       }
       return res;
   }
   string toString() const /*pure nothrow*/ {
       return format("(%(%02X:%))", this.toVLQ);
   }

}


void main() { // VLQ demo code.

   VLQ a = VLQ(0x7f),
       b = VLQ(0x4000),
       c;
   writefln("a:%8x = %s\nb:%8x = %s\nc:%8x = %s",
            a.value, a, b.value, b, c.value, c);
   // Some operations.
   c = (a + 1) * b;
   a = c - 1;
   b = VLQ().from(a.toVLQ);
   a <<= 1;
   // Convert ulong to octet sequence.
   writefln("\na:%8x = %s\nb:%8x = %s\nc:%8x = %s",
            a.value, a, b.value, b, c.value, c);
   // Write them to a binary file.
   std.file.write("vlqtest.bin", a.toVLQ ~ b.toVLQ ~ c.toVLQ);
   // Read them back.
   const buf = cast(ubyte[])std.file.read("vlqtest.bin");
   writefln("\nFile length: %d bytes.", buf.length);
   // Convert octet sequence to ulongs.
   foreach (immutable i, immutable v; VLQ.split(buf))
       writefln("%d:%8x = %s", i + 1, v, VLQ(v));

}</lang>

Output:
a:      7f = (7F)
b:    4000 = (81:80:00)
c:       0 = (00)

a:  3ffffe = (81:FF:FF:7E)
b:  1fffff = (FF:FF:7F)
c:  200000 = (81:80:80:00)

File length: 11 bytes.
1:  3ffffe = (81:FF:FF:7E)
2:  1fffff = (FF:FF:7F)
3:  200000 = (81:80:80:00)

Euphoria

<lang euphoria>function vlq_encode(integer n)

   sequence s
   s = {}
   while n > 0 do
       s = prepend(s, #80 * (length(s) > 0) + and_bits(n, #7F))
       n = floor(n / #80)
   end while
   if length(s) = 0 then
       s = {0}
   end if
   return s

end function

function vlq_decode(sequence s)

   integer n
   n = 0
   for i = 1 to length(s) do
       n *= #80
       n += and_bits(s[i], #7F)
       if not and_bits(s[i], #80) then
           exit
       end if
   end for
   return n

end function

function svlg(sequence s)

   sequence out
   out = ""
   for i = 1 to length(s) do
       out &= sprintf("#%02x:", {s[i]})
   end for
   return out[1..$-1]

end function

constant testNumbers = { #200000, #1FFFFF, 1, 127, 128 } sequence s for i = 1 to length(testNumbers) do

   s = vlq_encode(testNumbers[i])
   printf(1, "#%02x -> %s -> #%02x\n", {testNumbers[i], svlg(s), vlq_decode(s)})

end for</lang>

Output:

#200000 -> #81:#80:#80:#00 -> #200000
#1FFFFF -> #FF:#FF:#7F -> #1FFFFF
#01 -> #01 -> #01
#7F -> #7F -> #7F
#80 -> #81:#00 -> #80

Go

Go has an implementation of variable length quantities in the standard library. <lang go>package main

import (

   "fmt"
   "encoding/binary"

)

func main() {

   buf := make([]byte, binary.MaxVarintLen64)
   for _, x := range []int64{0x200000, 0x1fffff} {
       v := buf[:binary.PutVarint(buf, x)]
       fmt.Printf("%d encodes into %d bytes: %x\n", x, len(v), v)
       x, _ = binary.Varint(v)
       fmt.Println(x, "decoded")
   }

}</lang> Output required by task:

2097152 encodes into 4 bytes: 80808002
2097152 decoded
2097151 encodes into 4 bytes: feffff01
2097151 decoded

More output showing negative numbers, the roll over from one byte to two, and larger numbers of different lengths:

0 encodes into 1 bytes: 00
0 decoded
1 encodes into 1 bytes: 02
1 decoded
2 encodes into 1 bytes: 04
2 decoded
-1 encodes into 1 bytes: 01
-1 decoded
-2 encodes into 1 bytes: 03
-2 decoded
63 encodes into 1 bytes: 7e
63 decoded
64 encodes into 2 bytes: 8001
64 decoded
589723405834 encodes into 6 bytes: 94b888e4a922
589723405834 decoded
3679899543542109203 encodes into 9 bytes: a6d098dfe9c8d09166
3679899543542109203 decoded

Groovy

Solution: <lang groovy>final RADIX = 7 final MASK = 2**RADIX - 1

def octetify = { n ->

   def octets = []
   for (def i = n; i != 0; i >>>= RADIX) {
       octets << ((byte)((i & MASK) + (octets.empty ? 0 : MASK + 1)))
   }
   octets.reverse()

}

def deoctetify = { octets ->

   octets.inject(0) { long n, octet ->
       (n << RADIX) + ((int)(octet) & MASK)
   }

}</lang>

Test (samples borrowed from Java example): <lang groovy>def testNumbers = [ 0x200000, 0x1fffff, 1, 127, 128, 589723405834L ]

testNumbers.each { a ->

   def octets = octetify(a)
   octets.each { printf "0x%02x ", it }; println ()
   def a1 = deoctetify(octets)
   assert a1 == a

}</lang>

Output:

0x81 0x80 0x80 0x00 
0xff 0xff 0x7f 
0x01 
0x7f 
0x81 0x00 
0x91 0x94 0xf2 0x84 0x9c 0x0a

Icon and Unicon

<lang Icon>procedure main() every i := 2097152 | 2097151 | 1 | 127 | 128 | 589723405834 | 165 | 256 do

  write(image(i)," = ",string2hex(v := uint2vlq(i))," = ",vlq2uint(v))

end

procedure vlq2uint(s) #: decode a variable length quantity

  if *s > 0 then {
     i := 0
     s ? while h := ord(move(1)) do {
        if (pos(0) & h > 128) | (not pos(0) & h < 128) then fail
        i := 128 * i + h % 128
        }
     return i
     }

end

procedure uint2vlq(i,c) #: encode a whole number as a variable length quantity

  if "integer" == type(-1 < i) then 
     return if i = 0 then 
        char((/c := 0)) | ""
     else          
        uint2vlq(i/128,1) || char((i % 128) + ((/c := 0) | 128) )

end

procedure string2hex(s) #: convert a string to hex

  h := ""
  every i := ord(!s) do 
     h ||:= "0123456789abcdef"[i/16+1] || "0123456789abcdef"[i%16+1]
  return h

end</lang>

Output:

2097152 = 81808000 = 2097152
2097151 = ffff7f = 2097151
1 = 01 = 1
127 = 7f = 127
128 = 8100 = 128
589723405834 = 9194f2849c0a = 589723405834
165 = 8125 = 165
256 = 8200 = 256

J

<lang j>N=: 128x v2i=: (N&| N&#./.~ [: +/\ _1 |. N&>)@i.~&a. i2v=: a. {~ [:;}.@(N+//.@,:N&#.inv)&.> ifv=: v2i :. i2v vfi=: i2v :. v2i</lang>

ifv is an invertible function which gets an (unsigned, arbitrary precision) integer sequence from a variable-length quantity sequence. vfi is an invertable function which gets a variable-length quantity sequence from an unsigned integer sequence. av displays character code numbers corresponding to the characters in its argument.

Example use:

<lang j> require'convert'

  numbers=: 16b7f 16b4000 0 16b3ffffe 16b1fffff 200000
  av vlq=: vfi numbers

127 129 128 0 0 129 255 255 126 255 255 127 140 154 64

  av (vfi 1 2 3 4 5 6) +&.ifv vlq

129 0 129 128 2 3 130 128 128 2 129 128 128 4 140 154 70</lang>

Java

<lang java>public class VLQCode {

 public static byte[] encode(long n)
 {
   int numRelevantBits = 64 - Long.numberOfLeadingZeros(n);
   int numBytes = (numRelevantBits + 6) / 7;
   if (numBytes == 0)
     numBytes = 1;
   byte[] output = new byte[numBytes];
   for (int i = numBytes - 1; i >= 0; i--)
   {
     int curByte = (int)(n & 0x7F);
     if (i != (numBytes - 1))
       curByte |= 0x80;
     output[i] = (byte)curByte;
     n >>>= 7;
   }
   return output;
 }
 
 public static long decode(byte[] b)
 {
   long n = 0;
   for (int i = 0; i < b.length; i++)
   {
     int curByte = b[i] & 0xFF;
     n = (n << 7) | (curByte & 0x7F);
     if ((curByte & 0x80) == 0)
       break;
   }
   return n;
 }
 
 public static String byteArrayToString(byte[] b)
 {
   StringBuilder sb = new StringBuilder();
   for (int i = 0; i < b.length; i++)
   {
     if (i > 0)
       sb.append(", ");
     String s = Integer.toHexString(b[i] & 0xFF);
     if (s.length() < 2)
       s = "0" + s;
     sb.append(s);
   }
   return sb.toString();
 }
 
 public static void main(String[] args)
 {
   long[] testNumbers = { 2097152, 2097151, 1, 127, 128, 589723405834L };
   for (long n : testNumbers)
   {
     byte[] encoded = encode(n);
     long decoded = decode(encoded);
     System.out.println("Original input=" + n + ", encoded = [" + byteArrayToString(encoded) + "], decoded=" + decoded + ", " + ((n == decoded) ? "OK" : "FAIL"));
   }
 }

} </lang>

Output:

Original input=2097152, encoded = [81, 80, 80, 00], decoded=2097152, OK
Original input=2097151, encoded = [ff, ff, 7f], decoded=2097151, OK
Original input=1, encoded = [01], decoded=1, OK
Original input=127, encoded = [7f], decoded=127, OK
Original input=128, encoded = [81, 00], decoded=128, OK
Original input=589723405834, encoded = [91, 94, f2, 84, 9c, 0a], decoded=589723405834, OK

OCaml

<lang ocaml>let to_vlq n =

 let a, b = n lsr 7, n land 0x7F in
 let rec aux n acc =
   let x = (n land 0x7F) lor 0x80
   and xs = n lsr 7 in
   if xs > 0
   then aux xs (x::acc)
   else x::acc
 in
 aux a [b]

let to_num = List.fold_left (fun n x -> n lsl 7 + (x land 0x7F)) 0

let v_rep n =

 Printf.printf "%d ->" n;
 let seq = to_vlq n in
 List.iter (Printf.printf " 0x%02X") seq;
 let num = to_num seq in
 Printf.printf "-> %d\n%!" num;
 assert (n = num)

let _ =

 v_rep 0x200000;
 v_rep 0x1FFFFF

</lang>

Outputs:

$ ocaml variable_length.ml
2097152 -> 0x81 0x80 0x80 0x00 -> 2097152
2097151 -> 0xFF 0xFF 0x7F -> 2097151

Perl

The vlg_encode sub returns an array of octets in most -> least significant order. Simply reverse the array to reverse the order.

<lang perl> use warnings; use strict;

for my $testcase (

   0,   0xa,   123,   254,   255,   256,
   257, 65534, 65535, 65536, 65537, 0x1fffff,
   0x200000
 )

{

   my @vlq = vlq_encode($testcase);
   printf "%8s %12s %8s\n", $testcase, ( join ':', @vlq ), vlq_decode(@vlq);

}

sub vlq_encode {

   my @vlq;
   my $binary = sprintf "%s%b", 0 x 7, shift;
   $binary =~ s/(.{7})$//;
   @vlq = ( unpack 'H2', ( pack 'B8', '0' . $1 ) );
   while ( 0 + $binary ) {
       $binary =~ s/(.{7})$//;
       unshift @vlq, ( unpack 'H2', pack 'B8', '1' . $1 );
   }
   return @vlq;

}

sub vlq_decode {

   my $num;
   $num .= sprintf "%07b", hex(shift @_) & 0x7f while @_;
   return oct '0b' . $num;

} </lang>

Output:

       0           00        0
      10           0a       10
     123           7b      123
     254        81:7e      254
     255        81:7f      255
     256        82:00      256
     257        82:01      257
   65534     83:ff:7e    65534
   65535     83:ff:7f    65535
   65536     84:80:00    65536
   65537     84:80:01    65537
 2097151     ff:ff:7f  2097151
 2097152  81:80:80:00  2097152

Perl 6

vlq_encode() returns a string of characters whose ordinals are the encoded octets. vlq_decode() takes a string and returns a decimal number. <lang perl6>sub vlq_encode ($number is copy) {

   my $string = ;
   my $t = 0x7F +& $number;
   $number +>= 7;
   $string = $t.chr ~ $string;
   while ($number) {
      $t = 0x7F +& $number;
      $string = (0x80 +| $t).chr ~ $string;
      $number +>= 7; 
   }
   return $string;

}

sub vlq_decode ($string is copy) {

   my $number = '0b';
   for $string.ords -> $oct {
       $number ~= ($oct +& 0x7F).fmt("%07b");
   }
   return :2($number);

}

  1. test encoding and decoding

for (

   0,   0xa,   123,   254,   255,   256,
   257, 65534, 65535, 65536, 65537, 0x1fffff,
   0x200000
) -> $testcase {
   my $encoded = vlq_encode($testcase);
   printf "%8s %12s %8s\n", $testcase,
     ( join ':', $encoded.ords>>.fmt("%02X") ),
     vlq_decode($encoded);

}</lang>

Output:

       0           00        0
      10           0A       10
     123           7B      123
     254        81:7E      254
     255        81:7F      255
     256        82:00      256
     257        82:01      257
   65534     83:FF:7E    65534
   65535     83:FF:7F    65535
   65536     84:80:00    65536
   65537     84:80:01    65537
 2097151     FF:FF:7F  2097151
 2097152  81:80:80:00  2097152

PicoLisp

<lang PicoLisp>(de numToVlq (Num)

  (let Res (cons (& Num 127))
     (while (gt0 (setq Num (>> 7 Num)))
        (push 'Res (| 128 (& Num 127))) )
     Res ) )

(de vlqToNum (Vlq)

  (let Res 0
     (for N Vlq
        (setq Res (| (>> -7 Res) (& N 127))) ) ) )

(for Num (0 15 16 127 128 255 2097151 2097152)

  (let Vlq (numToVlq Num)
     (tab (12 12 12) Num (glue ":" (mapcar hex Vlq)) (vlqToNum Vlq)) ) )</lang>

Output:

           0           0           0
          15           F          15
          16          10          16
         127          7F         127
         128        81:0         128
         255       81:7F         255
     2097151    FF:FF:7F     2097151
     2097152  81:80:80:0     2097152

PL/I

<lang PL/I> test: procedure options(main);

  declare s character (20) varying;
  declare c character (1);
  declare v fixed binary (31);
  declare (i, k) fixed binary;
  get edit (s) (L);
  s = trim (s);
  v = 0;
  do i = 1 to length(s);
     c = substr(s, i, 1);
     k = index('0123456789abcdef', c);
     if k > 0 then v = v*16 + k - 1;
  end;
  put skip data (s, v);
  /* Convert back to hex */
  declare hex character(16) initial ('0123456789abcdef');
  declare hs character (20) initial ();
  declare d fixed binary;
  do i = length(hs) to 1 by -1 until (v = 0);
     d = mod(v, 16) + 1;
     substr(hs, i, 1) = substr(hex, d, 1);
     v = v/16;
  end;
  put skip list (hs);

end test; </lang> OUTPUT:

S='200000'              V=       2097152;
              200000 
S='1fffff'              V=       2097151;
              1fffff 

Python

The vlq format is computed in a form for printing. This could easily be changed to a series of 8 bit ASCII chars whose integer value corresponds to the vlq for saving or transmission.

When transmitting the Vlq, octets are sent from the rightmost of the Vlq first.

<lang python>def tobits(n, _group=8, _sep='_', _pad=False):

   'Express n as binary bits with separator'
   bits = '{0:b}'.format(n)[::-1]
   if _pad:
       bits = '{0:0{1}b}'.format(n,
                                 ((_group+len(bits)-1)//_group)*_group)[::-1]
       answer = _sep.join(bits[i:i+_group]
                                for i in range(0, len(bits), _group))[::-1]
       answer = '0'*(len(_sep)-1) + answer
   else:
       answer = _sep.join(bits[i:i+_group]
                          for i in range(0, len(bits), _group))[::-1]
   return answer

def tovlq(n):

   return tobits(n, _group=7, _sep='1_', _pad=True)

def toint(vlq):

   return int(.join(vlq.split('_1')), 2)    

def vlqsend(vlq):

   for i, byte in enumerate(vlq.split('_')[::-1]):
       print('Sent byte {0:3}: {1:#04x}'.format(i, int(byte,2)))</lang>


Sample Output The underscore separates groups of eight bits (octets), for readability <lang python>>>> for n in (254, 255, 256, 257, -2+(1<<16), -1+(1<<16), 1<<16, 1+(1<<16), 0x200000, 0x1fffff ):

   print('int: %7i bin: %26s vlq: %35s vlq->int: %7i' % (n, tobits(n,_pad=True), tovlq(n), toint(tovlq(n))))


int: 254 bin: 11111110 vlq: 00000001_11111110 vlq->int: 254 int: 255 bin: 11111111 vlq: 00000001_11111111 vlq->int: 255 int: 256 bin: 00000001_00000000 vlq: 00000010_10000000 vlq->int: 256 int: 257 bin: 00000001_00000001 vlq: 00000010_10000001 vlq->int: 257 int: 65534 bin: 11111111_11111110 vlq: 00000011_11111111_11111110 vlq->int: 65534 int: 65535 bin: 11111111_11111111 vlq: 00000011_11111111_11111111 vlq->int: 65535 int: 65536 bin: 00000001_00000000_00000000 vlq: 00000100_10000000_10000000 vlq->int: 65536 int: 65537 bin: 00000001_00000000_00000001 vlq: 00000100_10000000_10000001 vlq->int: 65537 int: 2097152 bin: 00100000_00000000_00000000 vlq: 00000001_10000000_10000000_10000000 vlq->int: 2097152 int: 2097151 bin: 00011111_11111111_11111111 vlq: 01111111_11111111_11111111 vlq->int: 2097151 >>> vlqsend(tovlq(0x200000)) Sent byte 0: 0x80 Sent byte 1: 0x80 Sent byte 2: 0x80 Sent byte 3: 0x01 >>> vlqsend(tovlq(0x1fffff)) Sent byte 0: 0xff Sent byte 1: 0xff Sent byte 2: 0x7f >>> </lang>


REXX

<lang rexx>/*REXX program to test displaying of octets. */

num1=x2d(200000)  ; say 'number1='num1 ' [in hex='d2x(num1)"]" num2=x2d(1fffff)  ; say 'number2='num2 ' [in hex='d2x(num2)"]" num3=2097172  ; say 'number3='num3 ' [in hex='d2x(num3)"]" num4=2097151  ; say 'number4='num4 ' [in hex='d2x(num4)"]"

                            say

onum1=octet(num1)  ; say 'number1 octet='onum1 onum2=octet(num2)  ; say 'number2 octet='onum2 onum3=octet(num3)  ; say 'number3 octet='onum3 onum4=octet(num4)  ; say 'number4 octet='onum4

                            say

bnum1=x2d(space(onum1,0)) ; say 'number1='bnum1 bnum2=x2d(space(onum2,0)) ; say 'number2='bnum2 bnum3=x2d(space(onum3,0)) ; say 'number3='bnum3 bnum4=x2d(space(onum4,0)) ; say 'number4='bnum4

                            say

if num1==bnum1 &,

  num2==bnum2 &,
  num3==bnum3 &,
  num4==bnum4    then say 'numbers are OK'
                 else say 'trouble in River City'

exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────OCTET subroutine────────────────────*/ octet: procedure; parse arg a,_; x=d2x(a) /*convert A to hex octet.*/

        do j=length(x) by -2 to 1           /*process little end first.*/
        _=substr(x,j-1,2,0) _      /*pad odd hexchars with an 0 on left*/
        end   /*j*/

return _</lang> output

number1=2097152   [in hex=200000]
number2=2097151   [in hex=1FFFFF]
number3=2097172   [in hex=200014]
number4=2097151   [in hex=1FFFFF]

number1 octet=20 00 00
number2 octet=1F FF FF
number3 octet=20 00 14
number4 octet=1F FF FF

number1=2097152
number2=2097151
number3=2097172
number4=2097151

numbers are OK

Ruby

Array#pack can encode the BER-compressed integer, which is identical to the variable-length quantity from the MIDI specification. String#unpack can decode it.

<lang ruby>[2097152, 2097151].each do |i|

 # Encode i => BER
 ber = [i].pack("w")
 hex = ber.unpack("C*").collect {|c| "%02x" % c}.join(":")
 printf "%s => %s\n", i, hex
 # Decode BER => j
 j = ber.unpack("w").first
 i == j or fail "BER not preserve integer"

end</lang>

2097152 => 81:80:80:00
2097151 => ff:ff:7f

Scala

<lang scala>object VlqCode {

 def encode(x:Long)={
   val result=scala.collection.mutable.Stack[Byte]()
   result push (x&0x7f).toByte
   var l = x >>> 7
   while(l>0){
     result push ((l&0x7f)|0x80).toByte
     l >>>= 7
   }
   result.toArray
 }
 
 def decode(a:Array[Byte])=a.foldLeft(0L)((r, b) => r<<7|b&0x7f)
 
 def toString(a:Array[Byte])=a map("%02x".format(_)) mkString("[", ", ", "]")  
 
 def test(x:Long)={
   val enc=encode(x)
   println("0x%x => %s => 0x%x".format(x, toString(enc), decode(enc)))
 }
 
 def main(args: Array[String]): Unit = {
   val xs=Seq(0, 0x7f, 0x80, 0x2000, 0x3fff, 0x4000, 0x1FFFFF, 0x200000, 0x8000000,
              0xFFFFFFF, 0xFFFFFFFFL, 0x842FFFFFFFFL, 0x0FFFFFFFFFFFFFFFL)
   xs foreach test
 }

}</lang> Output:

0x0 => [00] => 0x0
0x7f => [7f] => 0x7f
0x80 => [81, 00] => 0x80
0x2000 => [c0, 00] => 0x2000
0x3fff => [ff, 7f] => 0x3fff
0x4000 => [81, 80, 00] => 0x4000
0x1fffff => [ff, ff, 7f] => 0x1fffff
0x200000 => [81, 80, 80, 00] => 0x200000
0x8000000 => [c0, 80, 80, 00] => 0x8000000
0xfffffff => [ff, ff, ff, 7f] => 0xfffffff
0xffffffff => [8f, ff, ff, ff, 7f] => 0xffffffff
0x842ffffffff => [82, 88, af, ff, ff, ff, 7f] => 0x842ffffffff
0xfffffffffffffff => [8f, ff, ff, ff, ff, ff, ff, ff, 7f] => 0xfffffffffffffff

Seed7

The example below uses bigInteger numbers, since variable-length quantities are able to represent integer numbers of unlimited size. <lang seed7>$ include "seed7_05.s7i";

 include "bigint.s7i";

const func string: toSequence (in var bigInteger: number) is func

 result
   var string: sequence is "";
 begin
   sequence := str(chr(ord(number mod 128_)));
   number >>:= 7;
   while number <> 0_ do
     sequence := str(chr(ord(number mod 128_) + 128)) & sequence;
     number >>:= 7;
   end while;
 end func;

const func bigInteger: fromSequence (in string: sequence) is func

 result
   var bigInteger: number is 0_;
 local
   var integer: index is 1;
 begin
   while ord(sequence[index]) >= 128 do
     number <<:= 7;
     number +:= bigInteger conv (ord(sequence[index]) - 128);
     incr(index);
   end while;
   number <<:= 7;
   number +:= bigInteger conv ord(sequence[index]);
 end func;

const proc: main is func

 local
   const array bigInteger: testValues is [] (
       0_, 10_, 123_, 254_, 255_, 256_, 257_, 65534_, 65535_, 65536_, 65537_, 2097151_, 2097152_);
   var string: sequence is "";
   var bigInteger: testValue is 0_;
   var char: element is ' ';
 begin
   for testValue range testValues do
     sequence := toSequence(testValue);
     write("sequence from " <& testValue <& ": [ ");
     for element range sequence do
       write(ord(element) radix 16 lpad0 2 <& " ");
     end for;
     writeln("] back: " <& fromSequence(sequence));
   end for;
 end func;</lang>

Output:

sequence from 0: [ 00 ] back: 0
sequence from 10: [ 0a ] back: 10
sequence from 123: [ 7b ] back: 123
sequence from 254: [ 81 7e ] back: 254
sequence from 255: [ 81 7f ] back: 255
sequence from 256: [ 82 00 ] back: 256
sequence from 257: [ 82 01 ] back: 257
sequence from 65534: [ 83 ff 7e ] back: 65534
sequence from 65535: [ 83 ff 7f ] back: 65535
sequence from 65536: [ 84 80 00 ] back: 65536
sequence from 65537: [ 84 80 01 ] back: 65537
sequence from 2097151: [ ff ff 7f ] back: 2097151
sequence from 2097152: [ 81 80 80 00 ] back: 2097152

Tcl

<lang tcl>package require Tcl 8.5

proc vlqEncode number {

   if {$number < 0} {error "negative not supported"}
   while 1 {

lappend digits [expr {$number & 0x7f}] if {[set number [expr {$number >> 7}]] == 0} break

   }
   set out [format %c [lindex $digits 0]]
   foreach digit [lrange $digits 1 end] {

set out [format %c%s [expr {0x80+$digit}] $out]

   }
   return $out

} proc vlqDecode chars {

   set n 0
   foreach c [split $chars ""] {

scan $c %c c set n [expr {($n<<7) | ($c&0x7f)}] if {!($c&0x80)} break

   }
   return $n

}</lang> Demo code: <lang tcl>proc numtohex {num} {

   binary scan [string trimleft [binary format W $num] \0] H* hexEncoded
   regsub -all "..(?=.)" $hexEncoded "&:"

} proc strtohex {string} {

   binary scan $string H* hexEncoded
   regsub -all "..(?=.)" $hexEncoded "&:"

} foreach testcase {

   123
   254 255 256 257
   65534 65535 65536 65537
   2097152 2097151
   12345678901234566789

} {

   set encoded [vlqEncode $testcase]
   binary scan $encoded H* hexEncoded
   regsub -all {..(?=.)} $hexEncoded &: hexEncoded
   set decoded [vlqDecode $encoded]
   puts "$testcase ([numtohex $testcase]) ==>\

[strtohex $encoded] ([string length $encoded] bytes) ==>\ $decoded" }</lang> Output:

123 (7b) ==> 7b (1 bytes) ==> 123
254 (fe) ==> 81:7e (2 bytes) ==> 254
255 (ff) ==> 81:7f (2 bytes) ==> 255
256 (01:00) ==> 82:00 (2 bytes) ==> 256
257 (01:01) ==> 82:01 (2 bytes) ==> 257
65534 (ff:fe) ==> 83:ff:7e (3 bytes) ==> 65534
65535 (ff:ff) ==> 83:ff:7f (3 bytes) ==> 65535
65536 (01:00:00) ==> 84:80:00 (3 bytes) ==> 65536
65537 (01:00:01) ==> 84:80:01 (3 bytes) ==> 65537
2097152 (20:00:00) ==> 81:80:80:00 (4 bytes) ==> 2097152
2097151 (1f:ff:ff) ==> ff:ff:7f (3 bytes) ==> 2097151
12345678901234566789 (ab:54:a9:8c:eb:1f:06:85) ==> 81:ab:aa:aa:b1:ce:d8:fc:8d:05 (10 bytes) ==> 12345678901234566789