📄 mmd5.bas
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Attribute VB_Name = "mMD5"
' MODULE: CMD5
'*******************************************************************************
Option Explicit
Public Const BITS_TO_A_BYTE As Long = 8
Public Const BYTES_TO_A_WORD As Long = 4
Public Const BITS_TO_A_WORD As Long = BYTES_TO_A_WORD * BITS_TO_A_BYTE
Public m_lOnBits(0 To 30) As Long
Public m_l2Power(0 To 30) As Long
'*******************************************************************************
' Class_Initialize (SUB)
'
' DESCRIPTION:
' We will usually get quicker results by preparing arrays of bit patterns and
' powers of 2 ahead of time instead of calculating them every time, unless of
' course the methods are only ever getting called once per instantiation of the
' class.
'*******************************************************************************
Public Sub Initialize()
' Could have done this with a loop calculating each value, but simply
' assigning the values is quicker - BITS SET FROM RIGHT
m_lOnBits(0) = 1 ' 00000000000000000000000000000001
m_lOnBits(1) = 3 ' 00000000000000000000000000000011
m_lOnBits(2) = 7 ' 00000000000000000000000000000111
m_lOnBits(3) = 15 ' 00000000000000000000000000001111
m_lOnBits(4) = 31 ' 00000000000000000000000000011111
m_lOnBits(5) = 63 ' 00000000000000000000000000111111
m_lOnBits(6) = 127 ' 00000000000000000000000001111111
m_lOnBits(7) = 255 ' 00000000000000000000000011111111
m_lOnBits(8) = 511 ' 00000000000000000000000111111111
m_lOnBits(9) = 1023 ' 00000000000000000000001111111111
m_lOnBits(10) = 2047 ' 00000000000000000000011111111111
m_lOnBits(11) = 4095 ' 00000000000000000000111111111111
m_lOnBits(12) = 8191 ' 00000000000000000001111111111111
m_lOnBits(13) = 16383 ' 00000000000000000011111111111111
m_lOnBits(14) = 32767 ' 00000000000000000111111111111111
m_lOnBits(15) = 65535 ' 00000000000000001111111111111111
m_lOnBits(16) = 131071 ' 00000000000000011111111111111111
m_lOnBits(17) = 262143 ' 00000000000000111111111111111111
m_lOnBits(18) = 524287 ' 00000000000001111111111111111111
m_lOnBits(19) = 1048575 ' 00000000000011111111111111111111
m_lOnBits(20) = 2097151 ' 00000000000111111111111111111111
m_lOnBits(21) = 4194303 ' 00000000001111111111111111111111
m_lOnBits(22) = 8388607 ' 00000000011111111111111111111111
m_lOnBits(23) = 16777215 ' 00000000111111111111111111111111
m_lOnBits(24) = 33554431 ' 00000001111111111111111111111111
m_lOnBits(25) = 67108863 ' 00000011111111111111111111111111
m_lOnBits(26) = 134217727 ' 00000111111111111111111111111111
m_lOnBits(27) = 268435455 ' 00001111111111111111111111111111
m_lOnBits(28) = 536870911 ' 00011111111111111111111111111111
m_lOnBits(29) = 1073741823 ' 00111111111111111111111111111111
m_lOnBits(30) = 2147483647 ' 01111111111111111111111111111111
' Could have done this with a loop calculating each value, but simply
' assigning the values is quicker - POWERS OF 2
m_l2Power(0) = 1 ' 00000000000000000000000000000001
m_l2Power(1) = 2 ' 00000000000000000000000000000010
m_l2Power(2) = 4 ' 00000000000000000000000000000100
m_l2Power(3) = 8 ' 00000000000000000000000000001000
m_l2Power(4) = 16 ' 00000000000000000000000000010000
m_l2Power(5) = 32 ' 00000000000000000000000000100000
m_l2Power(6) = 64 ' 00000000000000000000000001000000
m_l2Power(7) = 128 ' 00000000000000000000000010000000
m_l2Power(8) = 256 ' 00000000000000000000000100000000
m_l2Power(9) = 512 ' 00000000000000000000001000000000
m_l2Power(10) = 1024 ' 00000000000000000000010000000000
m_l2Power(11) = 2048 ' 00000000000000000000100000000000
m_l2Power(12) = 4096 ' 00000000000000000001000000000000
m_l2Power(13) = 8192 ' 00000000000000000010000000000000
m_l2Power(14) = 16384 ' 00000000000000000100000000000000
m_l2Power(15) = 32768 ' 00000000000000001000000000000000
m_l2Power(16) = 65536 ' 00000000000000010000000000000000
m_l2Power(17) = 131072 ' 00000000000000100000000000000000
m_l2Power(18) = 262144 ' 00000000000001000000000000000000
m_l2Power(19) = 524288 ' 00000000000010000000000000000000
m_l2Power(20) = 1048576 ' 00000000000100000000000000000000
m_l2Power(21) = 2097152 ' 00000000001000000000000000000000
m_l2Power(22) = 4194304 ' 00000000010000000000000000000000
m_l2Power(23) = 8388608 ' 00000000100000000000000000000000
m_l2Power(24) = 16777216 ' 00000001000000000000000000000000
m_l2Power(25) = 33554432 ' 00000010000000000000000000000000
m_l2Power(26) = 67108864 ' 00000100000000000000000000000000
m_l2Power(27) = 134217728 ' 00001000000000000000000000000000
m_l2Power(28) = 268435456 ' 00010000000000000000000000000000
m_l2Power(29) = 536870912 ' 00100000000000000000000000000000
m_l2Power(30) = 1073741824 ' 01000000000000000000000000000000
End Sub
'*******************************************************************************
' LShift (FUNCTION)
'
' PARAMETERS:
' (In) - lValue - Long - The value to be shifted
' (In) - iShiftBits - Integer - The number of bits to shift the value by
'
' RETURN VALUE:
' Long - The shifted long integer
'
' DESCRIPTION:
' A left shift takes all the set binary bits and moves them left, in-filling
' with zeros in the vacated bits on the right. This function is equivalent to
' the << operator in Java and C++
'*******************************************************************************
Private Function LShift(ByVal lValue As Long, _
ByVal iShiftBits As Integer) As Long
' NOTE: If you can guarantee that the Shift parameter will be in the
' range 1 to 30 you can safely strip of this first nested if structure for
' speed.
'
' A shift of zero is no shift at all.
If iShiftBits = 0 Then
LShift = lValue
Exit Function
' A shift of 31 will result in the right most bit becoming the left most
' bit and all other bits being cleared
ElseIf iShiftBits = 31 Then
If lValue And 1 Then
LShift = &H80000000
Else
LShift = 0
End If
Exit Function
' A shift of less than zero or more than 31 is undefined
ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
Err.Raise 6
End If
' If the left most bit that remains will end up in the negative bit
' position (&H80000000) we would end up with an overflow if we took the
' standard route. We need to strip the left most bit and add it back
' afterwards.
If (lValue And m_l2Power(31 - iShiftBits)) Then
' (Value And OnBits(31 - (Shift + 1))) chops off the left most bits that
' we are shifting into, but also the left most bit we still want as this
' is going to end up in the negative bit marker position (&H80000000).
' After the multiplication/shift we Or the result with &H80000000 to
' turn the negative bit on.
LShift = ((lValue And m_lOnBits(31 - (iShiftBits + 1))) * _
m_l2Power(iShiftBits)) Or &H80000000
Else
' (Value And OnBits(31-Shift)) chops off the left most bits that we are
' shifting into so we do not get an overflow error when we do the
' multiplication/shift
LShift = ((lValue And m_lOnBits(31 - iShiftBits)) * _
m_l2Power(iShiftBits))
End If
End Function
'*******************************************************************************
' RShift (FUNCTION)
'
' PARAMETERS:
' (In) - lValue - Long - The value to be shifted
' (In) - iShiftBits - Integer - The number of bits to shift the value by
'
' RETURN VALUE:
' Long - The shifted long integer
'
' DESCRIPTION:
' The right shift of an unsigned long integer involves shifting all the set bits
' to the right and in-filling on the left with zeros. This function is
' equivalent to the >>> operator in Java or the >> operator in C++ when used on
' an unsigned long.
'*******************************************************************************
Private Function RShift(ByVal lValue As Long, _
ByVal iShiftBits As Integer) As Long
' NOTE: If you can guarantee that the Shift parameter will be in the
' range 1 to 30 you can safely strip of this first nested if structure for
' speed.
'
' A shift of zero is no shift at all
If iShiftBits = 0 Then
RShift = lValue
Exit Function
' A shift of 31 will clear all bits and move the left most bit to the right
' most bit position
ElseIf iShiftBits = 31 Then
If lValue And &H80000000 Then
RShift = 1
Else
RShift = 0
End If
Exit Function
' A shift of less than zero or more than 31 is undefined
ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
Err.Raise 6
End If
' We do not care about the top most bit or the final bit, the top most bit
' will be taken into account in the next stage, the final bit (whether it
' is an odd number or not) is being shifted into, so we do not give a jot
' about it
RShift = (lValue And &H7FFFFFFE) \ m_l2Power(iShiftBits)
' If the top most bit (&H80000000) was set we need to do things differently
' as in a normal VB signed long integer the top most bit is used to indicate
' the sign of the number, when it is set it is a negative number, so just
' deviding by a factor of 2 as above would not work.
' NOTE: (lValue And &H80000000) is equivalent to (lValue < 0), you could
' get a very marginal speed improvement by changing the test to (lValue < 0)
If (lValue And &H80000000) Then
' We take the value computed so far, and then add the left most negative
' bit after it has been shifted to the right the appropriate number of
' places
RShift = (RShift Or (&H40000000 \ m_l2Power(iShiftBits - 1)))
End If
End Function
'*******************************************************************************
' RShiftSigned (FUNCTION)
'
' PARAMETERS:
' (In) - lValue - Long -
' (In) - iShiftBits - Integer -
'
' RETURN VALUE:
' Long -
'
' DESCRIPTION:
' The right shift of a signed long integer involves shifting all the set bits to
' the right and in-filling on the left with the sign bit (0 if positive, 1 if
' negative. This function is equivalent to the >> operator in Java or the >>
' operator in C++ when used on a signed long integer. Not used in this class,
' but included for completeness.
'*******************************************************************************
Private Function RShiftSigned(ByVal lValue As Long, _
ByVal iShiftBits As Integer) As Long
' NOTE: If you can guarantee that the Shift parameter will be in the
' range 1 to 30 you can safely strip of this first nested if structure for
' speed.
'
' A shift of zero is no shift at all
If iShiftBits = 0 Then
RShiftSigned = lValue
Exit Function
' A shift of 31 will clear all bits if the left most bit was zero, and will
' set all bits if the left most bit was 1 (a negative indicator)
ElseIf iShiftBits = 31 Then
' NOTE: (lValue And &H80000000) is equivalent to (lValue < 0), you
' could get a very marginal speed improvement by changing the test to
' (lValue < 0)
If (lValue And &H80000000) Then
RShiftSigned = -1
Else
RShiftSigned = 0
End If
Exit Function
' A shift of less than zero or more than 31 is undefined
ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
Err.Raise 6
End If
' We get the same result by dividing by the appropriate power of 2 and
' rounding in the negative direction
RShiftSigned = Int(lValue / m_l2Power(iShiftBits))
End Function
'*******************************************************************************
' RotateLeft (FUNCTION)
'
' PARAMETERS:
' (In) - lValue - Long - Value to act on
' (In) - iShiftBits - Integer - Bits to move by
'
' RETURN VALUE:
' Long - Result
'
' DESCRIPTION:
' Rotates the bits in a long integer to the left, those bits falling off the
' left edge are put back on the right edge
'*******************************************************************************
Private Function RotateLeft(ByVal lValue As Long, _
ByVal iShiftBits As Integer) As Long
RotateLeft = LShift(lValue, iShiftBits) Or RShift(lValue, (32 - iShiftBits))
End Function
'*******************************************************************************
' AddUnsigned (FUNCTION)
'
' PARAMETERS:
' (In) - lX - Long - First value
' (In) - lY - Long - Second value
'
' RETURN VALUE:
' Long - Result
'
' DESCRIPTION:
' Adds two potentially large unsigned numbers without overflowing
'*******************************************************************************
Private Function AddUnsigned(ByVal lX As Long, _
ByVal lY As Long) As Long
Dim lX4 As Long
Dim lY4 As Long
Dim lX8 As Long
Dim lY8 As Long
Dim lResult As Long
lX8 = lX And &H80000000
lY8 = lY And &H80000000
lX4 = lX And &H40000000
lY4 = lY And &H40000000
lResult = (lX And &H3FFFFFFF) + (lY And &H3FFFFFFF)
If lX4 And lY4 Then
lResult = lResult Xor &H80000000 Xor lX8 Xor lY8
ElseIf lX4 Or lY4 Then
If lResult And &H40000000 Then
lResult = lResult Xor &HC0000000 Xor lX8 Xor lY8
Else
lResult = lResult Xor &H40000000 Xor lX8 Xor lY8
End If
Else
lResult = lResult Xor lX8 Xor lY8
End If
AddUnsigned = lResult
End Function
'*******************************************************************************
' F (FUNCTION)
'
' DESCRIPTION:
' MD5's F function
'*******************************************************************************
Private Function F(ByVal x As Long, _
ByVal y As Long, _
ByVal z As Long) As Long
F = (x And y) Or ((Not x) And z)
End Function
'*******************************************************************************
' G (FUNCTION)
'
' DESCRIPTION:
' MD5's G function
'*******************************************************************************
Private Function G(ByVal x As Long, _
ByVal y As Long, _
ByVal z As Long) As Long
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