Identifying the ENCRYPTION ALGORITHM used in advapi32.dll - winapi

Question

How to find which encryption is used in the following vb code
As far as i understood it is using
MD5 to hash the given key
MD5's result is obtained.
then use the resulting value as key for RC4 algorithm
The problem is i am not getting the same output when the above procedures are used separately.
The vb code gives the following output.
aa = !!S
a = !!
b = +
c = -
abc = !!P®
hello = W¡!!‘
A = 3
B = 0
C = 1
ABC = 3pŽ
Note: the value of b above is a "+" -like symbol in which the vertical line is longer.
The vb code is as follows.
Private Const MS_DEF_PROV = "Microsoft Base Cryptographic Provider v1.0"
Private Const PROV_RSA_FULL = 1
Private Const ALG_CLASS_DATA_ENCRYPT = 24576
Private Const ALG_CLASS_HASH = 32768
Private Const ALG_TYPE_ANY = 0
Private Const ALG_TYPE_BLOCK = 1536
Private Const ALG_TYPE_STREAM = 2048
Private Const ALG_SID_RC2 = 2
Private Const ALG_SID_RC4 = 1
Private Const ALG_SID_MD5 = 3
Private Const CALG_MD5 = ((ALG_CLASS_HASH Or ALG_TYPE_ANY) Or ALG_SID_MD5)
Private Const CALG_RC2 = ((ALG_CLASS_DATA_ENCRYPT Or ALG_TYPE_BLOCK) Or ALG_SID_RC2)
Private Const CALG_RC4 = ((ALG_CLASS_DATA_ENCRYPT Or ALG_TYPE_STREAM) Or ALG_SID_RC4)
Private Const ENCRYPT_ALGORITHM = CALG_RC4
Private Const ENCRYPT_BLOCK_SIZE = 1
Private Const CRYPT_EXPORTABLE = 1
Dim lHHash As Long
Dim lHkey As Long
Dim lResult As Long
Dim lHExchgKey As Long
Dim lHCryptprov As Long
Dim sContainer As String
Dim lCryptLength As Long
Dim lCryptBufLen As Long
Dim sCryptBuffer As String
On Error GoTo EncryptError
Dim sOutputBuffer As String
Dim sProvider
sOutputBuffer = ""
'Get handle to the default CSP
sProvider = MS_DEF_PROV & vbNullChar
If Len(PlainText) = 0 Then
DoCryptoEncrypt = ""
Exit Function
End If
If Not CBool(CryptAcquireContext(lHCryptprov, ByVal _
sContainer, ByVal sProvider, PROV_RSA_FULL, 0)) Then
' If there is no default key container then create one
' using Flags field
If GetLastError = 0 Then
If Not CBool(CryptAcquireContext(lHCryptprov, 0&, ByVal sProvider, PROV_RSA_FULL, CRYPT_NEWKEYSET)) Then
sOutputBuffer = PlainText
GoTo Finished
End If
End If
End If
'Create a hash object
If Not CBool(CryptCreateHash(lHCryptprov, CALG_MD5, 0, _
0, lHHash)) Then
GoTo Finished
End If
'Hash in the password text
If Not CBool(CryptHashData(lHHash, sPassword, _
Len(sPassword), 0)) Then
GoTo Finished
End If
'Create a session key from the hash object.
If Not CBool(CryptDeriveKey(lHCryptprov, _
ENCRYPT_ALGORITHM, lHHash, 0, lHkey)) Then
GoTo Finished
End If
'Destroy the hash object.
CryptDestroyHash (lHHash)
lHHash = 0
'Create a buffer for the CryptEncrypt function
lCryptLength = Len(PlainText)
lCryptBufLen = lCryptLength * 2
sCryptBuffer = String(lCryptBufLen, vbNullChar)
LSet sCryptBuffer = PlainText
'Encrypt the text data
If Not CBool(CryptEncrypt(lHkey, 0, 1, 0, sCryptBuffer, _
lCryptLength, lCryptBufLen)) Then
End If

Answer 1

The relevant code is:
If Not CBool(CryptDeriveKey(lHCryptprov, _
ENCRYPT_ALGORITHM, lHHash, 0, lHkey)) Then
GoTo Finished
End If
And ENCRYPT_ALGORITHM is defined as CALG_RC4. This constant is documented here: https://learn.microsoft.com/de-de/windows/desktop/SecCrypto/alg-id
So it's: RC4 stream encryption.

Answer 2

Your vb code works as follows.
it use MD5 to hash the given value.
then it creates a session key compatible with RC4 algorithm using MD5 hash value
creates encrypted value using RC4 as algorithm and session-key as key