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我的程序陷入了死胡同。我在内存中有一个由 DIB 位图（否BITMAPFILEHEADER）的 RGB 值组成的简单数组。该数组是在 C++ 中生成的，但我尝试在 VB.NET 中显示它。我不想使用 GDI+，因为我需要原始速度。

这是我的代码（文件中的图像没有标题，宽度：1920 和高度：100,24 位，总大小 6220804）：

Dim bData As Byte()
Dim br As BinaryReader = New BinaryReader(File.OpenRead("img1.bmp"))
bData = br.ReadBytes(br.BaseStream.Length) 'no headers just raw data


Dim g As Graphics = Me.CreateGraphics() 'System.Drawing.Graphics.FromImage(bmp) 'or PictureBox1.CreateGraphics()
Dim hdc As IntPtr = g.GetHdc()

Dim bmi As New BITMAPINFO
bmi.bmiheader = New BITMAPINFOHEADER

'Now we fill up the bmi (Bitmap information variable) with all the necessary data
bmi.bmiheader.biSize = 40 'Size, in bytes, of the header (always 40)
bmi.bmiheader.biPlanes = 1 'Number of planes (always one)
bmi.bmiheader.biBitCount = 24 'Bits per pixel (always 24 for image processing)
bmi.bmiheader.biCompression = 0 'Compression: none or RLE (always zero)
bmi.bmiheader.biWidth = 1920
bmi.bmiheader.biHeight = 100
bmi.bmiheader.biSizeImage = 6220804

Dim memHDC As IntPtr = CreateCompatibleDC(hdc)

StretchDIBits(memHDC, 0, 0, 1920, 100, 0, 0, 1920, 100, bData, bmi, 0, 13369376)   ' Copy RGB values on an intermediary HDC
BitBlt(hdc, 0, 0, 1920, 100, memHDC, 0, 0, 13369376)    'Print directly from the memHDC

这是我的结构：

<StructLayout(LayoutKind.Sequential)>
Structure RGBQUAD
    Public rgbBlue As Byte
    Public rgbGreen As Byte
    Public rgbRed As Byte
    Public rgbReserved As Byte
End Structure

<StructLayout(LayoutKind.Sequential)>
Private Class BITMAPINFOHEADER
    Public biSize As Int32
    Public biWidth As Int32
    Public biHeight As Int32
    Public biPlanes As Int16
    Public biBitCount As Int16
    Public biCompression As Int32
    Public biSizeImage As Int32
    Public biXPelsPerMeter As Int32
    Public biYPelsPerMeter As Int32
    Public biClrUsed As Int32
    Public biClrImportant As Int32
End Class

<StructLayout(LayoutKind.Sequential)>
Private Structure BITMAPINFO
    Dim bmiheader As BITMAPINFOHEADER
    Dim bmiColors As RGBQUAD
End Structure

我测试了几乎所有可能的变量组合、HDC 和图形。没有任何作用！我哪里失败了？

注意：StretchDIBits 和 BitBlt 似乎成功了

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1 回答 1

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我找到了解决方案。我认为问题出在 CreateCompatibleDC 创建一个逐个像素网格的事实。由于这个限制，我只是StretchDIBits在图片框的 HDC 上使用：

Dim bData As Byte()
Dim br As BinaryReader = New BinaryReader(File.OpenRead("img1_arr.bmp"))
bData = br.ReadBytes(br.BaseStream.Length)

Dim g As Graphics = PictureBox1.CreateGraphics() 'or Me.CreateGraphics()
Dim dsthdc As IntPtr = g.GetHdc()

Dim bmi As New BITMAPINFO
bmi.bmiheader = New BITMAPINFOHEADER

'Now we fill up the bmi (Bitmap information variable) with all the necessary data
bmi.bmiheader.biSize = 40 'Size, in bytes, of the header (always 40)
bmi.bmiheader.biPlanes = 1 'Number of planes (always one)
bmi.bmiheader.biBitCount = 24 'Bits per pixel (always 24 for image processing)
bmi.bmiheader.biCompression = 0 'Compression: none or RLE (always zero)
bmi.bmiheader.biWidth = 1920
bmi.bmiheader.biHeight = 1080
bmi.bmiheader.biSizeImage = 6220804


StretchDIBits(dsthdc, 0, 0, 1920, 1080, 0, 0, 1920, 1080, bData, bmi, 0, SRCCOPY)

当然，该示例仅出于测试目的使用固定值。它完美无缺。

于 2019-04-04T22:21:19.867 回答