为什么BitmapSource.Create引发ArgumentException？ [英] Why does BitmapSource.Create throw an ArgumentException?

问题描述

我想从原始数据创建了一个位图WPF中显示，通过使用图像和的BitmapSource：

I'm trying to get an bitmap created from raw data to show in WPF, by using an Image and a BitmapSource:

Int32[] data = new Int32[RenderHeight * RenderWidth];

for (Int32 i = 0; i < RenderHeight; i++)
{
    for (Int32 j = 0; j < RenderWidth; j++)
    {
        Int32 index = j + (i * RenderHeight);

        if (i + j % 2 == 0)
            data[index] = 0xFF0000;
        else
            data[index] = 0x00FF00;
    }
}

BitmapSource source = BitmapSource.Create(RenderWidth, RenderHeight, 96.0, 96.0, PixelFormats.Bgr32, null, data, 0);

RenderImage.Source = source;

然而，呼吁BitmapSource.Create引发ArgumentException，说：值没有在预期的范围内。这难道不是这样做的方式吗？我不能作出这一呼吁是否正确？

However the call to BitmapSource.Create throws an ArgumentException, saying "Value does not fall within the expected range". Is this not the way to do this? Am I not making that call properly?

推荐答案

您步幅不正确。步幅被分配用于所述一个扫描线的字节数
位图。这样，使用以下命令：

Your stride is incorrect. Stride is the number of bytes allocated for one scanline of the bitmap. Thus, use the following:

int stride = ((RenderWidth * 32 + 31) & ~31) / 8;

和替换最后一个参数（目前为 0 ）使用步幅如上定义。

and replace the last parameter (currently 0) with stride as defined above.

下面是神秘的步幅公式的解释：

Here is an explanation for the mysterious stride formula:

事实：Scanlines的必须在32位边界（参考）对齐

Fact: Scanlines must be aligned on 32-bit boundaries (reference).

每扫描线的字节数天真的公式将是：

The naive formula for the number of bytes per scanline would be:

(width * bpp) / 8

但是，这可能不会给我们一个32位的边界上对齐位图（宽* BPP）甚至可能没有被8整除了

But this might not give us a bitmap aligned on a 32-bit boundary and (width * bpp) might not even have been divisible by 8.

所以，我们做的是我们强迫我们的位图在一排至少有32位（我们假设宽度大于0 ）

So, what we do is we force our bitmap to have at least 32 bits in a row (we assume that width > 0):

width * bpp + 31

，然后我们说我们不关心的低位（位0--4），因为我们正试图在32位边界对齐：

and then we say that we don't care about the low-order bits (bits 0--4) because we are trying to align on 32-bit boundaries:

(width * bpp + 31) & ~31

和再除以8要回字节：

((width * bpp + 31) & ~31) / 8

的填充可以通过

int padding = stride - (((width * bpp) + 7) / 8)

天真的公式将是

stride - ((width * bpp) / 8)

但宽* BPP 可能不是一个字节边界上对齐，当它没有这个公式将在未来数由一个字节填充。 （想想使用1 BPP一个1个像素宽位图，跨距为4，天真的公式会说，填充是4，但在现实中却是3），所以我们加一点点地覆盖<$ C $的情况下C>宽* BPP 不是一个字节边界，然后我们得到上面给出正确的公式。

But width * bpp might not align on a byte boundary and when it doesn't this formula would over count the padding by a byte. (Think of a 1 pixel wide bitmap using 1 bpp. The stride is 4 and the naive formula would say that the padding is 4 but in reality it is 3.) So we add a little bit to cover the case that width * bpp is not a byte boundary and then we get the correct formula given above.

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