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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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