I want to port this simple hash algorithm to VB6.
I have come up with:
Public Function simpleHash(hashString As String) As Long
Dim hash As Long
Dim c As Long
Dim i As Integer
On Local Error Resume Next
hash = 5381
For i = 1 To Len(hashString)
c = AscW(Mid$(hashString, i, 1))
hash = hash * 33 + c
Next i
simpleHash = hash
End Function
The problem is, despite my On Error
statement which suppresses the Error 6: Overflow exceptions, the variable hash
is not updated any more if an overflow occurs.
How can I fix that or implement this algorithm differently?
I would question whether it makes sense to hash UTF-16LE ("Unicode") this way. It might make a lot more sense to convert your VB String to UTF-8 and then hash that.
While I can't find any test vectors for djb2 to validate my own implementation it seems to run quite quickly:
Private Type CURRENCY_CURRENCY
Value As Currency
End Type
Private Type CURRENCY_2LONGS
ValueLo As Long
ValueHi As Long
End Type
Public Function djb2(ByVal StringToHash As String) As Long
Dim C2L As CURRENCY_2LONGS
Dim CC As CURRENCY_CURRENCY
Dim I As Long
C2L.ValueLo = 5381
For I = 1 To Len(StringToHash)
LSet CC = C2L
CC.Value = CC.Value * 33@ + CCur(AscW(Mid$(StringToHash, I, 1))) / 10000@
LSet C2L = CC
C2L.ValueHi = 0
Next I
djb2 = C2L.ValueLo
End Function
I found a workaround using Currency and doing the overflow math myself (and it turns out that VB6 Mod
operator also won't work on Currency
):
Const pow2_32 As Currency = 2@ ^ 32
Const signed32Max As Currency = 2@ ^ 31 - 1
Public Function simpleHash(hashString As String) As Long
Dim hash As Currency
Dim i As Long
hash = 5381
For i = 1 To Len(hashString)
hash = hash * 33@ + CCur(AscW(Mid$(hashString, i, 1)))
While hash >= pow2_32
hash = hash - pow2_32
Wend
Next i
If hash > signed32Max Then
hash = hash - pow2_32
End If
simpleHash = CLng(hash)
End Function
However, see Bob77's answer for a solution that is almost twice as fast as mine.