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I am struggling to calculate the packet error code (PEC) of received data over I2C in order to know check if the data is valid.
PEC Definition
I used the code stated in a previous question but it does not work for me.
The data looks like this: 0x00, 0x07, 0x01, 0x12, 0x3b, 0xd5
PEC is 0xd5 which is based on the polynomial = x^8+ x^2+ x^1+ x^0 - 0x107
This works also fine with this calculator.
So my question is, where is the difference between the code from the website and the one from the linked question:
local function crc8(t)
local c = 0
for _, b in ipairs(t) do
for i = 0, 7 do
c = c >> 1 ~ ((c ~ b >> i) & 1) * 0xE0
end
end
return c
end
This definition of CRC uses reversed bits in all data bytes.
local function reverse(x)
-- reverse bits of a byte
local y = 0
for j = 1, 8 do
y = y * 2 + (x&1)
x = x >> 1
end
return y
end
local function crc8(t)
local c = 0
for _, b in ipairs(t) do
b = reverse(b)
for i = 0, 7 do
c = c >> 1 ~ ((c ~ b >> i) & 1) * 0xE0
end
end
c = reverse(c)
return c
end
print(tohex(crc8{0x00, 0x07, 0x01, 0x12, 0x3b})) --> 0xd5
I am trying to adapt the pure Lua implementation of the SecureHashAlgorithm found here for SHA2 512 instead of SHA2 256. When I try to use the adaptation, it does not give the correct answer.
Here is the adaptation:
--
-- UTILITY FUNCTIONS
--
-- transform a string of bytes in a string of hexadecimal digits
local function str2hexa (s)
local h = string.gsub(s, ".", function(c)
return string.format("%02x", string.byte(c))
end)
return h
end
-- transforms number 'l' into a big-endian sequence of 'n' bytes
--(coded as a string)
local function num2string(l, n)
local s = ""
for i = 1, n do
--most significant byte of l
local remainder = l % 256
s = string.char(remainder) .. s
--remove from l the bits we have already transformed
l = (l-remainder) / 256
end
return s
end
-- transform the big-endian sequence of eight bytes starting at
-- index 'i' in 's' into a number
local function s264num (s, i)
local n = 0
for i = i, i + 7 do
n = n*256 + string.byte(s, i)
end
return n
end
--
-- MAIN SECTION
--
-- FIRST STEP: INITIALIZE HASH VALUES
--(second 32 bits of the fractional parts of the square roots of the first 9th through 16th primes 23..53)
local HH = {}
local function initH512(H)
H = {0x6a09e667f3bcc908, 0xbb67ae8584caa73b, 0x3c6ef372fe94f82b, 0xa54ff53a5f1d36f1, 0x510e527fade682d1, 0x9b05688c2b3e6c1f, 0x1f83d9abfb41bd6b, 0x5be0cd19137e2179}
return H
end
-- SECOND STEP: INITIALIZE ROUND CONSTANTS
--(first 80 bits of the fractional parts of the cube roots of the first 80 primes 2..409)
local k = {
0x428a2f98d728ae22, 0x7137449123ef65cd, 0xb5c0fbcfec4d3b2f, 0xe9b5dba58189dbbc, 0x3956c25bf348b538,
0x59f111f1b605d019, 0x923f82a4af194f9b, 0xab1c5ed5da6d8118, 0xd807aa98a3030242, 0x12835b0145706fbe,
0x243185be4ee4b28c, 0x550c7dc3d5ffb4e2, 0x72be5d74f27b896f, 0x80deb1fe3b1696b1, 0x9bdc06a725c71235,
0xc19bf174cf692694, 0xe49b69c19ef14ad2, 0xefbe4786384f25e3, 0x0fc19dc68b8cd5b5, 0x240ca1cc77ac9c65,
0x2de92c6f592b0275, 0x4a7484aa6ea6e483, 0x5cb0a9dcbd41fbd4, 0x76f988da831153b5, 0x983e5152ee66dfab,
0xa831c66d2db43210, 0xb00327c898fb213f, 0xbf597fc7beef0ee4, 0xc6e00bf33da88fc2, 0xd5a79147930aa725,
0x06ca6351e003826f, 0x142929670a0e6e70, 0x27b70a8546d22ffc, 0x2e1b21385c26c926, 0x4d2c6dfc5ac42aed,
0x53380d139d95b3df, 0x650a73548baf63de, 0x766a0abb3c77b2a8, 0x81c2c92e47edaee6, 0x92722c851482353b,
0xa2bfe8a14cf10364, 0xa81a664bbc423001, 0xc24b8b70d0f89791, 0xc76c51a30654be30, 0xd192e819d6ef5218,
0xd69906245565a910, 0xf40e35855771202a, 0x106aa07032bbd1b8, 0x19a4c116b8d2d0c8, 0x1e376c085141ab53,
0x2748774cdf8eeb99, 0x34b0bcb5e19b48a8, 0x391c0cb3c5c95a63, 0x4ed8aa4ae3418acb, 0x5b9cca4f7763e373,
0x682e6ff3d6b2b8a3, 0x748f82ee5defb2fc, 0x78a5636f43172f60, 0x84c87814a1f0ab72, 0x8cc702081a6439ec,
0x90befffa23631e28, 0xa4506cebde82bde9, 0xbef9a3f7b2c67915, 0xc67178f2e372532b, 0xca273eceea26619c,
0xd186b8c721c0c207, 0xeada7dd6cde0eb1e, 0xf57d4f7fee6ed178, 0x06f067aa72176fba, 0x0a637dc5a2c898a6,
0x113f9804bef90dae, 0x1b710b35131c471b, 0x28db77f523047d84, 0x32caab7b40c72493, 0x3c9ebe0a15c9bebc,
0x431d67c49c100d4c, 0x4cc5d4becb3e42b6, 0x597f299cfc657e2a, 0x5fcb6fab3ad6faec, 0x6c44198c4a475817
}
-- THIRD STEP: PRE-PROCESSING (padding)
local function preprocess(toProcess, len)
--append a single '1' bit
--append K '0' bits, where K is the minimum number >= 0 such that L + 1 + K = 896mod1024
local extra = 128 - (len + 9) % 128
len = num2string(8 * len, 8)
toProcess = toProcess .. "\128" .. string.rep("\0", extra) .. len
assert(#toProcess % 128 == 0)
return toProcess
end
local function rrotate(rot, n)
return (rot >> n) | ((rot << 64 - n))
end
local function digestblock(msg, i, H)
local w = {}
for j = 1, 16 do w[j] = s264num(msg, i + (j - 1)*4) end
for j = 17, 80 do
local v = w[j - 15]
local s0 = rrotate(v, 1) ~ rrotate(v, 8) ~ (v >> 7)
v = w[j - 2]
w[j] = w[j - 16] + s0 + w[j - 7] + ((rrotate(v, 19) ~ rrotate(v, 61)) ~ (v >> 6))
end
local a, b, c, d, e, f, g, h = H[1], H[2], H[3], H[4], H[5], H[6], H[7], H[8]
for i = 1, 80 do
a, b, c, d, e, f, g, h = a , b , c , d , e , f , g , h
local s0 = rrotate(a, 28) ~ (rrotate(a, 34) ~ rrotate(a, 39))
local maj = ((a & b) ~ (a & c)) ~ (b & c)
local t2 = s0 + maj
local s1 = rrotate(e, 14) ~ (rrotate(e, 18) ~ rrotate(e, 41))
local ch = (e & f) ~ (~e & g)
local t1 = h + s1 + ch + k[i] + w[i]
h, g, f, e, d, c, b, a = g, f, e, d + t1, c, b, a, t1 + t2
end
H[1] = (H[1] + a)
H[2] = (H[2] + b)
H[3] = (H[3] + c)
H[4] = (H[4] + d)
H[5] = (H[5] + e)
H[6] = (H[6] + f)
H[7] = (H[7] + g)
H[8] = (H[8] + h)
end
local function finalresult512 (H)
-- Produce the final hash value:
return
str2hexa(num2string(H[1], 8)..num2string(H[2], 8)..num2string(H[3], 8)..num2string(H[4], 8)..
num2string(H[5], 8)..num2string(H[6], 8)..num2string(H[7], 8)..num2string(H[8], 8))
end
-- Returns the hash512 for the given string.
local function hash512 (msg)
msg = preprocess(msg, #msg)
local H = initH512(HH)
-- Process the message in successive 1024-bit (128 bytes) chunks:
for i = 1, #msg, 128 do
digestblock(msg, i, H)
end
return finalresult512(H)
end
Given hash512("a"):
Expect: 1f40fc92da241694750979ee6cf582f2d5d7d28e18335de05abc54d0560e0f5302860c652bf08d560252aa5e74210546f369fbbbce8c12cfc7957b2652fe9a75
Actual: e0b9623f2194cb81f2a62616a183edbe390be0d0b20430cadc3371efc237fa6bf7f8b48311f2fa249131c347fee3e8cde6acfdab286d648054541f92102cfc9c
I know that I am creating a message of the correct bit size (1024 bits) and also working in 1024-bit chunks, or at least I believe I am.
I am not sure if it has to do with the handling of the integers (the standard requires unsigned integers) or whether I made a mistake in one of the utility functions, or both. If it is indeed an issue with the handling of the integers, how would I go about taking care of the problem. I was able to resolve this when working on the 256-bit version of the adaptation by using mod 2^32 when working with numbers in the digestblock method. I attempted to do mod 2^64 and 2^63 with the 512-bit version and it does not correct the problem. I am stumped.
I should mention that I cannot use one of the many library implementations as I am using a sandboxed Lua that does not provide this access, which is why I need a pure lua implementation. Thanks in advance.
Unfortunately, after introducing integers in Lua 5.3 writing scripts for Lua becomes a more complicated task.
You must always think about transformations between integers and floating point numbers.
ALWAYS. Yes, that's boring.
One of your mistakes is an excellent example of this "dark corner of Lua".
local remainder = l % 256
s = string.char(remainder) .. s
--remove from l the bits we have already transformed
l = (l-remainder) / 256
Your value l is initially a 64-bit integer.
After cutting off its first byte l contains (64-8) = 56 bits, but now it's a floating point-number (with 53-bit precision, of course).
Possible solution: use l = l >> 8 or l = l // 256 instead of l = (l-remainder) / 256
Another mistake is using s264num(msg, i + (j - 1) * 4) instead of s264num(msg, i + (j - 1) * 8)
One more mistake is in the following line:
local extra = 128 - (len + 9) % 128
The correct code is
local extra = - (len + 17) % 128 + 8
(Please note that -a%m+b is not the same as b-a%m due to operator precedence)
After fixing these 3 mistakes your code works correctly.
I need set some bits in ByteData at position counted in bits.
How I can do this?
Eg.
var byteData = new ByteData(1024);
var bitData = new BitData(byteData);
// Offset in bits: 387
// Number of bits: 5
// Value: 3
bitData.setBits(387, 5, 3);
Yes it is quite complicated. I dont know dart, but these are the general steps you need to take. I will label each variable as a letter and also use a more complicated example to show you what happens when the bits overflow.
1. Construct the BitData object with a ByteData object (A)
2. Call setBits(offset (B), bits (C), value (D));
I will use example values of:
A: 11111111 11111111 11111111 11111111
B: 7
C: 10
D: 00000000 11111111
3. Rather than using an integer with a fixed length of bits, you could
use another ByteData object (D) containing your bits you want to write.
Also create a mask (E) containing the significant bits.
e.g.
A: 11111111 11111111 11111111 11111111
D: 00000000 11111111
E: 00000011 11111111 (2^C - 1)
4. As an extra bonus step, we can make sure the insignificant
bits are really zero by ANDing with the bitmask.
D = D & E
D 00000000 11111111
E 00000011 11111111
5. Make sure D and E contain at least one full zero byte since we want
to shift them.
D 00000000 00000000 11111111
E 00000000 00000011 11111111
6. Work out these two integer values:
F = The extra bit offset for the start byte: B mod 8 (e.g. 7)
G = The insignificant bits: size(D) - C (e.g. 14)
7. H = G-F which should not be negative here. (e.g. 14-7 = 7)
8. Shift both D and E left by H bits.
D 00000000 01111111 10000000
E 00000001 11111111 10000000
9. Work out first byte number (J) floor(B / 8) e.g. 0
10. Read the value of A at this index out and let this be K
K = 11111111 11111111 11111111
11. AND the current (K) with NOT E to set zeros for the new bits.
Then you can OR the new bits over the top.
L = (K & !E) | D
K & !E = 11111110 00000000 01111111
L = 11111110 01111111 11111111
12. Write L to the same place you read it from.
There is no BitData class, so you'll have to do some of the bit-pushing yourself.
Find the corresponding byte offset, read in some bytes, mask out the existing bits and set the new ones at the correct bit offset, then write it back.
The real complexity comes when you need to store more bits than you can read/write in a single operation.
For endianness, if you are treating the memory as a sequence of bits with arbitrary width, I'd go for little-endian. Endianness only really makes sense for full-sized (2^n-bit, n > 3) integers. A 5 bit integer as the one you are storing can't have any endianness, and a 37 bit integer also won't have any natural way of expressing an endianness.
You can try something like this code (which can definitely be optimized more):
import "dart:typed_data";
void setBitData(ByteBuffer buffer, int offset, int length, int value) {
assert(value < (1 << length));
assert(offset + length < buffer.lengthInBytes * 8);
int byteOffset = offset >> 3;
int bitOffset = offset & 7;
if (length + bitOffset <= 32) {
ByteData data = new ByteData.view(buffer);
// Can update it one read/modify/write operation.
int mask = ((1 << length) - 1) << bitOffset;
int bits = data.getUint32(byteOffset, Endianness.LITTLE_ENDIAN);
bits = (bits & ~mask) | (value << bitOffset);
data.setUint32(byteOffset, bits, Endianness.LITTLE_ENDIAN);
return;
}
// Split the value into chunks of no more than 32 bits, aligned.
do {
int bits = (length > 32 ? 32 : length) - bitOffset;
setBitData(buffer, offset, bits, value & ((1 << bits) - 1));
offset += bits;
length -= bits;
value >>= bits;
bitOffset = 0;
} while (length > 0);
}
Example use:
main() {
var b = new Uint8List(32);
setBitData(b.buffer, 3, 8, 255);
print(b.map((v)=>v.toRadixString(16)));
setBitData(b.buffer, 13, 6*4, 0xffffff);
print(b.map((v)=>v.toRadixString(16)));
setBitData(b.buffer, 47, 21*4, 0xaaaaaaaaaaaaaaaaaaaaa);
print(b.map((v)=>v.toRadixString(16)));
}
This has been bothering be for a while now, I cannot seem to find a pure Lua implementation of a popular hashing method like SHA256, SHA512, or Whirlpool. I need it as I will be hashing the password client side before sending it of to a server. Speed isn't a worry here, I don't care if it takes 10 or so seconds to hash 10,000 times, I will be running it on a thread.
I have tried a couple before, which seemed like they worked perfectly fine at first, but when I tried a different input strings (usually longer ones), the hash comes out as a totally incorrect hash output.
I am using the LuaJIT version of Love2D, so it already has the BitOp library implemented. If any of you know any good implementations of these hashing methods or any similar secure ones then please let me know!
Thank you!
UPDATE: Here are some results!
Firstly this is the test code I am using.
https://github.com/JustAPerson/LuaCrypt
INPUT: Test string
OUTPUT: a3e49d843df13c2e2a7786f6ecd7e0d184f45d718d1ac1a8a63e570466e489dd
EXPECTED: a3e49d843df13c2e2a7786f6ecd7e0d184f45d718d1ac1a8a63e570466e489dd
INPUT: This is a test string to hash
OUTPUT: 05b4ac920d4130cb9d9bb046cac7476f35d7404cf116dc8d6d4a113c3c79d904
EXPECTED: f70b476ff948472f8e4e52793a5a2779e636c20dd5336d3a8a4455374318db35
https://bitbucket.org/Boolsheet/sil/raw/tip/hash.lua
INPUT: Test string
OUTPUT: 8f1a5b37fbe986953c343d5b839b14843c6c29d47a6a7e52f263cd82ad6141a3
EXPECTED: a3e49d843df13c2e2a7786f6ecd7e0d184f45d718d1ac1a8a63e570466e489dd
INPUT: This is a test string to hash
OUTPUT: 167bf7b9000442419b3016a6e1edfcc7c8d40b5f0b80518a31ddb0bbd388e87ac
EXPECTED: f70b476ff948472f8e4e52793a5a2779e636c20dd5336d3a8a4455374318db35
I would recommend against using SHA256 for passwords. They are easy to bruteforce nowadays, and the way you are using them is vulnerable to replay attacks.
Also if you must use SHA256, use the version from OpenSSL if possible (especially if your program already depends on OpenSSL.)
But if you must use it (and cannot link with OpenSSL, but can use FFI) here is a LuaJIT version of SHA256 (only) that I am using in one of my projects.
local bit = require 'bit'
local ffi = require 'ffi'
local type = type
local band, bnot, bswap, bxor, rol, ror, rshift, tobit =
bit.band, bit.bnot, bit.bswap, bit.bxor, bit.rol, bit.ror, bit.rshift, bit.tobit
local min, max = math.min, math.max
local C = ffi.C
local istype, new, fill, copy, cast, sizeof, ffi_string =
ffi.istype, ffi.new, ffi.fill, ffi.copy, ffi.cast, ffi.sizeof, ffi.string
local sha256 = {}
ffi.cdef [[
void *malloc(size_t size);
void free(void *ptr);
]]
local ctHashState = ffi.typeof 'uint32_t[8]'
local cbHashState = ffi.sizeof(ctHashState)
local ctBlock = ffi.typeof 'uint32_t[64]'
local cbBlock = ffi.sizeof(ctBlock)
local ctpu8 = ffi.typeof 'uint8_t *'
local ctpcu8 = ffi.typeof 'const uint8_t *'
local ctpu32 = ffi.typeof 'uint32_t *'
local ctpu64 = ffi.typeof 'uint64_t *'
-- This struct is used by the 'preprocess' iterator function. It keeps track
-- of the end of the input string + the total input length in bits + a pointer
-- to the block buffer (where expansion takes place.)
local ctBlockIter
local cmtBlockIter = {}
function cmtBlockIter.__sub(a, b)
if istype(ctBlockIter, a) then a = a.limit end
if istype(ctBlockIter, b) then b = b.limit end
return a - b
end
function cmtBlockIter:__tostring()
return string.format("<ctBlockIter: limit=%s; keyLength=%s>",
tostring(self.base), tostring(self.keyLength))
end
ctBlockIter = ffi.metatype([[
struct {
const uint8_t *limit;
uint32_t *blockBuffer;
uint64_t keyLength;
}
]], cmtBlockIter)
-- Initial state of the hash
local init_h = new('const uint32_t[8]', {
0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a,
0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19
})
-- Constants used in the add step of the compression function
local k = new('const uint32_t[64]', {
0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5,
0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3,
0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc,
0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7,
0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13,
0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3,
0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5,
0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208,
0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2
})
-- Expand block from 512 to 2048 bits
local function expand(w)
for i = 16, 63 do
local s0 = bxor(ror(w[i-15], 7), ror(w[i-15], 18), rshift(w[i-15], 3))
local s1 = bxor(ror(w[i-2], 17), ror(w[i-2], 19), rshift(w[i-2], 10))
w[i] = w[i-16] + s0 + w[i-7] + s1
end
end
-- Process one expanded block and update the hash state
local function compress(hh, w)
local a, b, c, d, e, f, g, h =
hh[0],hh[1],hh[2],hh[3],hh[4],hh[5],hh[6],hh[7]
for i = 0, 63 do
local S1 = bxor(ror(e, 6), ror(e, 11), ror(e, 25))
local ch = bxor(band(e, f), band(bnot(e), g))
local t = tobit(h + S1 + ch + k[i] + w[i])
local S0 = bxor(ror(a, 2), ror(a, 13), ror(a, 22))
local maj = bxor(band(a, bxor(b, c)), band(b, c))
a, b, c, d, e, f, g, h =
tobit(t + S0 + maj),
a, b, c,
tobit(d + t),
e, f, g
end
hh[0],hh[1],hh[2],hh[3],hh[4],hh[5],hh[6],hh[7] =
hh[0]+a, hh[1]+b, hh[2]+c, hh[3]+d,
hh[4]+e, hh[5]+f, hh[6]+g, hh[7]+h
end
-- Take a 512-bit chunk from the input.
-- If it is the final chunk, also add padding
local keyLengthOfs = ffi.offsetof(ctBlockIter, 'keyLength')
local function nextBlock(state, input)
local w = state.blockBuffer
local cLen = min(state - input, 64)
if cLen < -8 then return nil end
fill(w, 256, 0)
copy(w, input, max(0, cLen))
if 0 <= cLen and cLen < 64 then
copy(cast(ctpu8, w)+cLen, '\128', 1)
end
for i = 0, 15 do w[i] = bswap(w[i]) end
if cLen <= (64-8-1) then
copy(cast(ctpu64, w) + 7, cast(ctpu8, state) + keyLengthOfs, 8)
w[14], w[15] = w[15], w[14]
end
input = input + 64
return input
end
-- Iterator that yields one block (possibly padded) at a time from the input
local function preprocess(input, len, w)
len = len or (type(input) == 'string' and #input or sizeof(input))
input = cast(ctpu8, input)
local it = new(ctBlockIter)
it.blockBuffer = w
it.limit = input+len
it.keyLength = len*8
return nextBlock, it, input
end
-- Compute a binary hash (32-byte binary string) from the input
function sha256.binFromBin(input, len)
local h = new(ctHashState)
local w = cast(ctpu32, C.malloc(cbBlock))
copy(h, init_h, cbHashState)
for _ in preprocess(input, len, w) do
expand(w)
compress(h, w)
end
for i = 0, 7 do h[i] = bswap(h[i]) end
C.free(w)
return ffi_string(h, 32)
end
local hexDigits = new('char[16]', "0123456789abcdef")
local hexOut = new('char[65]')
-- Compute the hash and convert to hexadecimal
function sha256.hexFromBin(input, len)
local h = new(ctHashState)
local w = cast(ctpu32, C.malloc(cbBlock))
copy(h, init_h, cbHashState)
for _ in preprocess(input, len, w) do
expand(w)
compress(h, w)
end
for i = 0, 7 do
local w = h[i]
for j = 0, 3 do
w = rol(w, 8)
hexOut[i*8 + j*2] = hexDigits[band(rshift(w, 4), 15)]
hexOut[i*8 + j*2 + 1] = hexDigits[band(w, 15)]
end
end
C.free(w)
return ffi_string(hexOut, 64)
end
return sha256
There is an implementation of SHA256 at the Lua User's Wiki. The page observes it is Lua 5.2. I would imagine that it would be practical to make that work in LuaJIT without too much trouble.
Do pay attention to the larger security issues surrounding passwords and authentication. The usual advice applies; rolling your own security rather than using an existing tested and supported implementation is not something to be done lightly.
Since you are using LuaJIT, you should be able to leverage its very powerful FFI capabilities to use crypto supplied on your native platform. That will likely require writing some FFI-flavored Lua that is platform specific to each platform on which your client expects to run, but from what I've seen by lurking in the LuaJIT mailing list that shouldn't be too painful.
I have a specific requirement to convert a stream of bytes into a character encoding that happens to be 6-bits per character.
Here's an example:
Input: 0x50 0x11 0xa0
Character Table:
010100 T
000001 A
000110 F
100000 SPACE
Output: "TAF "
Logically I can understand how this works:
Taking 0x50 0x11 0xa0 and showing as binary:
01010000 00010001 10100000
Which is "TAF ".
What's the best way to do this programmatically (pseudo code or c++). Thank you!
Well, every 3 bytes, you end up with four characters. So for one thing, you need to work out what to do if the input isn't a multiple of three bytes. (Does it have padding of some kind, like base64?)
Then I'd probably take each 3 bytes in turn. In C#, which is close enough to pseudo-code for C :)
for (int i = 0; i < array.Length; i += 3)
{
// Top 6 bits of byte i
int value1 = array[i] >> 2;
// Bottom 2 bits of byte i, top 4 bits of byte i+1
int value2 = ((array[i] & 0x3) << 4) | (array[i + 1] >> 4);
// Bottom 4 bits of byte i+1, top 2 bits of byte i+2
int value3 = ((array[i + 1] & 0xf) << 2) | (array[i + 2] >> 6);
// Bottom 6 bits of byte i+2
int value4 = array[i + 2] & 0x3f;
// Now use value1...value4, e.g. putting them into a char array.
// You'll need to decode from the 6-bit number (0-63) to the character.
}
Just in case if someone is interested - another variant that extracts 6-bit numbers from the stream as soon as they appear there. That is, results can be obtained even if less then 3 bytes are currently read. Would be useful for unpadded streams.
The code saves the state of the accumulator a in variable n which stores the number of bits left in accumulator from the previous read.
int n = 0;
unsigned char a = 0;
unsigned char b = 0;
while (read_byte(&byte)) {
// save (6-n) most significant bits of input byte to proper position
// in accumulator
a |= (b >> (n + 2)) & (077 >> n);
store_6bit(a);
a = 0;
// save remaining least significant bits of input byte to proper
// position in accumulator
a |= (b << (4 - n)) & ((077 << (4 - n)) & 077);
if (n == 4) {
store_6bit(a);
a = 0;
}
n = (n + 2) % 6;
}