color long value to seperate R,G,B values

[DiGiTaIErRoR]

I need to convert a long value of a color to the relative R G B values mathematically, since I don't know how, I leave it to this board's capible 'hands'.

Thank you,

[kedaman]

R = Color mod 256
G = int(Color / 256) mod 256
B = int(Color / &h10000)

[parksie]

Private Type ColourRGB
    lRed As Integer
    lGreen As Integer
    lBlue As Integer
End Type
Private Sub Form_Load()
    Dim z As ColourRGB
    z = ExtractColour(RGB(&HAA, &HBB, &HCC))
    Debug.Print Hex(z.lRed)
    Debug.Print Hex(z.lGreen)
    Debug.Print Hex(z.lBlue)
End Sub
Private Function ExtractColour(lColour As Long) As ColourRGB
    Dim tempC As ColourRGB
    tempC.lBlue = (lColour \ &H10000) And &HFF
    tempC.lGreen = (lColour \ &H100) And &HFF
    tempC.lRed = lColour And &HFF
    ExtractColour = tempC
End Function

- depending on where your colour came from, you may need to swap the Red and Blue values around (endian problems).

[markman]

while everyone is on the subject of colors, how is it possible to save the custom colors in a choosecolor dialog box when the user creates them even if they dont select the color as they click ok?

[Guv]

I have not checked the following, but it should be close to correct.

Dim Red As Integer
Dim Green As Integer
Dim Blue As Integer
Dim Tlong As Long

Blue = RGBvalue Mod 256 'Blue is remainder after divide by 256
Tlong = (RGBvalue - Blue)/256 'Tlong now contains Red & Blue values.
Green = Tlong Mod 256
Red = (Tlong - Green)/256

I am not sure what VB does with arithmetic using Integer & Long variables, but I think it does sensible integer arithmetic.

[kedaman]

well markman, depends what choosecolor dialog box you are using?

There's also a way to get the RGB values as you know the color is stored in a long and each value represent a byte

Type LngColor
    Color As Long
End Type
Type RGBColor
    Red As Byte
    Green As Byte
    Blue As Byte
    Dummy As Byte
End Type


Dim a As LngColor, b As RGBColor
    a.Color = 123456
    LSet b = a
    With b
      Debug.Print .Red
      Debug.Print .Green
      Debug.Print .Blue
    End With

lRed = lColor Mod &H100
lGreen = Int(lColor / &H100) Mod &H100
lBlue = Int(lColor / &H10000) Mod &H100

[kedaman]

meg, you don't need to use modulus for lblue