Public Function encrypt(ByVal message As Byte(), ByVal password As String) As Byte()
Dim passarr As Byte() = System.Text.Encoding.Default.GetBytes(password)
Randomize()
Dim rand As Integer = Int((255 - 0 + 1) * Rnd()) + 1
Dim outarr(message.Length) As Byte
Dim u As Integer
For i As Integer = 0 To message.Length - 1
outarr(i) += (message(i) Xor passarr(u)) Xor rand
If u = password.Length - 1 Then u = 0 Else u = u + 1
Next
outarr(message.Length) = 112 Xor rand
Return outarr
End Function
Questions I want to ask:
getbytes(password) its already declared as string ...why would you want to get bytes!!
what is the role of message here?
what are rand and outarr doing to message?
outarr(message.Length) = 112 Xor rand - I cannot understand this
The function is encrypting a string. Since it is performing arithmetic calculations on the message, you need a byte represenation (so you can do the arithmetics on the bytes).
outarr is the array of bytes that is the result of the encryption. The encryption is based on a random value rand, which is an integer between 0 and 255.
The 112 xor rand part, I believe, is there to be able to get the random number back from the encrypted code (by knowing the "secret" value 112)
Related
I'm trying to make a function that takes a three digit number and reverses it (543 into 345)
I can't take that value from a TextBox because I need it to use the three numbers trick to find a value.
RVal = ReverseDigits(Val)
Diff = Val - RVal
RDiff = ReverseDigits(Diff)
OVal = Diff + RDiff
543-345=198
198+891=1089
Then it puts 1089 in a TextBox
Function ReverseDigits(ByVal Value As Integer) As Integer
' Take input as abc
' Output is (c * 100 + b * 10 + a) = cba
Dim ReturnValue As Boolean = True
Dim Val As String = CStr(InputTextBox.Text)
Dim a As Char = Val(0)
Dim b As Char = Val(1)
Dim c As Char = Val(2)
Value = (c * 100) + (b * 10) + (a)
Return ReturnValue
End Function
I've tried this but can't figure out why it won't work.
You can convert the integer to a string, reverse the string, then convert back to an integer. You may want to enforce the three digit requirement. You can validate the argument before attempting conversion
Public Function ReverseDigits(value As Integer) As Integer
If Not (value > 99 AndAlso value < 1000) Then Throw New ArgumentException("value")
Return Integer.Parse(New String(value.ToString().Reverse().ToArray()))
End Function
My code is pretty simple and will also work for numbers that don't have three digits assuming you remove that validation. To see what's wrong with your code, there are a couple of things. See the commented lines which I changed. The main issue is using Val as a variable name, then trying to index the string like Val(0). Val is a built in function to vb.net and the compiler may interpret Val(0) as a function instead of indexing a string.
Function ReverseDigits(ByVal Value As Integer) As Integer
' Dim ReturnValue As Boolean = True
' Dim Val As String = CStr(InputTextBox.Text)
Dim s As String = CStr(Value)
Dim a As Char = s(0)
Dim b As Char = s(1)
Dim c As Char = s(2)
Value = Val(c) * 100 + Val(b) * 10 + Val(a)
'Return ReturnValue
Return Value
End Function
(Or the reduced version of your function, but I would still not hard-code the indices because it's limiting your function from expanding to more or less than 3 digits)
Public Function ReverseDigits(Value As Integer) As Integer
Dim s = CStr(Value)
Return 100 * Val(s(2)) + 10 * Val(s(1)) + Val(s(0))
End Function
And you could call the function like this
Dim inputString = InputTextBox.Text
Dim inputNumber = Integer.Parse(inputString)
Dim reversedNumber = ReverseDigits(inputNumber)
Bonus: If you really want to use use math to find the reversed number, here is a version which works for any number of digits
Public Function ReverseDigits(value As Integer) As Integer
Dim s = CStr(value)
Dim result As Integer
For i = 0 To Len(s) - 1
result += CInt(Val(s(i)) * (10 ^ i))
Next
Return result
End Function
Here's a method I wrote recently when someone else posted basically the same question elsewhere, probably doing the same homework:
Private Function ReverseNumber(input As Integer) As Integer
Dim output = 0
Do Until input = 0
output = output * 10 + input Mod 10
input = input \ 10
Loop
Return output
End Function
That will work on a number of any length.
With reference to this link
Calculate CRC32 of an String or Byte Array
I modified the code in order to calculate the CRC16 instead of CRC32, however I am getting wrong result, can some one point me where is the mistake?
Private Sub Main()
Crc16.ComputeChecksum(Encoding.UTF8.GetBytes("Some string"))
End Sub
Public Class CRC16
Shared table As UShort()
Shared Sub New()
Dim poly As UShort = &HA001US 'calculates CRC-16 using A001 polynomial (modbus)
table = New UShort(255) {}
Dim temp As UShort = 0
For i As UShort = 0 To table.Length - 1
temp = i
For j As Integer = 8 To 1 Step -1
If (temp And 1) = 1 Then
temp = CUShort((temp >> 1) Xor poly)
Else
temp >>= 1
End If
Next
table(i) = temp
Next
End Sub
Public Shared Function ComputeChecksum(ByVal bytes As Byte()) As UShort
Dim crc As UShort = &H0US ' The calculation start with 0x00
For i As Integer = 0 To bytes.Length - 1
Dim index As Byte = CByte(((crc) And &HFF) Xor bytes(i))
crc = CUShort((crc >> 8) Xor table(index))
Next
Return Not crc
End Function
End Class
Try this, it's working VB6 code for Instrument control. (sCommand is a temp string which contains all Bytes, Result is added to sCommand, Modbus is using LSB first, TextToString and StringToAscii are functions to convert a readable string "FF EE" into ASCII and back, thus they are not of interest here.):
Private Sub cmdCRC16_Click()
Dim sCommand As String
Dim x As Long
Dim y As Long
Dim lCRC As Long
sCommand = TextToString(txtASCII)
'Initial value
lCRC = 65535 '(&HFFFF results in Integer -1)
For x = 1 To Len(sCommand)
lCRC = lCRC Xor Asc(Mid(sCommand, x, 1))
For y = 1 To 8
If (lCRC Mod 2) > 0 Then
lCRC = (lCRC And 65534) / 2
lCRC = lCRC Xor 40961 '(&HA001 results in whatever negative integer)
Else
lCRC = (lCRC And 65534) / 2
End If
Next y
Next x
'Add CRC with LSB first
sCommand = sCommand + Chr(lCRC And 255)
sCommand = sCommand + Chr((lCRC And 65280) / 256)
txtASCII = StringToASCII(sCommand)
End Sub
I just came accross the same issue. Simple solution is to omit negation at the end, so just change your "Return Not crc" to "Return crc" and you be fine.
There are various variants of CRC-16, where "CRC-16" normally refers to the IBM variant, also called "ARC". It uses an XorOut value of zero. See Catalogue of parametrised CRC algorithms with 16 bits.
I've got a password generator in one of my scripts which generates 2 passwords and writes them to a database.
These passwords are later retrieved and set as the windows logon passwords for 2 users which are automatically created in a virtual machine.
I have a problem whereby my passwords are not always considered to be complex as far as windows is concerned.
I am using the following function to generate my passwords...
Public Function GeneratePassword(ByVal len As Integer) As String
Dim str As String = "1234567890qwertyuio1pa2sd3fg4hj5kl6zx7cv8bn9mQ0WE1RT2YU3IO4PA5SD6FG7HJ8KL9ZX0CVBNM"
Dim N As Integer = str.Length
Dim rnd As New Random((Now.Hour * 3600 + Now.Minute * 60 +
Now.Second) * 1000 + Now.Millisecond)
Dim sb As New StringBuilder
For l As Integer = 1 To len
sb.Append(str.Substring(rnd.Next(0, N), 1))
Next
Return sb.ToString
End Function
Which is then called like this...
Dim vm_password1 As String = GeneratePassword(10)
How would i best go about modifying this code to ensure that each password generated contains 1 uppercase character, 1 lowercase character and one number?
Any help appreciated! :)
Of course you can extend this further, but here's the jist
Public Function fnGenPw() As String
Dim RandomClass As New Random()
Const legalAlpha As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
Const legalNums As String = "1234567890"
Dim strbNewpw As New StringBuilder
For i As Integer = 0 To 2 'first 3 characters are alpha
Dim RandomNumber As Integer = RandomClass.Next(legalAlpha.Length - 1)
strbNewpw.Append(legalAlpha.Chars(RandomNumber))
Next
For i As Integer = 3 To 5 'characters 4,5,6 are numeric
Dim RandomNumber As Integer = RandomClass.Next(legalNums.Length - 1)
strbNewpw.Append(legalNums.Chars(RandomNumber))
Next
For i As Integer = 6 To 9 'characters 7,8,9,10 are alpha
Dim RandomNumber As Integer = RandomClass.Next(legalAlpha.Length - 1)
strbNewpw.Append(legalAlpha.Chars(RandomNumber))
Next
Return strbNewpw.ToString
End Function
I have a VB.Net application that goes through a series of processes to decode a string, one problem with this is that I have found a function that converts from binary to decimal for any number, but I cannot find a function that will convert a supplied number (in string format) into a binary string. For reference, the binary to decimal conversion function is below:
Public Function baseconv(d As String)
Dim N As Long
Dim Res As Long
For N = Len(d) To 1 Step -1
Res = Res + ((2 ^ (Len(d) - N)) * CLng(Mid(d, N, 1)))
Next N
Return Str(Res)
End Function
What i would do if the number is less than 16*1e18 <-> can hold in a Uint64 :
1) store the number inside an unsigned 64 bits integer : num.
2) then just loop on each bit to test it :
mask = 1<<63
do
if ( num AND mask ) then ->> push("1")
else ->> push("0")
mask = mask >> 1
until mask = 0
(where push builds the output with a string concatenation, or, if performance matters, a StringBuilder, or it can be a stream,... )
what about this?
Module Module1
Sub Main()
Console.WriteLine(Convert.ToString(2253483438943167 * 5, 2))
Console.ReadKey()
End Sub
End Module
Try using the System.Numerics.BigInteger class like this:
Dim d As String
d = "2253483438943167"
Dim bi As BigInteger
bi = BigInteger.Parse(d)
Dim ba() As Byte
ba = bi.ToByteArray()
Dim s As String
Dim N As Long
Dim pad as Char
pad = "0"c
For N = Len(ba) To 1 Step -1
s = s + Convert.ToString(ba(N)).PadLeft(8, pad)
Next N
How about
Dim foo As BigInteger = Long.MaxValue
foo += Long.MaxValue
foo += 2
Dim s As New System.Text.StringBuilder
For Each b As Byte In foo.ToByteArray.Reverse
s.Append(Convert.ToString(b, 2).PadLeft(8, "0"c))
Next
Debug.WriteLine(s.ToString.TrimStart("0"c))
'10000000000000000000000000000000000000000000000000000000000000000
I'm trying to convert some Java code to Excel and the required hashcode function generates an overflow error, instead of wrapping to the negative
Function FnGetStringHashCode(ByVal str As String) As Integer
Dim result, i
FnGetStringHashCode = 17
For i = 1 To Len(str)
Dim c, a
c = Mid(str, i, 1)
a = AscW(c)
FnGetStringHashCode = 31 * FnGetStringHashCode + a
Next i
End Function
Is there a way of doing this in Excel VBA?
Although there is no built-in way to do this, the computation is simple:
Public Function coerceLongToInt(toCoerce As Long) As Integer
Const MIN_INT As Long = -32768
Const MAX_INT As Long = 32767
Const NUM_INTS As Long = MAX_INT - MIN_INT + 1
Dim remainder As Long
remainder = toCoerce Mod NUM_INTS
If remainder > MAX_INT Then
coerceLongToInt = remainder - NUM_INTS
ElseIf remainder < MIN_INT Then
coerceLongToInt = remainder + NUM_INTS
Else
coerceLongToInt = remainder
End If
End Function
This is the behavior you want, right?
?coerceLongToInt(-32769)
32767
?coerceLongToInt(-32768)
-32768
?coerceLongToInt(-1)
-1
?coerceLongToInt(0)
0
?coerceLongToInt(1)
1
?coerceLongToInt(32767)
32767
?coerceLongToInt(32768)
-32768
You would use it like this:
Function FnGetStringHashCode(ByVal str As String) As Integer
Dim result, i
FnGetStringHashCode = 17
For i = 1 To Len(str)
Dim c, a
c = Mid(str, i, 1)
a = AscW(c)
FnGetStringHashCode = coerceLongToInt(31 * CLng(FnGetStringHashCode) + a)
Next i
End Function
You need the 'CLng' call in there to prevent VBA from raising an overflow error when it computes the intermediate value (31 * [some integer >= 1058]).
I have modified a little the script of ours. The main difference is returning type of your function. Now it returns Variant. As decimal is a subset of Variant, and it can store bigger numbers than long I think it is a good solution (see VBA data types) - I do not know is it possible to explicitly return Decimal. Here is the script
Function FnGetStringHashCode(ByVal str As String) As Variant
Dim tmp As Variant, c As String, a As Integer, i As Integer
tmp = 17
For i = 1 To Len(str)
c = Mid$(str, i, 1)
a = AscW(c)
tmp = 31 * tmp + a
Next i
FnGetStringHashCode = tmp
End Function
And a little test routine
Sub test()
Debug.Print CStr(FnGetStringHashCode("dawdaedae"))
End Sub