Visual Basic

29-Apr-09 21:15

Hi, I only have VB6 and need to write a program to parse mpeg4-asp frames (each frame is a "bitstream" ie a stream of sequential bits flowing across byte/word/longword boundaries).

Unfortunately VB6 has no unsigned data types, nor native bit operations, making masking and shifting left/right very difficult as you have to write your own functions to take into account (test for and fudge) the sign bit (eg bit 31 of the bit range 31..0 for long). Since any given file could have cumulative bitstreams of 4Gb, imaging how much life gets sucked out of the PC when using a per-bit home-grown function call with all the right masking and signbit testing/fixing complexity...

Does anyone know of system calls (eg kernel32) or some such which makes bitstream operations easy (eg rotate/shift longword/word/byte left and right, integer value of an arbitrarily positioned set of bits, etc) ?

Assume XP-SP3, and latest Java runtime if that helps, but no .NET since an objective is to be able to run on .NET-free PCs.

Mycroft Holmes29-Apr-09 21:56

Mycroft Holmes

29-Apr-09 21:56

Don't be an idiot - get VB.Net or C# express and use it. IT COSTS NOTHING and you will get a lot more support from the forums.

VB6 is dead and no longer supported, do not waste your time starting with VB6

Never underestimate the power of human stupidity
RAH

halsboss29-Apr-09 22:07

29-Apr-09 22:07

Thanks, however did you actually read the post which stated an objective "but no .NET since an objective is to be able to run on .NET-free PCs" ? My only other choice is free Java with the netbeans IDE and I already have a VB6 framework in place...

You see how complex it can get just with just a couple of operations, from the code:

Option Explicit
Option Base 0
Option Compare Binary
Sub Main()
    ' Define and create an object for bitstream operations
    Dim BitStream As BitStreamVB6
    Set BitStream = New BitStreamVB6
    MsgBox "just created a new BitStreamVB6 object by Set variable BitStream "

    Dim b(0 To 3) As Byte
    Dim s As String
    Dim i32 As Long
    Dim i16 As Integer
    Dim i As Long
    Dim j As Long
    Dim k As Long
    Dim l As Long
    Dim z As Long
    Dim lw As Long
    Dim stmp As String
    lw = &H80000000   ' 2^31 is the sign bit ?
    s = ""
    For z = 0 To 32
        s = s & " right(&H" & Hex(lw) & "," & z & ")=&H" & Hex(BitStream.i32_ShiftRight(lw, z)) & vbCrLf
    Next z
    MsgBox s
    lw = &H1&         ' 2^31 is the sign bit ?
    s = ""
    For z = 0 To 32
        s = s & " left(&H" & Hex(lw) & "," & z & ")=&H" & Hex(BitStream.i32_ShiftLeft(lw, z)) & vbCrLf
    Next z
    MsgBox s
End Sub

and the object BitStreamVB6

Option Explicit
Option Base 0
Option Compare Binary

' LITTLE-ENDIAN (x86 architecture) considerations :
' http://vbcity.com/forums/topic.asp?tid=7027
' text values                           D   C   B   A
' Bytes                                 3   2   1   0
' Integer Maths - Most significant bit 31           0 Least Significant Bit
' Appearance of string when printed (backward to storage as appears above)= A B C D
'
'ie bits         {31 30 29 28 27 26 25 24} | {23 22 21 20 19 18 17 16} | {15 14 13 12 11 10 09 08} | {07 06 05 04 03 02 01 00}
'   text values   ---------- D ----------     ---------- C ----------     ---------- B ----------     ---------- A ----------
'   byte number              3                           2                           1                           0
'
' However the sign bit causes OVERFLOW PROBLEMS in VB6 when masking and checking,
' so you have to check for it specifically and treat it separately
' eg the sign bit for LONG    it is bit 31 - base zero
'                 for INTEGER it is bit 15 - base zero
'                 for BYTE    it is bit 07 - base zero
'---------------------------------------------------------------------------------------------------
'
'byte   little-endian sequential bytes in the byte
' 0     07 06 05 04 03 02 01 00
' 1     15 14 13 12 11 10 09 08
' 2     23 22 21 20 19 18 17 16
' 3     31 30 29 28 27 26 25 24
'
' When bytes read from a file, they are byte 0,1,2,3,4,5,6...
'
' So, for an mpeg4 bitstream read from a file byte by byte,
' IS the starting bit at the LEFT or the RIGHT of each byte ?
' Need to check the C code of mpeg4modifier for the answer...
'
'---------------------------------------------------------------------------------------------------
'
' Little-endian based bit masks.
'
' To check a bit is set are return a boolean True or False, use
'   IsSet = (i32 AND i32_bit_31) = i32_bit_31 ' however in VB6 this returns a 16 byte integer FFFF=true 0000=false
' To check a bit is set are return a single bit in little-endian bit 0 (see above)
'   BitValue = IIf(((i32 And i32_bit_31) = i32_bit_31), i32_bit_00, 0)
'
Const i32_bit_00 As Long = &H1&
Const i32_bit_01 As Long = &H2&
Const i32_bit_02 As Long = &H4&
Const i32_bit_03 As Long = &H8&
Const i32_bit_04 As Long = &H10&
Const i32_bit_05 As Long = &H20&
Const i32_bit_06 As Long = &H40&
Const i32_bit_07 As Long = &H80&
Const i32_bit_08 As Long = &H100&
Const i32_bit_09 As Long = &H200&
Const i32_bit_10 As Long = &H400&
Const i32_bit_11 As Long = &H800&
Const i32_bit_12 As Long = &H1000&
Const i32_bit_13 As Long = &H2000&
Const i32_bit_14 As Long = &H4000&
Const i32_bit_15 As Long = &H8000&
Const i32_bit_16 As Long = &H10000
Const i32_bit_17 As Long = &H20000
Const i32_bit_18 As Long = &H40000
Const i32_bit_19 As Long = &H80000
Const i32_bit_20 As Long = &H100000
Const i32_bit_21 As Long = &H200000
Const i32_bit_22 As Long = &H400000
Const i32_bit_23 As Long = &H800000
Const i32_bit_24 As Long = &H1000000
Const i32_bit_25 As Long = &H2000000
Const i32_bit_26 As Long = &H4000000
Const i32_bit_27 As Long = &H8000000
Const i32_bit_28 As Long = &H10000000
Const i32_bit_29 As Long = &H20000000
Const i32_bit_30 As Long = &H40000000
Const i32_bit_31 As Long = &H80000000

Const i16_bit_00 As Integer = &H1
Const i16_bit_01 As Integer = &H2
Const i16_bit_02 As Integer = &H4
Const i16_bit_03 As Integer = &H8
Const i16_bit_04 As Integer = &H10
Const i16_bit_05 As Integer = &H20
Const i16_bit_06 As Integer = &H40
Const i16_bit_07 As Integer = &H80
Const i16_bit_08 As Integer = &H100
Const i16_bit_09 As Integer = &H200
Const i16_bit_10 As Integer = &H400
Const i16_bit_11 As Integer = &H800
Const i16_bit_12 As Integer = &H1000
Const i16_bit_13 As Integer = &H2000
Const i16_bit_14 As Integer = &H4000
Const i16_bit_15 As Integer = &H8000

Const i08_bit_00 As Byte = &H1
Const i08_bit_01 As Byte = &H2
Const i08_bit_02 As Byte = &H4
Const i08_bit_03 As Byte = &H8
Const i08_bit_04 As Byte = &H10
Const i08_bit_05 As Byte = &H20
Const i08_bit_06 As Byte = &H40
Const i08_bit_07 As Byte = &H80

Const i32_mask_byte0 As Long = &HFF&        ' *** CAREFUL HERE - if you leave the trailing "&" off then it all falls to pieces !!!!!!!!!!!!!!!!
Const i32_mask_byte1 As Long = &HFF00&      ' *** CAREFUL HERE - if you leave the trailing "&" off then it all falls to pieces !!!!!!!!!!!!!!!!
Const i32_mask_byte2 As Long = &HFF0000
Const i32_mask_byte3_nosignbit As Long = &H7F000000    ' notice this excludes the sign bit at bit07 (bit31 if the long)

Const i16_mask_byte0 As Long = &HFF
Const i16_mask_byte1_nosignbit As Long = &H7F00         ' notice this excludes the sign bit at bit07 (bit15 if the int)

Const i08_mask_byte0_nosignbit As Long = &H7F         ' notice this excludes the sign bit at bit07 (bit07 of the byte)

Private i32_bitarray(0 To 31) As Long
Private i16_bitarray(0 To 15) As Integer
Private i08_bitarray(0 To 7) As Byte

Private tmp_long As Long
Private signbit_long As Long
Private tmp_int As Integer
Private signbit_int As Integer
Private tmp_byte As Byte
Private signbit_byte As Byte

Private Sub Class_Initialize()
' Initialise bitarrays ready for use in masking by (zero-based) "bit number"
' MsgBox " in Class_Initialize in BitStreamVB6"
i32_bitarray(0) = i32_bit_00
i32_bitarray(1) = i32_bit_01
i32_bitarray(2) = i32_bit_02
i32_bitarray(3) = i32_bit_03
i32_bitarray(4) = i32_bit_04
i32_bitarray(5) = i32_bit_05
i32_bitarray(6) = i32_bit_06
i32_bitarray(7) = i32_bit_07
i32_bitarray(8) = i32_bit_08
i32_bitarray(9) = i32_bit_09
i32_bitarray(10) = i32_bit_10
i32_bitarray(11) = i32_bit_11
i32_bitarray(12) = i32_bit_12
i32_bitarray(13) = i32_bit_13
i32_bitarray(14) = i32_bit_14
i32_bitarray(15) = i32_bit_15
i32_bitarray(16) = i32_bit_16
i32_bitarray(17) = i32_bit_17
i32_bitarray(18) = i32_bit_18
i32_bitarray(19) = i32_bit_19
i32_bitarray(20) = i32_bit_20
i32_bitarray(21) = i32_bit_21
i32_bitarray(22) = i32_bit_22
i32_bitarray(23) = i32_bit_23
i32_bitarray(24) = i32_bit_24
i32_bitarray(25) = i32_bit_25
i32_bitarray(26) = i32_bit_26
i32_bitarray(27) = i32_bit_27
i32_bitarray(28) = i32_bit_28
i32_bitarray(29) = i32_bit_29
i32_bitarray(30) = i32_bit_30
i32_bitarray(31) = i32_bit_31

i16_bitarray(0) = i16_bit_00
i16_bitarray(1) = i16_bit_01
i16_bitarray(2) = i16_bit_02
i16_bitarray(3) = i16_bit_03
i16_bitarray(4) = i16_bit_04
i16_bitarray(5) = i16_bit_05
i16_bitarray(6) = i16_bit_06
i16_bitarray(7) = i16_bit_07
i16_bitarray(8) = i16_bit_08
i16_bitarray(9) = i16_bit_09
i16_bitarray(10) = i16_bit_10
i16_bitarray(11) = i16_bit_11
i16_bitarray(12) = i16_bit_12
i16_bitarray(13) = i16_bit_13
i16_bitarray(14) = i16_bit_14
i16_bitarray(15) = i16_bit_15

i08_bitarray(0) = i08_bit_00
i08_bitarray(1) = i08_bit_01
i08_bitarray(2) = i08_bit_02
i08_bitarray(3) = i08_bit_03
i08_bitarray(4) = i08_bit_04
i08_bitarray(5) = i08_bit_05
i08_bitarray(6) = i08_bit_06
i08_bitarray(7) = i08_bit_07

'MsgBox "Class_Initialize: i32 masks[0..3]=" & Hex(i32_mask_byte0) & " " & Hex(i32_mask_byte1) & " " & Hex(i32_mask_byte2) & " " & Hex(i32_mask_byte3_nosignbit)
End Sub

Private Sub Class_Terminate()
' MsgBox " in Class_Terminate in BitStreamVB6"
End Sub

Sub i32_to_bytearray(ByRef src_i32 As Long, ByRef dst_i8() As Byte)
' Copy up to 4 bytes from an i32 into a byte array
    dst_i8(0) = (src_i32 And i32_mask_byte0)
    dst_i8(1) = (src_i32 And i32_mask_byte1) \ i32_bit_08
    dst_i8(2) = (src_i32 And i32_mask_byte2) \ i32_bit_16
    dst_i8(3) = (src_i32 And i32_mask_byte3_nosignbit) \ i32_bit_24           ' notice it is masked by 7F and not FF due to the sign bit issue
    If (src_i32 And i32_bit_31) = i32_bit_31 Then               ' sign bit was set, ensure bit 7 (base 0) is also set in the retrieved byte too
        dst_i8(3) = dst_i8(3) Or i08_bit_07
    End If
End Sub

Sub i32_to_string(ByRef src_i32 As Long, ByRef dst_ch As String)
' Copy 4 bytes from an i32 into a string
    Dim stringToReturn As String
    Dim ch1 As String * 1
    Dim longwordUpperByte As Long
    dst_ch = ""
    ch1 = Chr(src_i32 And i32_mask_byte0)
    dst_ch = dst_ch & ch1
    ch1 = Chr((src_i32 And i32_mask_byte1) \ i32_bit_08)
    dst_ch = dst_ch & ch1
    ch1 = Chr((src_i32 And i32_mask_byte2) \ i32_bit_16)
    dst_ch = dst_ch & ch1
    longwordUpperByte = (src_i32 And i32_mask_byte3_nosignbit) \ i32_bit_24   ' notice it is masked by 7F and not FF due to the sign bit issue
    If (src_i32 And i32_bit_31) = i32_bit_31 Then               ' sign bit was set, ensure bit 7 (base 0) is also set in the retrieved byte too
        longwordUpperByte = longwordUpperByte Or i32_bit_07
    End If
    ch1 = Chr(longwordUpperByte)
    dst_ch = dst_ch & ch1
End Sub

Sub i16_to_bytearray(ByRef src_i16 As Integer, ByRef dst_i8() As Byte)
' Copy 2 bytes from an i16 into a byte array
    dst_i8(0) = (src_i16 And i16_mask_byte0)
    dst_i8(1) = (src_i16 And i16_mask_byte1_nosignbit) \ i16_bit_08              ' notice it is masked by 7F and not FF due to the sign bit issue
    If (src_i16 And i16_bit_15) = i16_bit_15 Then               ' sign bit was set, ensure bit 7 (base 0) is also set in the retrieved byte too
        dst_i8(1) = dst_i8(1) Or i08_bit_07
    End If
End Sub

Sub i16_to_string(ByRef src_i16 As Integer, ByRef dst_ch As String)
' Copy 4 bytes from an i32 into a string
    Dim ch1 As String * 1
    Dim wordUpperByte As Integer
    dst_ch = ""
    ch1 = Chr(src_i16 And i16_mask_byte0)
    dst_ch = dst_ch & ch1
    wordUpperByte = (src_i16 And i16_mask_byte1_nosignbit) \ i16_bit_08           ' notice it is masked by 7F and not FF due to the sign bit issue
    If (src_i16 And i16_bit_15) = i16_bit_15 Then               ' sign bit was set, ensure bit 7 (base 0) is also set in the retrieved byte too
        wordUpperByte = wordUpperByte Or i16_bit_07
    End If
    ch1 = Chr(wordUpperByte)
    dst_ch = dst_ch & ch1
End Sub

Function i32_GetBit(ByRef src_i32 As Long, ByVal src_bitno As Long) As Long
    If src_bitno < 0 Or src_bitno > 31 Then
        Err.Raise 5, "i32_is_bit_set", "bit number not in range [0..31)"
    End If
    i32_GetBit = IIf(((src_i32 And i32_bitarray(src_bitno)) = i32_bitarray(src_bitno)), 1, 0) ' RETURN - IF 1 BIR CLEAR OR 1 IF BIT SET
End Function

Function i16_GetBit(ByRef src_i16 As Integer, ByVal src_bitno As Long) As Long
    If src_bitno < 0 Or src_bitno > 15 Then
        Err.Raise 5, "i16_is_bit_set", "bit number not in range [0..15)"
    End If
    i16_GetBit = IIf(((src_i16 And i16_bitarray(src_bitno)) = i16_bitarray(src_bitno)), 1, 0) ' RETURN - IF 1 BIR CLEAR OR 1 IF BIT SET
End Function

Function i08_GetBit(ByRef src_i08 As Byte, ByVal src_bitno As Long) As Long
    If src_bitno < 0 Or src_bitno > 7 Then
        Err.Raise 5, "i08_is_bit_set", "bit number not in range [0..7)"
    End If
    i08_GetBit = IIf(((src_i08 And i08_bitarray(src_bitno)) = i08_bitarray(src_bitno)), 1, 0) ' RETURN - IF 1 BIR CLEAR OR 1 IF BIT SET
End Function

Function i32_ShiftRight(ByRef src_i32 As Long, ByVal src_nobits As Long) As Long
    If src_nobits < 0 Or src_nobits > 32 Then
        Err.Raise 5, "i32_ShiftRight", "bit number not in range [0..32)"
    End If
    If src_nobits = 0 Then  ' no shifting, zero bitshift requested :)
        i32_ShiftRight = src_i32
        Exit Function
    End If
    If src_nobits = 32 Then ' full longword bitshift requested, hence will be zero :)
        i32_ShiftRight = &H0&
        Exit Function
    End If
    ' the damn sign bit stuffs things up
    signbit_long = src_i32 And i32_bit_31   ' save the sign bit, we need to move it separately
    If src_nobits = 31 Then ' sign bit requested via a shiftright by 31 bits, so handle the special case because VB6 overflows if you don't
        i32_ShiftRight = IIf((signbit_long = i32_bit_31), 1, 0)  ' RETURN - IF 1 BIR CLEAR OR 1 IF BIT SET
        Exit Function
    End If
    tmp_long = src_i32 And (Not i32_bit_31) ' make a variable excluding the sign bit
    tmp_long = tmp_long / (2 ^ src_nobits)       ' shift bits right by simple division
    'now put back the sign bit also shifted right by src_nobits bits
    If signbit_long = i32_bit_31 Then
        signbit_long = i32_bitarray(31 - src_nobits) ' use a lookup to do it instead of division because VB6 hates signs
    Else
        signbit_long = &H0&
    End If
    i32_ShiftRight = tmp_long Or signbit_long
End Function

Function i32_ShiftLeft(ByRef src_i32 As Long, ByVal src_nobits As Long) As Long
    Static bit0 As Long
    Static i As Long
    Static bitmask As Long
    Static topbit As Long
    Static aLong As Long
    Static no_saved_bits As Long
    If src_nobits < 0 Or src_nobits > 32 Then
        Err.Raise 5, "i32_ShiftLeft", "bit number not in range [0..32)"
    End If
    If src_nobits = 0 Then  ' no shifting, zero bitshift requested :)
        i32_ShiftLeft = src_i32
        Exit Function
    End If
    If src_nobits = 32 Then ' full longword bitshift requested, hence will be zero :)
        i32_ShiftLeft = &H0&
        Exit Function
    End If
    ' the damn sign bit stuffs things up/
    ' calc how many bits are to be saved (from bit0) and create a bitmask
    no_saved_bits = 31 - src_nobits ' remember this is zero-based so 0 result means there is 1 bit to be saved
    bitmask = 0
    For i = 0 To no_saved_bits - 1 ' no_saved_bits=0 means move 31 bits to the left, and this FOR LOOP gets skipped :)
        bitmask = bitmask Or i32_bitarray(i)
    Next i
    ' mask them in (ie mask out the bits to be lost), shift them left, and put in the sign bit
    aLong = src_i32 And bitmask
    aLong = aLong * (2& ^ src_nobits)
    topbit = IIf((src_i32 And i32_bitarray(no_saved_bits)) = i32_bitarray(no_saved_bits), i32_bit_31, 0)
    aLong = aLong Or topbit
    i32_ShiftLeft = aLong
End Function

Mycroft Holmes29-Apr-09 22:21

Mycroft Holmes

29-Apr-09 22:21

Sorry, like many here I get as far as seeing VB6 and stop reading. You are posting to a site that is VERY .net centric so I doudt you will get a satisfactory response.

Do you really expect someone to read through the code dump.

Never underestimate the power of human stupidity
RAH

halsboss30-Apr-09 1:20

30-Apr-09 1:20

OK thanks for clarifying that it's a .net-centric site. I'd hoped the forum was able to help, but never mind, it was worth a try. No not expected anyone to read the code other than prompting a bite (pun intentional) from a techo who'd be interested in something esoteric - which is the focus of the problem, really.

Oh well... probably means off to learn Java. Although I do recall a long time ago someone said something about inline assembler in VB6 Smile | :)

I'll Google it but don't hope for anything since long searching on VB6 bit operations yielded next to nothing.

Not aware of any "ordinary" Win32 calls available to do bitwise shift/rotate/compare operations ?

Dave Kreskowiak30-Apr-09 13:16

Dave Kreskowiak

30-Apr-09 13:16

halsboss wrote:
Not aware of any "ordinary" Win32 calls available to do bitwise shift/rotate/compare operations ?

There are none considering Win32 is written in C and Intel x86 Assembler and those languages handle all that stuff natively.

VB6 does not natively support inline assembler. It requires an IDE plugin or some other "hack" to get working. Google for "vb6 inline asm" for clues.

You'll have to write the functions to do it in VB6 yourself. Bitwise operations are just small math problems. Or, you could write equivilent functions in C and compile that to a .DLL which your VB6 code could import and use.

Oh, BTW, bitwise operations are also native in C# and VB.NET.

A guide to posting questions on CodeProject[^]

Dave Kreskowiak
Microsoft MVP
Visual Developer - Visual Basic
2006, 2007, 2008

CPallini30-Apr-09 1:48

30-Apr-09 1:48

Why don't you use C++, for instance?
Or cannot you build a C DLL for parsing frames and make it available to your VB6 app?

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

halsboss8-May-09 21:54

8-May-09 21:54

no C, stuck with VB6, have to code my own bitstream functions by the looks. damnation.

CPallini9-May-09 1:41

9-May-09 1:41

Why Cannot you use C the way I suggested (i.e. to make a DLL and then call it from the 'damned' VB6 app)?
Smile | :)

halsboss9-May-09 2:06

9-May-09 2:06

That's a really good suggestion. I don't have either of a compiler nor C skills. If I did then I'd jump at your suggestion. If you have some hints (links or something) on how to go about acquiring a free compiler, using it to create a DLL callable from VB6, and a place to learn C then I'm happy to give it a go.

Bitstream isn't rocket science, it's just very tedious in VB doing that with a byte array where the byte array goes from [0..n] and each bit goes (MSB->LSB) [7..0] but the bitstream bits come off in reverse from within the bytes eg bit7 1st as bitstream bit #1 then bit6 2nd as bitstream bit #2 etc... Now if VB had unsigned Longs and come good equivalence statement like the old fortran77 (showing my age, that's where I did most of my work) then it's be a little easier for me.

The challenge is to yield requested bits from the bitstream (up to 32 at a time) in a VB Long integer as a proper usable integer - a fair bit of byte masking and shifting in that. (eg request an arbitrary 30 bits starting anywhere in the bitstream to yield them in a VB long justified to become the integer value of those 30 bits).

Suggested code in any language welcomed Smile | :)

I'm not sure about copying Java code though, because that's a big-endian Longs system and I don't know the effect of that (if any).

CPallini9-May-09 4:06

9-May-09 4:06

There's Microsoft Visual C++ Express Edition [^], a free IDE for C/C++ developement!
Its wizard may create a skeleton DLL project for you.
Smile | :)

halsboss11-May-09 0:07

11-May-09 0:07

Strike me pink, that's good news ! I hope it doesn't require NET runtime for the resulting .DLL/.exe though ?

CPallini11-May-09 0:23

11-May-09 0:23

halsboss wrote:
I hope it doesn't require NET runtime for the resulting .DLL/.exe though ?

It doesn't: you may write 100% native code with it (you may also write managed C++, i.e. .NET code, if you like to, with it).
Smile | :)

halsboss11-May-09 3:15

11-May-09 3:15

Thankyou very much. Off to download and install it and see if I can generate a dll callable from VB6. I'll have to do some reading 'cause I don't know anything about objects let alone how to poke them into life from a C++ DLL from VB6 ... big learning curve Smile | :)

Oh, and does this microsoft link [^] mean that I just can't give the .exe and .DLL to someone ? ie I HAVE to install the runtime ? Kinda defeats the purpose since the VB6 runtime comes with XP an up or so I read. I might be better off hacking up very slow VB6 functions to manipulate bits, if it does need a C++ runtime installed.

how to find ascii value?

JC.KaNNaN29-Apr-09 20:35

JC.KaNNaN

29-Apr-09 20:35

hi
i am new developer in vvb.net.

wen ever i press f1 to f12 keys some window ill show????

Yours,
KaNNaN

-----------------------------------------------------------------
"Success is When Ur Signature Becomes An Autograph"

Mail To : foreverkans@gmail.com

_Damian S_29-Apr-09 20:53

_Damian S_

29-Apr-09 20:53

Google the keypress event... I found this when I did:

MSDN function you use[^]

There are plenty of code examples out there!

Knowledge is knowing that the tomato is a fruit. Wisdom is not putting it in fruit salad!!

Booger Mobile - Camp Quality esCarpade 2010

Baran M30-Apr-09 1:00

Baran M

30-Apr-09 1:00

Private Sub Form1_KeyDown(ByVal sender As Object, ByVal e As System.Windows.Forms.KeyEventArgs) Handles Me.KeyDown

If e.KeyCode = Keys.F1 Then
Form2.Show()
ElseIf e.KeyCode = Keys.F2 Then
Form3.Show()
End If

End Sub

Is that you have asked. Smile | :)

JC.KaNNaN3-May-09 23:58

JC.KaNNaN

3-May-09 23:58

thank u....

Yours,
KaNNaN

-----------------------------------------------------------------
"Success is When Ur Signature Becomes An Autograph"

Mail To : foreverkans@gmail.com