What is the real purpose of Base64 encoding? - encoding

Why do we have Base64 encoding? I am a beginner and I really don't understand why would you obfuscate the bytes into something else (unless it is encryption). In one of the books I read Base64 encoding is useful when binary transmission is not possible. Eg. When we post a form it is encoded. But why do we convert bytes into letters? Couldn't we just convert bytes into string format with a space in between? For example, 00000001 00000004? Or simply 0000000100000004 without any space because bytes always come in pair of 8?

Base64 is a way to encode binary data into an ASCII character set known to pretty much every computer system, in order to transmit the data without loss or modification of the contents itself.
For example, mail systems cannot deal with binary data because they expect ASCII (textual) data. So if you want to transfer an image or another file, it will get corrupted because of the way it deals with the data.
Note: base64 encoding is NOT a way of encrypting, nor a way of compacting data. In fact a base64 encoded piece of data is 1.333… times bigger than the original datapiece. It is only a way to be sure that no data is lost or modified during the transfer.

Base64 is a mechanism to enable representing and transferring binary data over mediums that allow only printable characters.It is most popular form of the “Base Encoding”, the others known in use being Base16 and Base32.
The need for Base64 arose from the need to attach binary content to emails like images, videos or arbitrary binary content . Since SMTP [RFC 5321] only allowed 7-bit US-ASCII characters within the messages, there was a need to represent these binary octet streams using the seven bit ASCII characters...
Hope this answers the Question

Base64 is a more or less compact way of transmitting (encoding, in fact, but with goal of transmitting) any kind of binary data.
See http://en.wikipedia.org/wiki/Base64
"The general rule is to choose a set of 64 characters that is both part of a subset common to most encodings, and also printable."
That's a very general purpose and the common need is not to waste more space than needed.
Historically, it's based on the fact that there is a common subset of (almost) all encodings used to store chars into bytes and that a lot of the 2^8 possible bytes risk loss or transformations during simple data transfer (for example a copy-paste-emailsend-emailreceive-copy-paste sequence).
(please redirect upvote to Brian's comment, I just make it more complete and hopefully more clear).

For data transmission, data can be textual or non-text(binary) like image, video, file etc.
As we know, during transmission only a stream of data(textual/printable characters) can be sent or received, hence we need a way encode non-text data like image, video, file.
Binary and ASCII representation of non-text(image, video, file) is easily obtainable.
Such non-text(binary) represenation is encoded in textual format such that each ASCII character takes one out of sixty four(A-Z, a-z, 0-9, + and /) possible character set.
Table 1: The Base 64 Alphabet
Value Encoding Value Encoding Value Encoding Value Encoding
0 A 17 R 34 i 51 z
1 B 18 S 35 j 52 0
2 C 19 T 36 k 53 1
3 D 20 U 37 l 54 2
4 E 21 V 38 m 55 3
5 F 22 W 39 n 56 4
6 G 23 X 40 o 57 5
7 H 24 Y 41 p 58 6
8 I 25 Z 42 q 59 7
9 J 26 a 43 r 60 8
10 K 27 b 44 s 61 9
11 L 28 c 45 t 62 +
12 M 29 d 46 u 63 /
13 N 30 e 47 v
14 O 31 f 48 w (pad) =
15 P 32 g 49 x
16 Q 33 h 50 y
These sixty four character set is called Base64 and encoding a given data into this character set having sixty four allowed characters is called Base64 encoding.
Let us take examples of few ASCII characters when encoded to Base64.
1 ==> MQ==
12 ==> MTI=
123 ==> MTIz
1234 ==> MTIzNA==
12345 ==> MTIzNDU=
123456 ==> MTIzNDU2
Here few points are to be noted:
Base64 encoding occurs in size of 4 characters. Because an ASCII character can take any out of 256 characters, which needs 4 characters of Base64 to cover. If the given ASCII value is represented in lesser character then rest of characters are padded with =.
= is not part of base64 character set. It is used for just padding.
Hence, one can see that the Base64 encoding is not encryption but just a way to transform any given data into a stream of printable characters which can be transmitted over network.

Related

Is this a bug in the passlib base64 encoding?

I am trying to decode an re-encode a bytesytring using passlibs base64 encoding:
from passlib.utils import binary
engine = binary.Base64Engine(binary.HASH64_CHARS)
s2 = engine.encode_bytes(engine.decode_bytes(b"1111111111111111111111w"))
print(s2)
This prints b'1111111111111111111111A' which is of course not what I expected. The last character is different.
Where is my mistake? Is this a bug?
No, it's not a bug.
In all variants of Base64, every encoded character represents just 6 bits and depending on the number of bytes encoded you can end up with 0, 2 or 4 insignificant bits on the end.
In this case the encoded string 1111111111111111111111w is 23 characters long, that means 23*6 = 138 bits which can be decoded to 17 bytes (136 bits) + 2 insignifcant bits.
The encoding you use here is not Base64 but Hash64
Base64 character map used by a number of hash formats; the ordering is wildly different from the standard base64 character map.
In the character map
HASH64_CHARS = u("./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz")
we find A on index 12 (001100) and w on index 60 (111100)
Now the 'trick' here is, that
binary.Base64Engine(binary.HASH64_CHARS) has a default parameter big=False, which means that the encoding is done in little endian format by default.
In your example it means that w is 001111 and A is 001100. During decoding the last two bits are cut off as they are not needed as explained above. When you encode it again, A is taken as the first character in the character map that can be used two encode 0011 plus two insignifiant bits.

How can U+203C be represented in (226, 128, 188) in swift chapter Strings and Characters?

When I read The Swift Programming Language Strings and Characters. I don't know how U+203C (means !!) can represented by (226, 128, 188) in utf-8.
How did it happen ?
I hope you already know how UTF-8 reserves certain bits to indicate that the Unicode character occupies several bytes. (This website can help).
First, write 0x203C in binary:
0x230C = 10000000111100
So this character takes 16 bits to represent. Due to the "header bits" in the UTF-8 encoding scheme, it would take 3 bytes to encode it:
0x230C = 10 000000 111100
1st byte 2nd byte 3rd byte
-------- -------- --------
header 1110 10 10
actual data 10 000000 111100
-------------------------------------------
full byte 11100010 10000000 10111100
decimal 226 128 188

UTF-8 in decimal

Is representing UTF-8 encoding in decimals even possible? I think only values till 255 would be correct, am I right?
As far as I know, we can only represent UTF-8 in hex or binary form.
I think it is possible. Let's look at an example:
The Unicode code point for ∫ is U+222B.
Its UTF-8 encoding is E2 88 AB, in hexadecimal representation. In octal, this would be 342 210 253. In decimal, it would be 226 136 171. That is, if you represent each byte separately.
If you look at the same 3 bytes as a single number, you have E288AB in hexadecimal; 70504253 in octal; and 14846123 in decimal.

How to convert a large number to base 36 using DC or other

I am trying to represent the maximum 64-bit unsigned value in different bases.
For base 2 (binary) it would be 64 1's:
1111111111111111111111111111111111111111111111111111111111111111
For base 16 (hex) it would be 16 F's
FFFFFFFFFFFFFFFF
For base 10 (decimal) it would be:
18446744073709551615
I'm trying to get the representation of this value in base 36 (it uses 0-9 and A-Z). There are many online base converters, but they all fail to produce the correct representation because they are limited by 64-bit math.
Does anyone know how to use DC (which is an extremely hard to use string math processors that can handle numbers of unlimited magnitude) and know how to do this conversion? Either that or can anyone tell me how I can perform this conversion with a calculator that won't fail due to integer roll-over?
I mad a quick test with ruby:
i = 'FFFFFFFFFFFFFFFF'.to_i(16)
puts i #18446744073709551615
puts i.to_s(36) #3w5e11264sgsf
You may also use larger numbers:
i = 'FFFFFFFFFFFFFFFF'.to_i(16) ** 16
puts i
puts i.to_s(36)
result:
179769313486231590617005494896502488139538923424507473845653439431848569886227202866765261632299351819569917639009010788373365912036255753178371299382143631760131695224907130882552454362167933328609537509415576609030163673758148226168953269623548572115351901405836315903312675793605327103910016259918212890625
1a1e4vngailcqaj6ud31s2kk9s94o3tyofvllrg4rx6mxa0pt2sc06ngjzleciz7lzgdt55aedc9x92w0w2gclhijdmj7le6osfi1w9gvybbfq04b6fm705brjo535po1axacun6f7013c4944wa7j0yyg93uzeknjphiegfat0ojki1g5pt5se1ylx93knpzbedn29
A short explanation what happens with big numbers:
Normal numbers are Fixnums. If you get larger numbers, the number becomes a Bignum:
small = 'FFFFFFF'.to_i(16)
big = 'FFFFFFFFFFFFFFFF'.to_i(16) ** 16
puts "%i is a %s" % [ small, small.class ]
puts "%i\n is a %s" % [ big, big.class ]
puts "%i^2 is a %s" % [ small, (small ** 2).class ]
Result:
268435455 is a Fixnum
179769313486231590617005494896502488139538923424507473845653439431848569886227202866765261632299351819569917639009010788373365912036255753178371299382143631760131695224907130882552454362167933328609537509415576609030163673758148226168953269623548572115351901405836315903312675793605327103910016259918212890625
is a Bignum
268435455^2 is a Bignum
From the documentation of Bignum:
Bignum objects hold integers outside the range of Fixnum. Bignum objects are created automatically when integer calculations would otherwise overflow a Fixnum. When a calculation involving Bignum objects returns a result that will fit in a Fixnum, the result is automatically converted.
It can be done with dc, but the output is not extremely useful.
$ dc
36
o
16
i
FFFFFFFFFFFFFFFF
p
03 32 05 14 01 01 02 06 04 28 16 28 15
Here's the explanation:
Entering a number by itself pushes that number
o pops the stack and sets the output radix.
i pops the stack and sets the input radix.
p prints the top number on the stack, in the current output radix. However, dc prints any output with a higher radix than 16 as binary (not ASCII).
In dc, the commands may be all put on the same line, like so:
$ dc
36o16iFFFFFFFFFFFFFFFFp
03 32 05 14 01 01 02 06 04 28 16 28 15
Get any language that can handle arbitrarily large integers. Ruby, Python, Haskell, you name it.
Implement the basic step: modulo 36 gives you the next digit, division by 36 gives you the number with the last digit cut out.
Map the digits to characters the way you like. For instance, '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'[digit] is fine by me. Append digits to the result as you produce them.
???
Return the concatenated string of digits. Profit!

How does LLV encoding work?

I'm working with a binary protocol that uses LLV to encode some variables.
I was given an example below which is used to specify a set of 5 chars to display.
F1 F0 F5 4C 69 6E 65 31
the F1 is specific to my device, it indicates display text on line one. The f0 and f5 I'm not sure about, the rest looks like ASCII text.
Anyone know how this encoding works exactly?
LLV is referenced in this protocol spec. pasted below, but doesn't seem to be defined in there.
http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBIQFjAA&url=http%3A%2F%2Fwww.terminalhersteller.de%2FDownload%2FPA00P016_03_en.pdf&ei=yUFPTOSzH432tgON5PjuBw&usg=AFQjCNGjS_y264qKIRCSJQpdhlSXWtiadw&sig2=jMGtIwd42dozDSq7ub844w
Since the F1 is device-specific, this leaves the rest as F0 F5 ..., and this looks like an LLVAR sequence, in which the first two bytes specify the length of the rest (decimal 05 here). My guess would be that the whole data represents F1 "Line1", which looks quite reasonable.
By the way, LLVAR stands for "VARiable length with two decimal digits specifying the length". With three decimal digits for the length, it's LLLVAR.