2

我正在使用Scintilla（不是ScintillaNET），我正在尝试设置自动完成列表中使用的图像。Scintilla 需要 RGBA 格式的字节/像素字符串。

Scintilla 对此 RGBA 格式的描述：（引用自文档）

RGBA 格式允许半透明，每个像素都有一个 alpha 值。它比 XPM 更简单，功能更强大。

数据是一个 4 字节像素值序列，从顶行的像素开始，最左边的像素在前，然后是后续行的像素。由于对齐原因，线之间没有间隙。

每个像素依次由​​一个红色字节、一个绿色字节、一个蓝色字节和一个阿尔法字节组成。颜色字节不预乘 alpha 值。也就是说，25% 不透明的全红色像素将是 [FF, 00, 00, 3F]

我希望我误解了上面解释的格式，文档不是很清楚。

我的转换方法：

我编写了这个函数来将 PNG 转换为这个字节串：

public static string ConvertFromPNG(Bitmap PNG)
{
    string rgba = "";

    int pixWidth = PNG.Width;
    int pixHeight = PNG.Height;

    for (var y = 0; y < pixHeight; y++)
    {
        for (var x = 0; x < pixWidth; x++)
        {
            Color pix = PNG.GetPixel(x, y);

            rgba += pix.R.ToString("X2") + pix.G.ToString("X2") + pix.B.ToString("X2") + pix.A.ToString("X2");
        }
    }

    return rgba;
}

结果图像：

但 Scintilla 只是将图像显示为灰色框：

自动完成框.

有问题的图像是Microsoft 的 Babel 图像之一的副本：通天塔图标.

我知道这组字节是正确的，因为有趣的是，如果我将它们格式化并缩小它们，我可以看到图像的轮廓：

字节（在图像编辑器中缩小）

生成的字节数：

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
4

2 回答 2

2

您将字节格式化为十六进制值并将它们放在一个字符串中。Scintilla 只想要字节本身。

我怀疑您想创建一个字节数组而不是字符串，并将 , , 和 的值直接存储到其中pix.R，pix.G而pix.B不是pix.A将它们格式化为字符串。

于 2013-03-12T16:16:49.040 回答
0

接受 JasonD 的回答后，我更新了我的代码并且它正在工作。如果它对其他人有帮助，这里是工作代码：

public static byte[] ConvertFromPNG(Bitmap PNG)
{
    byte[] rgbaB = new byte[4 * (PNG.Width * PNG.Height)];

    int i = 0;

    for (var y = 0; y < PNG.Height; y++)
    {
        for (var x = 0; x < PNG.Width; x++)
        {
            Color pix = PNG.GetPixel(x, y);

            rgbaB[i++] = pix.R;
            rgbaB[i++] = pix.G;
            rgbaB[i++] = pix.B;
            rgbaB[i++] = pix.A;
        }
    }

    return rgbaB;
}
于 2013-03-12T17:58:32.013 回答