倾斜位图,RGB565 C#的步幅计算 [英] Slanted bitmap, stride calculation for RGB565 C#
问题描述
我的一些结果图像是倾斜的,有些则不是。
Some of my resulting images are slanted, some are not.
预期结果:(529x22)
Expected Result: (529x22)
实际结果:(529x22)
Actual Result: (529x22)
不介意不同的图像尺寸,这些都是截图。它们都是529x22。
Don't mind the different image sizes, these are screenshots. They are both 529x22.
我正在使用的代码,我刚从SO的问题答案得到了这个。
The code I am using, I just got this from an answer on a question here at SO.
// some other method
byte[] pixels = new byte[size - 16];
Array.Copy(this.data, offset, pixels, 0, pixels.Length);
this.ByteToImage(w, h, pixels);
// builds the pixels to a image
private Bitmap ByteToImage(int w, int h, byte[] pixels)
{
var bmp = new Bitmap(w, h, PixelFormat.Format16bppRgb565);
var BoundsRect = new Rectangle(0, 0, bmp.Width, bmp.Height);
BitmapData bmpData = bmp.LockBits(BoundsRect,
ImageLockMode.WriteOnly,
bmp.PixelFormat);
// bytes => not using this because it gives error
// eg. pixel.Length = 16032, bytes = 16064
int bytes = bmpData.Stride * bmp.Height;
Marshal.Copy(pixels, 0, bmpData.Scan0, pixels.Length);
bmp.UnlockBits(bmpData);
return bmp;
}
我很困惑因为有些工作正常,而不是倾斜。但其他人则倾斜。我错过了什么?
I'm confused because some works ok, not slanted. But others are slanted. What did I miss?
如评论和答案中所述,问题在于我如何计算步幅。我仍然对如何做到这一点感到困惑,但我试过这个:
As stated in the comments and answers, the problem is how I'm calculating stride. I'm still confused on how to do it but I tried this:
public static void RemovePadding(this Bitmap bitmap)
{
int bytesPerPixel = Image.GetPixelFormatSize(bitmap.PixelFormat) / 8;
BitmapData bitmapData = bitmap.LockBits(new Rectangle(0, 0, bitmap.Width, bitmap.Height), ImageLockMode.ReadOnly, bitmap.PixelFormat);
var pixels = new byte[bitmapData.Width * bitmapData.Height * bytesPerPixel];
for (int row = 0; row < bitmapData.Height; row++)
{
var dataBeginPointer = IntPtr.Add(bitmapData.Scan0, row * bitmapData.Stride);
Marshal.Copy(dataBeginPointer, pixels, row * bitmapData.Width * bytesPerPixel, bitmapData.Width * bytesPerPixel);
}
Marshal.Copy(pixels, 0, bitmapData.Scan0, pixels.Length);
bitmap.UnlockBits(bitmapData);
}
但结果是(更倾斜):
推荐答案
这似乎在这里工作:
private Bitmap ByteToImage(int w, int h, byte[] pixels)
{
var bmp = new Bitmap(w, h, PixelFormat.Format16bppRgb565);
byte bpp = 2;
var BoundsRect = new Rectangle(0, 0, bmp.Width, bmp.Height);
BitmapData bmpData = bmp.LockBits(BoundsRect,
ImageLockMode.WriteOnly,
bmp.PixelFormat);
// copy line by line:
for (int y = 0; y < h; y++ )
Marshal.Copy(pixels, y * w * bpp, bmpData.Scan0 + bmpData.Stride * y, w * bpp);
bmp.UnlockBits(bmpData);
return bmp;
}
我使用循环将每行数据放在正确的位置。数据不包括填充,但目标地址必须这样做。
I use a loop to place each row of data at the right spot. The data do not include the padding, but the target address must do so.
因此我们需要将数据访问乘以实际 width * bytePerPixel
但是目标地址 .aspx?f = 255& MSPPError = -2147217396rel =nofollow> Stride
,即扫描线的长度,填充到下一个倍数四个字节。对于 width = 300
, stride = 300
, width = 301
它是 stride = 304
..
Therefore we need to multiply the data access by the actual width * bytePerPixel
but the target adress by the Stride
, i.e. the length of the scanline, padded to the next multiple of four bytes. For width=300
it is stride=300
, for width=301
it is stride=304
..
一步移动所有像素数据只能在那里工作没有填充,即当宽度是 4
的倍数时。
Moving all pixel data in one step can only work when there is no padding, i.e. when the width is a multiple of 4
.
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