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Setting Pixel Color in VB 2005I am using VB 2005 and want to generate a graphic in a picture box control
where the color of each pixel is determined by particular values calculated for each pixel. These values are manipulated in such a way that I come up with three values (each in the range 0-155) and I want to use these for the RGB values of the color of the associated pixel. My question is: In VB 2005 how do I set a particular pixel to a color with a given set of RGB values? In VB 6 I used the PSET function but it is not available in VB 2005. Thanks. "fripper" <yo***@indiana.edu> schrieb: Create a new 'Bitmap' object of appropriate size. Then you can use its >I am using VB 2005 and want to generate a graphic in a picture box control >where the color of each pixel is determined by particular values calculated >for each pixel. These values are manipulated in such a way that I come up >with three values (each in the range 0-155) and I want to use these for the >RGB values of the color of the associated pixel. My question is: In VB >2005 how do I set a particular pixel to a color with a given set of RGB >values? In VB 6 I used the PSET function but it is not available in VB >2005. 'SetPixel' method to change the color of a certain pixel. The bitmap object can be assigned to a picturebox' 'Image' property. -- M S Herfried K. Wagner M V P <URL:http://dotnet.mvps.org/> V B <URL:http://classicvb.org/petition/> fripper wrote:
> I am using VB 2005 and want to generate a graphic in a picture box control Dim C as Color = Color.FromArgb(IntAlpha, IntRed, IntGreen, IntBlue)> where the color of each pixel is determined by particular values calculated > for each pixel. These values are manipulated in such a way that I come up > with three values (each in the range 0-155) and I want to use these for the > RGB values of the color of the associated pixel. My question is: In VB > 2005 how do I set a particular pixel to a color with a given set of RGB > values? In VB 6 I used the PSET function but it is not available in VB > 2005. > > Thanks. > > Chris
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