Android Question small function can not convert to B4A

[Sinan Tuzcu]

Hi,

I have a function or a sub in VB.NET but can not convert in B4A.
Can anyboady help

VB.NET

  Private Function Send(ByRef Klartext As String) As Boolean
        Dim idx As Byte
        Dim tmpByte As Byte
        Dim tmpChar As String
        Dim Textsammeln As String = ""

        For idx = 1 To CByte(Len(Klartext.Trim))
            tmpChar = Mid(Klartext.Trim, idx, 1)
            tmpByte = CByte(Asc(tmpChar) + 3)
            tmpChar = Chr(tmpByte)
            Mid(Klartext, idx, 1) = tmpChar
            Textsammeln = Trim(Klartext)
        Next idx

        Return Senden(Textsammeln)
    End Function

B4A
This I have try it:

For idx = 1 To Klartext.Trim.Length
    tmpChar = Klartext.SubString2(idx, 1).Trim
    tmpByte = Asc(tmpChar) + 3
    tmpChar = Chr(tmpByte)
    Klartext.SubString2(idx, 1) = tmpChar
    Textsammeln =  Textsammeln & tmpChar.Trim
Next
    Return Send(Textsammeln)
 End Sub

get compiler error regarding Substring2

 

[DonManfred]

get compiler error regarding Substring2

Substring2 needs 2 parameters. BeginIndex and EndIndex.
EndIndex must be higher than BeginIndex!

 

[Sinan Tuzcu]

in this row, I get unecpected type

Klartext.SubString(idx) = tmpChar

but why?

 

[Brian Dean]

Substring cannot appear on the left side of an equation.

 

[OliverA]

Try

    Dim idx As Int
    Dim tmpByte As Byte
    Dim tmpChar As String
    
    Dim sb As StringBuilder
    sb.Initialize
    For idx = 1 To Klartext.Trim.Length
        tmpChar = Klartext.SubString2(idx, 1)
        tmpByte = Asc(tmpChar) + 3
        sb.Append(Chr(tmpByte))
    Next
    Return Send(sb.ToString)

BTW, reading up on CByte (https://www.oreilly.com/library/view/vbnet-language-in/0596003080/re32.html), your original code makes a lot of assumptions:
1) The length of Klartext is between 0 to 255 (inclusive) characters. If it is over 255 characters, CByte would bomb out
2) The unicode value returned by the first Mid call is between 0 to 252 (inclusive), otherwise, the next call to CByte is going to bomb out
Note: The above two points are only valid if the linked documentation is valid. I'm not a VB.Net user

As to the new code you tried:

tmpChar = Klartext.SubString2(idx, 1).Trim

does not equal

tmpChar = Mid(Klartext.Trim, idx, 1)

You are calling Trim AFTER extracting the substring in the new code. That could lead to a slight bug in cases where white spaces are located inside the string (not at the end)

Strings in B4X are immutable (https://www.javatpoint.com/immutable-string). That's why

Klartext.SubString2(idx, 1) = tmpChar

does not work and why you should not build strings via & (since you are creating a new string with each &, not appending to an existing string), but via StringBuilder

 

[Geezer]

If i remember correctly, the substring and substring2 start with 0 ... so 0 to length - 1

 

[Sinan Tuzcu]

Try

    Dim idx As Int
    Dim tmpByte As Byte
    Dim tmpChar As String
   
    Dim sb As StringBuilder
    sb.Initialize
    For idx = 1 To Klartext.Trim.Length
        tmpChar = Klartext.SubString2(idx, 1)
        tmpByte = Asc(tmpChar) + 3
        sb.Append(Chr(tmpByte))
    Next
    Return Send(sb.ToString)

BTW, reading up on CByte (https://www.oreilly.com/library/view/vbnet-language-in/0596003080/re32.html), your original code makes a lot of assumptions:
1) The length of Klartext is between 0 to 255 (inclusive) characters. If it is over 255 characters, CByte would bomb out
2) The unicode value returned by the first Mid call is between 0 to 252 (inclusive), otherwise, the next call to CByte is going to bomb out
Note: The above two points are only valid if the linked documentation is valid. I'm not a VB.Net user

As to the new code you tried:

tmpChar = Klartext.SubString2(idx, 1).Trim

does not equal

tmpChar = Mid(Klartext.Trim, idx, 1)

You are calling Trim AFTER extracting the substring in the new code. That could lead to a slight bug in cases where white spaces are located inside the string (not at the end)

Strings in B4X are immutable (https://www.javatpoint.com/immutable-string). That's why

Klartext.SubString2(idx, 1) = tmpChar

does not work and why you should not build strings via & (since you are creating a new string with each &, not appending to an existing string), but via StringBuilder

Thanks,

but the code in VB.NET works very well

 Private Function Send(ByRef Klartext As String) As Boolean
        Dim idx As Byte
        Dim tmpByte As Byte
        Dim tmpChar As String
        Dim Textsammeln As String = ""

        For idx = 1 To CByte(Len(Klartext.Trim))
            tmpChar = Mid(Klartext.Trim, idx, 1)
            tmpByte = CByte(Asc(tmpChar) - 4)
            tmpChar = Chr(tmpByte)
            Mid(Klartext, idx, 1) = tmpChar
            Textsammeln = Trim(Klartext)
        Next idx

        Return Senden(Textsammeln)
    End Function

and your posted code is not equal with this.

 

[OliverA]

your posted code is not equal with this

Then you need to provide us with some inputs and expected outputs

 

[Sinan Tuzcu]

Hi,

if I send this so then it works very well

'"1|255|" & CRLF <-- each charecter are minus 4
Send(Chr(45) & Chr(120) & Chr(46) & Chr(49) & Chr(49) & Chr(120)  & CRLF)


private Sub Send(msg As String)
    Dim Buffer() As Byte
    msg = msg & CRLF
  
    Buffer = msg.GetBytes("Windows-1252") 

    AStreams.Write2(Buffer,0,Buffer.Length)
End Sub

but if I do this with this code, then not works

SendComando("1|" & "255"  & "|" & CRLF)

Private Sub SendComando(Klartext As String)
     Dim idx As Int
    Dim tmpByte As Byte
    Dim tmpChar As String
  
    Dim sb As StringBuilder
    sb.Initialize
    For idx = 0 To Klartext.Trim.Length-1
        tmpChar = Klartext.trim.SubString(idx)
        tmpByte = Asc(tmpChar.Trim) - 4
        sb.Append(Chr(tmpByte))
    Next
    Return Send(sb.ToString.Trim)

End Sub

private Sub Send(msg As String)
    Dim Buffer() As Byte
    msg = msg & CRLF
  
    Buffer = msg.GetBytes("Windows-1252") 

    AStreams.Write2(Buffer,0,Buffer.Length)
End Sub

 

[OliverA]

Send(Chr(45) & Chr(120) & Chr(46) & Chr(49) & Chr(49) & Chr(120) & CRLF)

The above <> the below

SendComando("1|" & "255" & "|" & CRLF)

In the first quoted code, you did not change the CRLF values
But in the second quoted code, the SendComando will change the CRLF (by subtracting 4 from CR and 4 from LF)
Note: Other issues may exist, but this is the first one that stand out without looking at the code

 

[OliverA]

Private Sub SendComando(Klartext As String)
    Dim idx As Int
    Dim tmpByte As Byte
    Dim tmpChar As String
  
    Dim sb As StringBuilder
    sb.Initialize
    For idx = 0 To Klartext.Trim.Length-1
        tmpChar = Klartext.trim.SubString(idx)
        tmpByte = Asc(tmpChar.Trim) - 4
        Log(tmpByte)
        sb.Append(Chr(tmpByte))
    Next
    Send(sb.ToString.Trim)

End Sub

private Sub Send(msg As String)
    Dim Buffer() As Byte
    msg = msg & CRLF
  
    Buffer = msg.GetBytes("Windows-1252")
    Log("--Buffer--")
    For x = 0 To Buffer.Length - 1
        Log(Buffer(x))
    Next
End Sub

Using the above code and calling it with

SendComando("1|" &  "255" & "|")

Gives me

45
120
46
49
49
120
--Buffer--
45
120
46
49
49
120
10

So to me it looks like the conversion is correct. The only thing that may not produce what you expect is the CRLF. In my case, it only produces an LF. If you always want to add CR and LF to your message, then use

private Sub Send(msg As String)
    Dim Buffer() As Byte
    msg = $"${msg}${Chr(13)}${Chr(10)}"$ 'Start using Smart String Literals: https://www.b4x.com/android/forum/threads/b4x-smart-string-literal.50135/
  
    Buffer = msg.GetBytes("Windows-1252")
    Log("--Buffer--")
    For x = 0 To Buffer.Length - 1
        Log(Buffer(x))
    Next
End Sub

And then the output changes to

45
120
46
49
49
120
--Buffer--
45
120
46
49
49
120
13
10