VB6.CopyArray Question
I have tried to converet Fresle's CGenericRegistration and CMD5 classes from VB6 to VB.Net 2005. (www.frez.co.uk/registration.zip) The wizard converts the TestRegistration program, which includes the classes and two others, and runs fine when AscB is replaced with Asc, although the results are not the same. The wizard inserts the following code in the ConvertToWordArray function:
ConvertToWordArray = vb6.copyarray (lWordArray)
This works great when run in the IDE but when the classes are put into a class library this line gives a "VB6 not defined" type error.
The following import statement is in both the TestRegistration program and the Class Library project:
Imports VB = Microsoft.VisualBasic
I have been unable to find where the VB6.CopyArray method is located or what method is available to duplicate the functionality.
Any help would be greatly appreciated. Both VB6 and VB.Net 2005 versions follow.
**************************
VB6 code:
Private Function ConvertToWordArray(sMessag
Dim lMessageLength As Long
Dim lNumberOfWords As Long
Dim lWordArray() As Long
Dim lBytePosition As Long
Dim lByteCount As Long
Dim lWordCount As Long
Dim lChar 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
lChar = AscB(Mid(sMessage, lByteCount + 1, 1))
lWordArray(lWordCount) = lWordArray(lWordCount) Or LShift(lChar, 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
**************************
VB.Net 2005 Code:
Private Function ConvertToWordArray(ByRef sMessage As String) As Integer()
Dim lMessageLength As Integer
Dim lNumberOfWords As Integer
Dim lWordArray() As Integer
Dim lBytePosition As Integer
Dim lByteCount As Integer
Dim lWordCount As Integer
Dim lChar As Integer
Const MODULUS_BITS As Integer = 512
Const CONGRUENT_BITS As Integer = 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
' lChar = AscB(Mid(sMessage, lByteCount + 1, 1)) AscB no longer supported.
lChar = Asc(Mid(sMessage, lByteCount + 1, 1))
lWordArray(lWordCount) = lWordArray(lWordCount) Or LShift(lChar, 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(&H80s, 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 = VB6.CopyArray(lWordArray)
End Function
ConvertToWordArray = vb6.copyarray (lWordArray)
This works great when run in the IDE but when the classes are put into a class library this line gives a "VB6 not defined" type error.
The following import statement is in both the TestRegistration program and the Class Library project:
Imports VB = Microsoft.VisualBasic
I have been unable to find where the VB6.CopyArray method is located or what method is available to duplicate the functionality.
Any help would be greatly appreciated. Both VB6 and VB.Net 2005 versions follow.
**************************
VB6 code:
Private Function ConvertToWordArray(sMessag
Dim lMessageLength As Long
Dim lNumberOfWords As Long
Dim lWordArray() As Long
Dim lBytePosition As Long
Dim lByteCount As Long
Dim lWordCount As Long
Dim lChar 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
lChar = AscB(Mid(sMessage, lByteCount + 1, 1))
lWordArray(lWordCount) = lWordArray(lWordCount) Or LShift(lChar, 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
**************************
VB.Net 2005 Code:
Private Function ConvertToWordArray(ByRef sMessage As String) As Integer()
Dim lMessageLength As Integer
Dim lNumberOfWords As Integer
Dim lWordArray() As Integer
Dim lBytePosition As Integer
Dim lByteCount As Integer
Dim lWordCount As Integer
Dim lChar As Integer
Const MODULUS_BITS As Integer = 512
Const CONGRUENT_BITS As Integer = 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
' lChar = AscB(Mid(sMessage, lByteCount + 1, 1)) AscB no longer supported.
lChar = Asc(Mid(sMessage, lByteCount + 1, 1))
lWordArray(lWordCount) = lWordArray(lWordCount) Or LShift(lChar, 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(&H80s, 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 = VB6.CopyArray(lWordArray)
End Function
ASKER
Hilwaaa,
Thanks for the suggestion. Unfortunately, that did not resolve the problem, but I am sure you are on the right track. Going down to the VisualBasic.Compatibility.
I did some more testing this morning and discovered what I should have discovered at first:
ConvertToWordArray = lWordArray
and
ConvertToWordArray = VB6.CopyArray(lWordArray)
yielded exactly the same ConvertToWordArray result. That meant the problem was really in the calling program. I found it and all is now well.
I will close this question even though it means I may never know where VB6.CopyArray really exists.
Thanks for the suggestion. Unfortunately, that did not resolve the problem, but I am sure you are on the right track. Going down to the VisualBasic.Compatibility.
I did some more testing this morning and discovered what I should have discovered at first:
ConvertToWordArray = lWordArray
and
ConvertToWordArray = VB6.CopyArray(lWordArray)
yielded exactly the same ConvertToWordArray result. That meant the problem was really in the calling program. I found it and all is now well.
I will close this question even though it means I may never know where VB6.CopyArray really exists.
ASKER CERTIFIED SOLUTION
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ASKER
Hillwaaa:
I stand corrected. It did not appear in my VB 2005 Express IDE. My professional edition is in transit. Thanks for your help.
I stand corrected. It did not appear in my VB 2005 Express IDE. My professional edition is in transit. Thanks for your help.
Ah - that's great. Good luck with the rest of your conversion! (they can be fun :)
ASKER
I stand corrected again. I found it in the Express IDE. The imports statement should be:
Imports VB6 = Microsoft.VisualBasic.Comp
Then VB6.CopyArray will work (or you could use the fully-qualified name rather than the alias VB6)
Thanks for the encouragement. The original program is about 40,000 lines. I will try to re-write much of it in a simpler form. Great fun.
Imports VB6 = Microsoft.VisualBasic.Comp
Then VB6.CopyArray will work (or you could use the fully-qualified name rather than the alias VB6)
Thanks for the encouragement. The original program is about 40,000 lines. I will try to re-write much of it in a simpler form. Great fun.
You will probably need to use the following import instead (this gives you the VB6 stuff)
Imports Microsoft.VisualBasic.Comp
note that you will need to add a project reference to the "Microsoft Visual Basic .NET Compatibility Runtime".
Cheers!