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8 Mar 2013

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great simple stuff, well presented  with a few nuggets like flags attribute that I didn't know about!





Thank you!
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





once we have boolean algebra tool then this enables us to;
1. make assembly code simulators for various microcontrollers
2. once code has been debugged then download to device target to run on actual hardware platform.
let's get to work folks
cheers!





Please help me with conversion to C# this vb6 code.
Option Explicit
' Visual Basic MD5 Implementation
' Robert Hubley and David Midkiff (mdj2023@hotmail.com)
' Standard MD5 implementation optimised for the Visual Basic environment.
' Conforms to all standards and can be used in digital signature or password
' protection related schemes.
Private Const OFFSET_4 = 4294967296#
Private Const MAXINT_4 = 2147483647
Private State(4) As Long
Private ByteCounter As Long
Private ByteBuffer(63) As Byte
Private Const S11 = 7
Private Const S12 = 12
Private Const S13 = 17
Private Const S14 = 22
Private Const S21 = 5
Private Const S22 = 9
Private Const S23 = 14
Private Const S24 = 20
Private Const S31 = 4
Private Const S32 = 11
Private Const S33 = 16
Private Const S34 = 23
Private Const S41 = 6
Private Const S42 = 10
Private Const S43 = 15
Private Const S44 = 21
Property Get RegisterA() As String
RegisterA = State(1)
End Property
Property Get RegisterB() As String
RegisterB = State(2)
End Property
Property Get RegisterC() As String
RegisterC = State(3)
End Property
Property Get RegisterD() As String
RegisterD = State(4)
End Property
Public Function DigestStrToHexStr(SourceString As String) As String
MD5Init
MD5Update Len(SourceString), StringToArray(SourceString)
MD5Final
DigestStrToHexStr = GetValues
End Function
Public Function DigestFileToHexStr(InFile As String) As String
On Error GoTo errorHandler
GoSub begin
errorHandler:
DigestFileToHexStr = ""
Exit Function
begin:
Dim FileO As Integer
FileO = FreeFile
Call FileLen(InFile)
Open InFile For Binary Access Read As #FileO
MD5Init
Do While Not eof(FileO)
Get #FileO, , ByteBuffer
If Loc(FileO) < LOF(FileO) Then
ByteCounter = ByteCounter + 64
MD5Transform ByteBuffer
End If
Loop
ByteCounter = ByteCounter + (LOF(FileO) Mod 64)
Close #FileO
MD5Final
DigestFileToHexStr = GetValues
End Function
Private Function StringToArray(InString As String) As Byte()
Dim i As Integer, bytBuffer() As Byte
ReDim bytBuffer(Len(InString))
For i = 0 To Len(InString)  1
bytBuffer(i) = Asc(Mid$(InString, i + 1, 1))
Next i
StringToArray = bytBuffer
End Function
Public Function GetValues() As String
GetValues = LongToString(State(1)) & LongToString(State(2)) & LongToString(State(3)) & LongToString(State(4))
End Function
Private Function LongToString(Num As Long) As String
Dim a As Byte, b As Byte, c As Byte, D As Byte
a = Num And &HFF&
If a < 16 Then LongToString = "0" & Hex(a) Else LongToString = Hex(a)
b = (Num And &HFF00&) \ 256
If b < 16 Then LongToString = LongToString & "0" & Hex(b) Else LongToString = LongToString & Hex(b)
c = (Num And &HFF0000) \ 65536
If c < 16 Then LongToString = LongToString & "0" & Hex(c) Else LongToString = LongToString & Hex(c)
If Num < 0 Then D = ((Num And &H7F000000) \ 16777216) Or &H80& Else D = (Num And &HFF000000) \ 16777216
If D < 16 Then LongToString = LongToString & "0" & Hex(D) Else LongToString = LongToString & Hex(D)
End Function
Public Sub MD5Init()
ByteCounter = 0
State(1) = UnsignedToLong(1732584193#)
State(2) = UnsignedToLong(4023233417#)
State(3) = UnsignedToLong(2562383102#)
State(4) = UnsignedToLong(271733878#)
End Sub
Public Sub MD5Final()
Dim dblBits As Double, padding(72) As Byte, lngBytesBuffered As Long
padding(0) = &H80
dblBits = ByteCounter * 8
lngBytesBuffered = ByteCounter Mod 64
If lngBytesBuffered <= 56 Then MD5Update 56  lngBytesBuffered, padding Else MD5Update 120  ByteCounter, padding
padding(0) = UnsignedToLong(dblBits) And &HFF&
padding(1) = UnsignedToLong(dblBits) \ 256 And &HFF&
padding(2) = UnsignedToLong(dblBits) \ 65536 And &HFF&
padding(3) = UnsignedToLong(dblBits) \ 16777216 And &HFF&
padding(4) = 0
padding(5) = 0
padding(6) = 0
padding(7) = 0
MD5Update 8, padding
End Sub
Public Sub MD5Update(InputLen As Long, InputBuffer() As Byte)
Dim II As Integer, i As Integer, j As Integer, k As Integer, lngBufferedBytes As Long, lngBufferRemaining As Long, lngRem As Long
lngBufferedBytes = ByteCounter Mod 64
lngBufferRemaining = 64  lngBufferedBytes
ByteCounter = ByteCounter + InputLen
If InputLen >= lngBufferRemaining Then
For II = 0 To lngBufferRemaining  1
ByteBuffer(lngBufferedBytes + II) = InputBuffer(II)
Next II
MD5Transform ByteBuffer
lngRem = (InputLen) Mod 64
For i = lngBufferRemaining To InputLen  II  lngRem Step 64
For j = 0 To 63
ByteBuffer(j) = InputBuffer(i + j)
Next j
MD5Transform ByteBuffer
Next i
lngBufferedBytes = 0
Else
i = 0
End If
For k = 0 To InputLen  i  1
ByteBuffer(lngBufferedBytes + k) = InputBuffer(i + k)
Next k
End Sub
Private Sub MD5Transform(Buffer() As Byte)
Dim x(16) As Long, a As Long, b As Long, c As Long, D As Long
a = State(1)
b = State(2)
c = State(3)
D = State(4)
Decode 64, x, Buffer
FF a, b, c, D, x(0), S11, 680876936
FF D, a, b, c, x(1), S12, 389564586
FF c, D, a, b, x(2), S13, 606105819
FF b, c, D, a, x(3), S14, 1044525330
FF a, b, c, D, x(4), S11, 176418897
FF D, a, b, c, x(5), S12, 1200080426
FF c, D, a, b, x(6), S13, 1473231341
FF b, c, D, a, x(7), S14, 45705983
FF a, b, c, D, x(8), S11, 1770035416
FF D, a, b, c, x(9), S12, 1958414417
FF c, D, a, b, x(10), S13, 42063
FF b, c, D, a, x(11), S14, 1990404162
FF a, b, c, D, x(12), S11, 1804603682
FF D, a, b, c, x(13), S12, 40341101
FF c, D, a, b, x(14), S13, 1502002290
FF b, c, D, a, x(15), S14, 1236535329
GG a, b, c, D, x(1), S21, 165796510
GG D, a, b, c, x(6), S22, 1069501632
GG c, D, a, b, x(11), S23, 643717713
GG b, c, D, a, x(0), S24, 373897302
GG a, b, c, D, x(5), S21, 701558691
GG D, a, b, c, x(10), S22, 38016083
GG c, D, a, b, x(15), S23, 660478335
GG b, c, D, a, x(4), S24, 405537848
GG a, b, c, D, x(9), S21, 568446438
GG D, a, b, c, x(14), S22, 1019803690
GG c, D, a, b, x(3), S23, 187363961
GG b, c, D, a, x(8), S24, 1163531501
GG a, b, c, D, x(13), S21, 1444681467
GG D, a, b, c, x(2), S22, 51403784
GG c, D, a, b, x(7), S23, 1735328473
GG b, c, D, a, x(12), S24, 1926607734
HH a, b, c, D, x(5), S31, 378558
HH D, a, b, c, x(8), S32, 2022574463
HH c, D, a, b, x(11), S33, 1839030562
HH b, c, D, a, x(14), S34, 35309556
HH a, b, c, D, x(1), S31, 1530992060
HH D, a, b, c, x(4), S32, 1272893353
HH c, D, a, b, x(7), S33, 155497632
HH b, c, D, a, x(10), S34, 1094730640
HH a, b, c, D, x(13), S31, 681279174
HH D, a, b, c, x(0), S32, 358537222
HH c, D, a, b, x(3), S33, 722521979
HH b, c, D, a, x(6), S34, 76029189
HH a, b, c, D, x(9), S31, 640364487
HH D, a, b, c, x(12), S32, 421815835
HH c, D, a, b, x(15), S33, 530742520
HH b, c, D, a, x(2), S34, 995338651
II a, b, c, D, x(0), S41, 198630844
II D, a, b, c, x(7), S42, 1126891415
II c, D, a, b, x(14), S43, 1416354905
II b, c, D, a, x(5), S44, 57434055
II a, b, c, D, x(12), S41, 1700485571
II D, a, b, c, x(3), S42, 1894986606
II c, D, a, b, x(10), S43, 1051523
II b, c, D, a, x(1), S44, 2054922799
II a, b, c, D, x(8), S41, 1873313359
II D, a, b, c, x(15), S42, 30611744
II c, D, a, b, x(6), S43, 1560198380
II b, c, D, a, x(13), S44, 1309151649
II a, b, c, D, x(4), S41, 145523070
II D, a, b, c, x(11), S42, 1120210379
II c, D, a, b, x(2), S43, 718787259
II b, c, D, a, x(9), S44, 343485551
State(1) = LongOverflowAdd(State(1), a)
State(2) = LongOverflowAdd(State(2), b)
State(3) = LongOverflowAdd(State(3), c)
State(4) = LongOverflowAdd(State(4), D)
End Sub
Private Sub Decode(Length As Integer, OutputBuffer() As Long, InputBuffer() As Byte)
Dim intDblIndex As Integer, intByteIndex As Integer, dblSum As Double
For intByteIndex = 0 To Length  1 Step 4
dblSum = InputBuffer(intByteIndex) + InputBuffer(intByteIndex + 1) * 256# + InputBuffer(intByteIndex + 2) * 65536# + InputBuffer(intByteIndex + 3) * 16777216#
OutputBuffer(intDblIndex) = UnsignedToLong(dblSum)
intDblIndex = intDblIndex + 1
Next intByteIndex
End Sub
Private Function FF(a As Long, b As Long, c As Long, D As Long, x As Long, S As Long, ac As Long) As Long
a = LongOverflowAdd4(a, (b And c) Or (Not (b) And D), x, ac)
a = LongLeftRotate(a, S)
a = LongOverflowAdd(a, b)
End Function
Private Function GG(a As Long, b As Long, c As Long, D As Long, x As Long, S As Long, ac As Long) As Long
a = LongOverflowAdd4(a, (b And D) Or (c And Not (D)), x, ac)
a = LongLeftRotate(a, S)
a = LongOverflowAdd(a, b)
End Function
Private Function HH(a As Long, b As Long, c As Long, D As Long, x As Long, S As Long, ac As Long) As Long
a = LongOverflowAdd4(a, b Xor c Xor D, x, ac)
a = LongLeftRotate(a, S)
a = LongOverflowAdd(a, b)
End Function
Private Function II(a As Long, b As Long, c As Long, D As Long, x As Long, S As Long, ac As Long) As Long
a = LongOverflowAdd4(a, c Xor (b Or Not (D)), x, ac)
a = LongLeftRotate(a, S)
a = LongOverflowAdd(a, b)
End Function
Function LongLeftRotate(value As Long, Bits As Long) As Long
Dim lngSign As Long, lngI As Long
Bits = Bits Mod 32
If Bits = 0 Then LongLeftRotate = value: Exit Function
For lngI = 1 To Bits
lngSign = value And &HC0000000
value = (value And &H3FFFFFFF) * 2
value = value Or ((lngSign < 0) And 1) Or (CBool(lngSign And &H40000000) And &H80000000)
Next
LongLeftRotate = value
End Function
Private Function LongOverflowAdd(Val1 As Long, Val2 As Long) As Long
Dim lngHighWord As Long, lngLowWord As Long, lngOverflow As Long
lngLowWord = (Val1 And &HFFFF&) + (Val2 And &HFFFF&)
lngOverflow = lngLowWord \ 65536
lngHighWord = (((Val1 And &HFFFF0000) \ 65536) + ((Val2 And &HFFFF0000) \ 65536) + lngOverflow) And &HFFFF&
LongOverflowAdd = UnsignedToLong((lngHighWord * 65536#) + (lngLowWord And &HFFFF&))
End Function
Private Function LongOverflowAdd4(Val1 As Long, Val2 As Long, val3 As Long, val4 As Long) As Long
Dim lngHighWord As Long, lngLowWord As Long, lngOverflow As Long
lngLowWord = (Val1 And &HFFFF&) + (Val2 And &HFFFF&) + (val3 And &HFFFF&) + (val4 And &HFFFF&)
lngOverflow = lngLowWord \ 65536
lngHighWord = (((Val1 And &HFFFF0000) \ 65536) + ((Val2 And &HFFFF0000) \ 65536) + ((val3 And &HFFFF0000) \ 65536) + ((val4 And &HFFFF0000) \ 65536) + lngOverflow) And &HFFFF&
LongOverflowAdd4 = UnsignedToLong((lngHighWord * 65536#) + (lngLowWord And &HFFFF&))
End Function
Private Function UnsignedToLong(value As Double) As Long
If value < 0 Or value >= OFFSET_4 Then Error 6
If value <= MAXINT_4 Then UnsignedToLong = value Else UnsignedToLong = value  OFFSET_4
End Function
Private Function LongToUnsigned(value As Long) As Double
If value < 0 Then LongToUnsigned = value + OFFSET_4 Else LongToUnsigned = value
End Function





No. Nobody is going to sit down and convert what looks like a whole VB6 project you found on the internet and don't understand to C#. It's not a good use of our time, especially as they don;t work in the same way, and a direct "translation" would not produce a good C# result.
You need it, you learn VB6 and / or C# and convert it.
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...






i'm reading the very first section. at one point there's this:
"For example 110100112 is 21110. And 14310 is 100011112." shortly after that is, "Stepbystep explanation on converting 78310 to a binary number:..."
can you explain what those subscripted numbers are? (i think it should be explained in the article, the first time you use them. i can already convert dec to bin and some other basics but i'm not a math(s) person and have no idea what those small numbers are doing there.) thanks.





Those subscript numbers indicate the base of the number. I thought that it was a common way to express a base, or am I wrong in that?
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





Expanding my previous reply: I updated the article. I explained the subscript numbers the first time I'm using them.
The quick brown ProgramFOX jumps right over the Lazy<Dog> .






Wow Just Wow ...!!!!!!!!!!!





Thanks
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





Because binary representation and calculations are still (and will always be, until quantum computers will actually exist) extremely important in Computer Science yet they are topics often overlooked both by CS students and selftaught developers.
I always admire and respect the fundamental articles, especially those which include math  the real basis of Computer Science (which in fact has been a branch of Math courses for quite some time).
GCS d s/++ a C++++ U+++ P L E W++ N++ o+ K w+++ O? M V? PS+ PE Y+ PGP t++ 5? X R++ tv b+ DI+++ D++ G e++>+++ h ++>+++ y+++* Weapons extension: ma k++ F+2 X
If you think 'goto' is evil, try writing an Assembly program without JMP.  TNCaver





Thank you!
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





The code must be return only one character. Code above are allright, but it can generate 3 characters to return only one. I want to remove unnecessary processing.
I understand that processing is insignificant, but wish I could do to understand more about this type of emun flag (for example, instead of generating only one character, each item can have a large processing code block, it is necessary to optimize the algorithm).
[Flags]
public enum CharType
{
Number = 0x0,
Upper = 0x1,
Lower = 0x2
}
public CharGenerator(CharType charType)
{
this. _random = new Random();
}
public char genarateRandomChar(CharType charType)
{
string temp = string.Empty;
if (charType.HasFlag(CharType.Number))
{
temp += this.gerarCharNumber();
}
if (charType.HasFlag(CharType.Upper))
{
temp += this.gerarCharUpper();
}
if (charType.HasFlag(CharType.Lower))
{
temp += this.gerarCharLower();
}
return temp[_random.Next(0, temp.Length)];
}
Yitzhak Stone
andrade@hotmail.com.br





I'm not entirely sure what you want to do, but if I look at your code, I'm not seeing a way to optimize it at a first glance.
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





Hey ProgramFOX,
Thanks for the great article! This was very informative and the extent of content covered was perfect.
I saw in the part about signed numbers, it looks like the binary form of the negative number should have a 1 in the 1st digit (the first digit has a 0 in explanation step 5, but it has a 1 in a previous step).





Thanks for your comment! I have tried to look for the exact position of the possible typo, but I couldn't find the part you talked about. Do you want to copy a small snippet of the article containing the possible typo, so I can review it? Thanks!
(Also, sorry for my late reply  I was on vacation and didn't have internet access there)
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





Hi,
altough knowing bit operations, I rarely used them. What was absolutly new for me: the left and right shift Operator preserving the sign!
Thanks a lot.
What I tried next is BigInteger: I found the bit shift Operators works there the same way





Thank you!
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





An excellent article, thank you.
I was completely unaware of the Flags attribute in C#. It's very much the way I used to code things in C. Stacking flags as powers of 2 into a byte and "anding" or "oring" as appropriate always made for much more readable code than using a bunch of separate flags. Really glad to know we can do that with a C# enum. You've kind of made my day.





Thank you!
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





The animations you included in the sections about shifting are outstanding, and add significantly to the educational value of the article.
Although I am fairly comfortable operating on bits, I read your article with interest, and in case it offered some useful tidbit. My effort was rewarded with your XOR swap algorithm and some useful, though not all that surprising, information (from the discussion) about the shift quirks exhibited by signed integers. For what it may be worth, it would never occur to me to use a signed type for a variable upon which I intend to perform any kind of bit operations. The Common Language Runtime has a complete set of unsigned types that are ideal for such uses.





Thank you!
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





Nice, this is one of the best bitwise operator tutorials.
max = a ^ ((a ^ b) & (a < b));
// maximum kudos





Thanks!
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





Easy && usefull !
Thank you!





Thanks!
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





Very good, just one small "oops":
If we have a decimal number, 783 for example, then we convert it to a decimal binary number using this way:
Brent





Well spotted, thanks! Fixed.
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





very good article! thx
bug there is mistake when taking about "Circular Right Shift" —— (and for an Int32, the formula is a >> n  a << (8  n).)
not 8 , it's 32.





Good spot, thanks! Fixed
The quick brown ProgramFOX jumps right over the Lazy<Dog> .






.NET uses arithmetic bit shifts. This means that for a right shift on a signed number the leftmost bit is preserved while shifting, it does NOT shift in zeroes if the number is a negative number.
It actually behaves a bit different in VB.NET and C# depending on the type of the number, both have excellent explanations on MSDN.
For example in VB.NET:
Using SByte to lower the amount of binary digits, but works on other signed integers too.
SByte.MaxValue is represented by: 0111 1111. Doing a left shift with 1 bit on that results in 1111 1110 which is 2 in decimal. Right shifting that with 1 bit results in 1111 1111 which is 1 in decimal.
MsgBox(SByte.MaxValue << 1 >> 1)
If you put this code in the event handler of a button and click the button, you'll get 1 in a pop up.





Thanks for your comment! I believe I have added a notice for this:
Quote: Important note: if you've a signed data type, then the sign will be preserved. However, I added that one for left shift too, and that seems to be incorrect, so I removed it there and kept it only for the right shift.
If the notice doesn't actually complete the explanation, please let me know.
The quick brown ProgramFOX jumps right over the Lazy<Dog> .





You're correct it's different for left and right shifting.
A left shift does exactly what you expect and turns Integer.MaxValue(all 1's except leftmost bit) into 2 (all 1's except the rightmost bit). If you then rightshift it, it becomes 1 (all 1's, no exception). This confused the hell out of me in the beginning until I read the MSDN texts.





Try this in the VB.Net IDE:
? 4 AND 1 = 0
False
? (4 AND 1) = 0
True
I cannot understand the logic here.
Why is the AND operator treated as a logical in the first case, and bitwise operator in the second? The parenthetical grouping shouldn't affect the outcome of this operation.
A nasty logic bomb is waiting to happen to VB developers here... trust me, it happened to me.





This might be better posted elsewhere.
I don't think you interpret the first one correctly. It looks, to me, as 1 = 0 yields 0 , then (bitwise) 4 AND 0 yields 0 .





That is because operator precedence. That can also happen in C#.
First one is
4 AND (1=0) ==> 4 and (false) ==> false
So you cant do bit operation with false so use the AND as a logical
Second is
(4 AND 1) = 0 ==> 0 = 0 ==> true.
Now AND have two numbers and can do logical operation.





I'm a seasoned programmer, and made this mistake.
Generally when one wants several conditions to be true, one will AND each of the terms together and then compare to some result.
IMO, it is quite counterintuitive that the equality comparison take precedence over the AND terms.
modified 11Jun15 12:35pm.





Im not saying if that is right or wrong, just try to answer to your question regarding what is the logic behind different result. Now if you want suggest behaviour improvement for VS we need move to another forum





A sensible article that I actually like 5'ed.








nice






Great, now I can yell at my PC when it misses a bit,
No excuses anymore for you Z620!!!
[Great article ProgramFox]





Maimonides wrote: [Great article ProgramFox]
Thank you!





I've also rotated the last 2 digits of a number of length n.
(number >> 2) + ((number & 0x3) << (2 * (n  1)))
This could be generalized to x by changing the first 2 to x, 0x3 to 2^x1, and the last 2 to 2^(x1) (I think).
modified 23Jul13 13:01pm.







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