2014-03-26 1 views
5

Excel에서 SHA256 해시 함수를 만들려고합니다. 나는 this stackoverflow thread에 근거하여 가까운 것을 얻었다. Frez 라이브러리 (찾은 here)를 클래스 모듈로로드하고 워크 시트에서 사용할 함수를 만들었습니다. 불행히도, 함수를 사용할 때 결과는 예상 해시 된 문자열 대신 항상 "0"입니다. 나는 그것을 어떻게 고칠 수 있을지에 대한 아이디어를 찾고있다. 여기 SHA-256 라이브러리를 Excel로 이식하는 방법

모듈을 만들 수있는 코드이며, 그 아래에 클래스 모듈의 코드입니다 :

Public Function SHA256function(sMessage As String) 

    Dim clsX As CSHA256 
    Set clsX = New CSHA256 

    SHA256 = clsX.SHA256(sMessage) 

    Set clsX = Nothing 

End Function 

클래스 모듈 :

'******************************************************************************* 
' MODULE:  CSHA256 
' FILENAME:  CSHA256.cls 
' AUTHOR:  Phil Fresle 
' CREATED:  10-Apr-2001 
' COPYRIGHT: Copyright 2001 Phil Fresle. All Rights Reserved. 
' 
' DESCRIPTION: 
' This class is used to generate a SHA-256 'digest' or 'signature' of a string. 
' The SHA-256 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 SHA-256 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 SHA-256 for 
' your passwords, the random characters will make it harder to dictionary attack. 
' 
' *** CAUTION *** 
' See the comment attached to the SHA256 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: [email protected] 
' 
' MODIFICATION HISTORY: 
' 10-Apr-2001 Phil Fresle  Initial Version 
'******************************************************************************* 
Option Explicit 


Private m_lOnBits(30) As Long 
Private m_l2Power(30) As Long 
Private K(63)   As Long 


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 


'******************************************************************************* 
' Class_Initialize (SUB) 
'******************************************************************************* 
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 

    ' Just put together the K array once 
    K(0) = &H428A2F98 
    K(1) = &H71374491 
    K(2) = &HB5C0FBCF 
    K(3) = &HE9B5DBA5 
    K(4) = &H3956C25B 
    K(5) = &H59F111F1 
    K(6) = &H923F82A4 
    K(7) = &HAB1C5ED5 
    K(8) = &HD807AA98 
    K(9) = &H12835B01 
    K(10) = &H243185BE 
    K(11) = &H550C7DC3 
    K(12) = &H72BE5D74 
    K(13) = &H80DEB1FE 
    K(14) = &H9BDC06A7 
    K(15) = &HC19BF174 
    K(16) = &HE49B69C1 
    K(17) = &HEFBE4786 
    K(18) = &HFC19DC6 
    K(19) = &H240CA1CC 
    K(20) = &H2DE92C6F 
    K(21) = &H4A7484AA 
    K(22) = &H5CB0A9DC 
    K(23) = &H76F988DA 
    K(24) = &H983E5152 
    K(25) = &HA831C66D 
    K(26) = &HB00327C8 
    K(27) = &HBF597FC7 
    K(28) = &HC6E00BF3 
    K(29) = &HD5A79147 
    K(30) = &H6CA6351 
    K(31) = &H14292967 
    K(32) = &H27B70A85 
    K(33) = &H2E1B2138 
    K(34) = &H4D2C6DFC 
    K(35) = &H53380D13 
    K(36) = &H650A7354 
    K(37) = &H766A0ABB 
    K(38) = &H81C2C92E 
    K(39) = &H92722C85 
    K(40) = &HA2BFE8A1 
    K(41) = &HA81A664B 
    K(42) = &HC24B8B70 
    K(43) = &HC76C51A3 
    K(44) = &HD192E819 
    K(45) = &HD6990624 
    K(46) = &HF40E3585 
    K(47) = &H106AA070 
    K(48) = &H19A4C116 
    K(49) = &H1E376C08 
    K(50) = &H2748774C 
    K(51) = &H34B0BCB5 
    K(52) = &H391C0CB3 
    K(53) = &H4ED8AA4A 
    K(54) = &H5B9CCA4F 
    K(55) = &H682E6FF3 
    K(56) = &H748F82EE 
    K(57) = &H78A5636F 
    K(58) = &H84C87814 
    K(59) = &H8CC70208 
    K(60) = &H90BEFFFA 
    K(61) = &HA4506CEB 
    K(62) = &HBEF9A3F7 
    K(63) = &HC67178F2 
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 


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


'******************************************************************************* 
' Ch (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function 
'******************************************************************************* 
Private Function Ch(ByVal x As Long, _ 
        ByVal y As Long, _ 
        ByVal z As Long) As Long 
    Ch = ((x And y) Xor ((Not x) And z)) 
End Function 


'******************************************************************************* 
' Maj (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function 
'******************************************************************************* 
Private Function Maj(ByVal x As Long, _ 
        ByVal y As Long, _ 
        ByVal z As Long) As Long 
    Maj = ((x And y) Xor (x And z) Xor (y And z)) 
End Function 


'******************************************************************************* 
' S (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function (rotate right) 
'******************************************************************************* 
Private Function S(ByVal x As Long, _ 
        ByVal n As Long) As Long 
    S = (RShift(x, (n And m_lOnBits(4))) Or LShift(x, (32 - (n And m_lOnBits(4))))) 
End Function 


'******************************************************************************* 
' R (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function (just a right shift) 
'******************************************************************************* 
Private Function R(ByVal x As Long, _ 
        ByVal n As Long) As Long 
    R = RShift(x, CInt(n And m_lOnBits(4))) 
End Function 


'******************************************************************************* 
' Sigma0 (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function 
'******************************************************************************* 
Private Function Sigma0(ByVal x As Long) As Long 
    Sigma0 = (S(x, 2) Xor S(x, 13) Xor S(x, 22)) 
End Function 


'******************************************************************************* 
' Sigma1 (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function 
'******************************************************************************* 
Private Function Sigma1(ByVal x As Long) As Long 
    Sigma1 = (S(x, 6) Xor S(x, 11) Xor S(x, 25)) 
End Function 


'******************************************************************************* 
' Gamma0 (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function 
'******************************************************************************* 
Private Function Gamma0(ByVal x As Long) As Long 
    Gamma0 = (S(x, 7) Xor S(x, 18) Xor R(x, 3)) 
End Function 


'******************************************************************************* 
' Gamma1 (FUNCTION) 
' 
' DESCRIPTION: 
' SHA-256 function 
'******************************************************************************* 
Private Function Gamma1(ByVal x As Long) As Long 
    Gamma1 = (S(x, 17) Xor S(x, 19) Xor R(x, 10)) 
End Function 


'******************************************************************************* 
' 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 SHA-256 rules (similar to MD5 routine). 
'******************************************************************************* 
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 
    Dim lByte   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 lNumberOfWords 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, unlike MD5 we put the high-order 
    ' (most significant) byte first in each word. 
    lBytePosition = 0 
    lByteCount = 0 
    Do Until lByteCount >= lMessageLength 
     ' Each word is 4 bytes 
     lWordCount = lByteCount \ BYTES_TO_A_WORD 

     lBytePosition = (3 - (lByteCount Mod BYTES_TO_A_WORD)) * BITS_TO_A_BYTE 

     ' NOTE: This is where we are using just the first byte of each unicode 
     ' character, you may want to make the change here, or to the SHA256 method 
     ' so it accepts a byte array. 
     lByte = AscB(Mid(sMessage, lByteCount + 1, 1)) 

     lWordArray(lWordCount) = lWordArray(lWordCount) Or LShift(lByte, lBytePosition) 
     lByteCount = lByteCount + 1 
    Loop 


    ' Terminate according to SHA-256 rules with a 1 bit, zeros and the length in 
    ' bits stored in the last two words 
    lWordCount = lByteCount \ BYTES_TO_A_WORD 
    lBytePosition = (3 - (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 last word. Any bits shifted off the left edge 
    ' need to be put in the penultimate word, we can work out which bits by shifting 
    ' right the length by 29 bits. 
    lWordArray(lNumberOfWords - 1) = LShift(lMessageLength, 3) 
    lWordArray(lNumberOfWords - 2) = RShift(lMessageLength, 29) 

    ConvertToWordArray = lWordArray 
End Function 


'******************************************************************************* 
' SHA256 (FUNCTION) 
' 
' PARAMETERS: 
' (In/Out) - sMessage - String - Message to digest 
' 
' RETURN VALUE: 
' String - The digest 
' 
' DESCRIPTION: 
' Takes a string and uses the SHA-256 digest to produce a signature for it. 
' 
' 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 SHA256(sMessage As String) As String 
    Dim HASH(7) As Long 
    Dim M()  As Long 
    Dim W(63) As Long 
    Dim a  As Long 
    Dim b  As Long 
    Dim c  As Long 
    Dim d  As Long 
    Dim e  As Long 
    Dim f  As Long 
    Dim g  As Long 
    Dim h  As Long 
    Dim i  As Long 
    Dim j  As Long 
    Dim T1  As Long 
    Dim T2  As Long 

    ' Initial hash values 
    HASH(0) = &H6A09E667 
    HASH(1) = &HBB67AE85 
    HASH(2) = &H3C6EF372 
    HASH(3) = &HA54FF53A 
    HASH(4) = &H510E527F 
    HASH(5) = &H9B05688C 
    HASH(6) = &H1F83D9AB 
    HASH(7) = &H5BE0CD19 

    ' Preprocessing. Append padding bits and length and convert to words 
    M = ConvertToWordArray(sMessage) 

    ' Main loop 
    For i = 0 To UBound(M) Step 16 
     a = HASH(0) 
     b = HASH(1) 
     c = HASH(2) 
     d = HASH(3) 
     e = HASH(4) 
     f = HASH(5) 
     g = HASH(6) 
     h = HASH(7) 

     For j = 0 To 63 
      If j < 16 Then 
       W(j) = M(j + i) 
      Else 
       W(j) = AddUnsigned(AddUnsigned(AddUnsigned(Gamma1(W(j - 2)), _ 
        W(j - 7)), Gamma0(W(j - 15))), W(j - 16)) 
      End If 

      T1 = AddUnsigned(AddUnsigned(AddUnsigned(AddUnsigned(h, Sigma1(e)), _ 
       Ch(e, f, g)), K(j)), W(j)) 
      T2 = AddUnsigned(Sigma0(a), Maj(a, b, c)) 

      h = g 
      g = f 
      f = e 
      e = AddUnsigned(d, T1) 
      d = c 
      c = b 
      b = a 
      a = AddUnsigned(T1, T2) 
     Next 

     HASH(0) = AddUnsigned(a, HASH(0)) 
     HASH(1) = AddUnsigned(b, HASH(1)) 
     HASH(2) = AddUnsigned(c, HASH(2)) 
     HASH(3) = AddUnsigned(d, HASH(3)) 
     HASH(4) = AddUnsigned(e, HASH(4)) 
     HASH(5) = AddUnsigned(f, HASH(5)) 
     HASH(6) = AddUnsigned(g, HASH(6)) 
     HASH(7) = AddUnsigned(h, HASH(7)) 
    Next 

    ' Output the 256 bit digest 
    SHA256 = LCase(Right("00000000" & Hex(HASH(0)), 8) & _ 
     Right("00000000" & Hex(HASH(1)), 8) & _ 
     Right("00000000" & Hex(HASH(2)), 8) & _ 
     Right("00000000" & Hex(HASH(3)), 8) & _ 
     Right("00000000" & Hex(HASH(4)), 8) & _ 
     Right("00000000" & Hex(HASH(5)), 8) & _ 
     Right("00000000" & Hex(HASH(6)), 8) & _ 
     Right("00000000" & Hex(HASH(7)), 8)) 
End Function 

감사합니다.

+0

Frez 라이브러리에 링크가 다운되었습니다. 새 것을 추가 할 수 있습니까? – Hastur

답변

6

당신은 아무것도 반환하지 않는, 그냥 변경 :

SHA256 = clsX.SHA256(sMessage) 

에 :

SHA256function = clsX.SHA256(sMessage) 
+3

그게 있다면,'Option Explicit'에 대한 또 다른 승리로 ch ch ... – AakashM

+0

고마워, Alex K. 그 트릭을. – mlattis

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