How Can I Create an Optimal Bitmap for 203 DPI in VB6?

How Can I Create an Optimal Bitmap for 203 DPI in VB6?

The sample code below uses a pre-set (custom) module width, sets the DPI and then calculates the total width of the symbol. The principle works for other printers, too.

The OCX method BCWidthHdc2 can be used to calculate the width of the barcode. If a module width has been set, use a dummy value for the [in] nWidth. If you calculate the width in unit=mm, convert it to Pixels for the SaveImage function. If you create a 2D code, use the BCHeightHd2 function for the height.

Dim widthPx As Long
Dim heightPx As Long
Dim heightMM As Double
TBarCode111.Dpi = 203
TBarCode111.SizeMode = eSizeMode_CustomModuleWidth  ' =1
TBarCode111.ModuleWidth = "250"
heightPx = 100
heightMM = heightPx * (25.4 / TBarCode111.Dpi)
widthPx = TBarCode111.BCWidthHdc2(0, 1000, heightMM * 1000, eMUPixel)
TBarCode111.SaveImage "c:\temp\test.bmp", eIMBmp, widthPx, heightPx, 203, 203