Access - Encriptar contraseñas con cmd5

Private Sub Comando28_Click()

Call GenerarHashPassword(Texto32)

End Sub

Option Explicit

'*******************************************************************************

' MODULE:       CMD5

' FILENAME:     C:\My Code\vb\md5\CMD5.cls

' AUTHOR:       Phil Fresle

' CREATED:      16-Feb-2001

' COPYRIGHT:    Copyright 2001 Frez Systems Limited. All Rights Reserved.

'

' DESCRIPTION:

' Derived from the RSA Data Security, Inc. MD5 Message-Digest Algorithm,

' as set out in the memo RFC1321.

'

' This class is used to generate an MD5 'digest' or 'signature' of a string. The

' MD5 algorithm is one of the industry standard methods for generating digital

' signatures. It is generically known as a digest, digital signature, one-way

' encryption, hash or checksum algorithm. A common use for MD5 is for password

' encryption as it is one-way in nature, that does not mean that your passwords

' are not free from a dictionary attack. If you are using the

' routine for passwords, you can make it a little more secure by concatenating

' some known random characters to the password before you generate the signature

' and on subsequent tests, so even if a hacker knows you are using MD5 for

' your passwords, the random characters will make it harder to dictionary attack.

'

' *** CAUTION ***

' See the comment attached to the MD5 method below regarding use on systems

' with different character sets.

'

' This is 'free' software with the following restrictions:

'

' You may not redistribute this code as a 'sample' or 'demo'. However, you are free

' to use the source code in your own code, but you may not claim that you created

' the sample code. It is expressly forbidden to sell or profit from this source code

' other than by the knowledge gained or the enhanced value added by your own code.

'

' Use of this software is also done so at your own risk. The code is supplied as

' is without warranty or guarantee of any kind.

'

' Should you wish to commission some derivative work based on this code provided

' here, or any consultancy work, please do not hesitate to contact us.

'

' Web Site:  http://www.frez.co.uk

' E-mail:    sales@frez.co.uk

'

' MODIFICATION HISTORY:

' 1.0       16-Feb-2001

'           Phil Fresle

'           Initial Version

'*******************************************************************************

Private Const BITS_TO_A_BYTE  As Long = 8

Private Const BYTES_TO_A_WORD As Long = 4

Private Const BITS_TO_A_WORD  As Long = BYTES_TO_A_WORD * BITS_TO_A_BYTE

Private m_lOnBits(0 To 30) As Long

Private 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.

'*******************************************************************************

Private Sub Class_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

    G = (x And z) Or (y And (Not z))

End Function

'*******************************************************************************

' H (FUNCTION)

'

' DESCRIPTION:

' MD5's H function

'*******************************************************************************

Private Function H(ByVal x As Long, _

                   ByVal y As Long, _

                   ByVal z As Long) As Long

    H = (x Xor y Xor z)

End Function

'*******************************************************************************

' I (FUNCTION)

'

' DESCRIPTION:

' MD5's I function

'*******************************************************************************

Private Function i(ByVal x As Long, _

                   ByVal y As Long, _

                   ByVal z As Long) As Long

    i = (y Xor (x Or (Not z)))

End Function

'*******************************************************************************

' FF (SUB)

'

' DESCRIPTION:

' MD5's FF procedure

'*******************************************************************************

Private Sub FF(a As Long, _

               ByVal b As Long, _

               ByVal c As Long, _

               ByVal d As Long, _

               ByVal x As Long, _

               ByVal s As Long, _

               ByVal ac As Long)

    a = AddUnsigned(a, AddUnsigned(AddUnsigned(F(b, c, d), x), ac))

    a = RotateLeft(a, s)

    a = AddUnsigned(a, b)

End Sub

'*******************************************************************************

' GG (SUB)

'

' DESCRIPTION:

' MD5's GG procedure

'*******************************************************************************

Private Sub GG(a As Long, _

               ByVal b As Long, _

               ByVal c As Long, _

               ByVal d As Long, _

               ByVal x As Long, _

               ByVal s As Long, _

               ByVal ac As Long)

    a = AddUnsigned(a, AddUnsigned(AddUnsigned(G(b, c, d), x), ac))

    a = RotateLeft(a, s)

    a = AddUnsigned(a, b)

End Sub

'*******************************************************************************

' HH (SUB)

'

' DESCRIPTION:

' MD5's HH procedure

'*******************************************************************************

Private Sub hh(a As Long, _

               ByVal b As Long, _

               ByVal c As Long, _

               ByVal d As Long, _

               ByVal x As Long, _

               ByVal s As Long, _

               ByVal ac As Long)

    a = AddUnsigned(a, AddUnsigned(AddUnsigned(H(b, c, d), x), ac))

    a = RotateLeft(a, s)

    a = AddUnsigned(a, b)

End Sub

'*******************************************************************************

' II (SUB)

'

' DESCRIPTION:

' MD5's II procedure

'*******************************************************************************

Private Sub II(a As Long, _

               ByVal b As Long, _

               ByVal c As Long, _

               ByVal d As Long, _

               ByVal x As Long, _

               ByVal s As Long, _

               ByVal ac As Long)

    a = AddUnsigned(a, AddUnsigned(AddUnsigned(i(b, c, d), x), ac))

    a = RotateLeft(a, s)

    a = AddUnsigned(a, b)

End Sub

'*******************************************************************************

' ConvertToWordArray (FUNCTION)

'

' PARAMETERS:

' (In/Out) - sMessage - String - String message

'

' RETURN VALUE:

' Long() - Converted message as long array

'

' DESCRIPTION:

' Takes the string message and puts it in a long array with padding according to

' the MD5 rules.

'*******************************************************************************

Private Function ConvertToWordArray(sMessage As String) As Long()

    Dim lMessageLength  As Long

    Dim lNumberOfWords  As Long

    Dim lWordArray()    As Long

    Dim lBytePosition   As Long

    Dim lByteCount      As Long

    Dim lWordCount      As Long

    Const MODULUS_BITS      As Long = 512

    Const CONGRUENT_BITS    As Long = 448

    lMessageLength = Len(sMessage)

    ' Get padded number of words. Message needs to be congruent to 448 bits,

    ' modulo 512 bits. If it is exactly congruent to 448 bits, modulo 512 bits

    ' it must still have another 512 bits added. 512 bits = 64 bytes

    ' (or 16 * 4 byte words), 448 bits = 56 bytes. This means lMessageSize must

    ' be a multiple of 16 (i.e. 16 * 4 (bytes) * 8 (bits))

    lNumberOfWords = (((lMessageLength + _

        ((MODULUS_BITS - CONGRUENT_BITS) \ BITS_TO_A_BYTE)) \ _

        (MODULUS_BITS \ BITS_TO_A_BYTE)) + 1) * _

        (MODULUS_BITS \ BITS_TO_A_WORD)

    ReDim lWordArray(lNumberOfWords - 1)

    ' Combine each block of 4 bytes (ascii code of character) into one long

    ' value and store in the message. The high-order (most significant) bit of

    ' each byte is listed first. However, the low-order (least significant) byte

    ' is given first in each word.

    lBytePosition = 0

    lByteCount = 0

    Do Until lByteCount >= lMessageLength

        ' Each word is 4 bytes

        lWordCount = lByteCount \ BYTES_TO_A_WORD

        ' The bytes are put in the word from the right most edge

        lBytePosition = (lByteCount Mod BYTES_TO_A_WORD) * BITS_TO_A_BYTE

        lWordArray(lWordCount) = lWordArray(lWordCount) Or _

            LShift(AscB(Mid(sMessage, lByteCount + 1, 1)), lBytePosition)

        lByteCount = lByteCount + 1

    Loop

    ' Terminate according to MD5 rules with a 1 bit, zeros and the length in

    ' bits stored in the last two words

    lWordCount = lByteCount \ BYTES_TO_A_WORD

    lBytePosition = (lByteCount Mod BYTES_TO_A_WORD) * BITS_TO_A_BYTE

    ' Add a terminating 1 bit, all the rest of the bits to the end of the

    ' word array will default to zero

    lWordArray(lWordCount) = lWordArray(lWordCount) Or _

        LShift(&H80, lBytePosition)

    ' We put the length of the message in bits into the last two words, to get

    ' the length in bits we need to multiply by 8 (or left shift 3). This left

    ' shifted value is put in the first word. Any bits shifted off the left edge

    ' need to be put in the second word, we can work out which bits by shifting

    ' right the length by 29 bits.

    lWordArray(lNumberOfWords - 2) = LShift(lMessageLength, 3)

    lWordArray(lNumberOfWords - 1) = RShift(lMessageLength, 29)

    ConvertToWordArray = lWordArray

End Function

'*******************************************************************************

' WordToHex (FUNCTION)

'

' PARAMETERS:

' (In) - lValue - Long - Long value to convert

'

' RETURN VALUE:

' String - Hex value to return

'

' DESCRIPTION:

' Takes a long integer and due to the bytes reverse order it extracts the

' individual bytes and converts them to hex appending them for an overall hex

' value

'*******************************************************************************

Private Function WordToHex(ByVal lValue As Long) As String

    Dim lByte As Long

    Dim lCount As Long

    For lCount = 0 To 3

        lByte = RShift(lValue, lCount * BITS_TO_A_BYTE) And _

            m_lOnBits(BITS_TO_A_BYTE - 1)

        WordToHex = WordToHex & Right("0" & Hex(lByte), 2)

    Next

End Function

'*******************************************************************************

' MD5 (FUNCTION)

'

' PARAMETERS:

' (In/Out) - sMessage - String - String to be digested

'

' RETURN VALUE:

' String - The MD5 digest

'

' DESCRIPTION:

' This function takes a string message and generates an MD5 digest for it.

' sMessage can be up to the VB string length limit of 2^31 (approx. 2 billion)

' characters.

'

' NOTE: Due to the way in which the string is processed the routine assumes a

' single byte character set. VB passes unicode (2-byte) character strings, the

' ConvertToWordArray function uses on the first byte for each character. This

' has been done this way for ease of use, to make the routine truely portable

' you could accept a byte array instead, it would then be up to the calling

' routine to make sure that the byte array is generated from their string in

' a manner consistent with the string type.

'*******************************************************************************

Public Function MD5(sMessage As String) As String

    Dim x() As Long

    Dim k   As Long

    Dim AA  As Long

    Dim BB  As Long

    Dim CC  As Long

    Dim DD  As Long

    Dim a   As Long

    Dim b   As Long

    Dim c   As Long

    Dim d   As Long

    Const S11 As Long = 7

    Const S12 As Long = 12

    Const S13 As Long = 17

    Const S14 As Long = 22

    Const S21 As Long = 5

    Const S22 As Long = 9

    Const S23 As Long = 14

    Const S24 As Long = 20

    Const S31 As Long = 4

    Const S32 As Long = 11

    Const S33 As Long = 16

    Const S34 As Long = 23

    Const S41 As Long = 6

    Const S42 As Long = 10

    Const S43 As Long = 15

    Const S44 As Long = 21

    ' Steps 1 and 2.  Append padding bits and length and convert to words

    x = ConvertToWordArray(sMessage)

    ' Step 3.  Initialise

    a = &H67452301

    b = &HEFCDAB89

    c = &H98BADCFE

    d = &H10325476

    ' Step 4.  Process the message in 16-word blocks

    For k = 0 To UBound(x) Step 16

        AA = a

        BB = b

        CC = c

        DD = d

        ' The hex number on the end of each of the following procedure calls is

        ' an element from the 64 element table constructed with

        ' T(i) = Int(4294967296 * Abs(Sin(i))) where i is 1 to 64.

        '

        ' However, for speed we don't want to calculate the value every time.

        FF a, b, c, d, x(k + 0), S11, &HD76AA478

        FF d, a, b, c, x(k + 1), S12, &HE8C7B756

        FF c, d, a, b, x(k + 2), S13, &H242070DB

        FF b, c, d, a, x(k + 3), S14, &HC1BDCEEE

        FF a, b, c, d, x(k + 4), S11, &HF57C0FAF

        FF d, a, b, c, x(k + 5), S12, &H4787C62A

        FF c, d, a, b, x(k + 6), S13, &HA8304613

        FF b, c, d, a, x(k + 7), S14, &HFD469501

        FF a, b, c, d, x(k + 8), S11, &H698098D8

        FF d, a, b, c, x(k + 9), S12, &H8B44F7AF

        FF c, d, a, b, x(k + 10), S13, &HFFFF5BB1

        FF b, c, d, a, x(k + 11), S14, &H895CD7BE

        FF a, b, c, d, x(k + 12), S11, &H6B901122

        FF d, a, b, c, x(k + 13), S12, &HFD987193

        FF c, d, a, b, x(k + 14), S13, &HA679438E

        FF b, c, d, a, x(k + 15), S14, &H49B40821

        GG a, b, c, d, x(k + 1), S21, &HF61E2562

        GG d, a, b, c, x(k + 6), S22, &HC040B340

        GG c, d, a, b, x(k + 11), S23, &H265E5A51

        GG b, c, d, a, x(k + 0), S24, &HE9B6C7AA

        GG a, b, c, d, x(k + 5), S21, &HD62F105D

        GG d, a, b, c, x(k + 10), S22, &H2441453

        GG c, d, a, b, x(k + 15), S23, &HD8A1E681

        GG b, c, d, a, x(k + 4), S24, &HE7D3FBC8

        GG a, b, c, d, x(k + 9), S21, &H21E1CDE6

        GG d, a, b, c, x(k + 14), S22, &HC33707D6

        GG c, d, a, b, x(k + 3), S23, &HF4D50D87

        GG b, c, d, a, x(k + 8), S24, &H455A14ED

        GG a, b, c, d, x(k + 13), S21, &HA9E3E905

        GG d, a, b, c, x(k + 2), S22, &HFCEFA3F8

        GG c, d, a, b, x(k + 7), S23, &H676F02D9

        GG b, c, d, a, x(k + 12), S24, &H8D2A4C8A

        hh a, b, c, d, x(k + 5), S31, &HFFFA3942

        hh d, a, b, c, x(k + 8), S32, &H8771F681

        hh c, d, a, b, x(k + 11), S33, &H6D9D6122

        hh b, c, d, a, x(k + 14), S34, &HFDE5380C

        hh a, b, c, d, x(k + 1), S31, &HA4BEEA44

        hh d, a, b, c, x(k + 4), S32, &H4BDECFA9

        hh c, d, a, b, x(k + 7), S33, &HF6BB4B60

        hh b, c, d, a, x(k + 10), S34, &HBEBFBC70

        hh a, b, c, d, x(k + 13), S31, &H289B7EC6

        hh d, a, b, c, x(k + 0), S32, &HEAA127FA

        hh c, d, a, b, x(k + 3), S33, &HD4EF3085

        hh b, c, d, a, x(k + 6), S34, &H4881D05

        hh a, b, c, d, x(k + 9), S31, &HD9D4D039

        hh d, a, b, c, x(k + 12), S32, &HE6DB99E5

        hh c, d, a, b, x(k + 15), S33, &H1FA27CF8

        hh b, c, d, a, x(k + 2), S34, &HC4AC5665

        II a, b, c, d, x(k + 0), S41, &HF4292244

        II d, a, b, c, x(k + 7), S42, &H432AFF97

        II c, d, a, b, x(k + 14), S43, &HAB9423A7

        II b, c, d, a, x(k + 5), S44, &HFC93A039

        II a, b, c, d, x(k + 12), S41, &H655B59C3

        II d, a, b, c, x(k + 3), S42, &H8F0CCC92

        II c, d, a, b, x(k + 10), S43, &HFFEFF47D

        II b, c, d, a, x(k + 1), S44, &H85845DD1

        II a, b, c, d, x(k + 8), S41, &H6FA87E4F

        II d, a, b, c, x(k + 15), S42, &HFE2CE6E0

        II c, d, a, b, x(k + 6), S43, &HA3014314

        II b, c, d, a, x(k + 13), S44, &H4E0811A1

        II a, b, c, d, x(k + 4), S41, &HF7537E82

        II d, a, b, c, x(k + 11), S42, &HBD3AF235

        II c, d, a, b, x(k + 2), S43, &H2AD7D2BB

        II b, c, d, a, x(k + 9), S44, &HEB86D391

        a = AddUnsigned(a, AA)

        b = AddUnsigned(b, BB)

        c = AddUnsigned(c, CC)

        d = AddUnsigned(d, DD)

    Next

    ' Step 5.  Output the 128 bit digest

    MD5 = LCase(WordToHex(a) & WordToHex(b) & WordToHex(c) & WordToHex(d))

End Function

Private Sub cmdLogin_Click()

    Dim ConstraseñaHash As String

    Static Intentos As Integer

    ContraseñaHash = Nz(DLookUp("ContraseñaHash", "Usuarios", "Usuario = '" & Nz(Me.txtUsuario) & "'*NOT FOUND*"))

    Intentos = Intentos + 1

    If ContraseñaHash <> "*NOT FOUND*" Then 'Se ha encontrado el usuario

        If GenerarHashPassword(Me.txtContraseña) = ContraseñaHash Then 'Las contraseñas coinciden

            'Lo que haya que hacer para validar el login

            Intentos = 0 'Reestablecemos la variable por si acaso.

            DoCmd.Close AcForm, Me.Name 'Cerramos el formulario

        Else

            If Intentos = 3 Then

                MsgBox "Login incorrecto."

                Docmd.Quit

            Else

                MsgBox "Usuario y/o contraseña incorrectos. Dispone de " & 3 - Intentos & " intentos."

                Me.txtUsuario = Null

                Me.txtContraseña = Null

                Me.txtUsuario.SetFocus

            End If

        End If

    End If

End Sub