Alpha在C中混合2种RGBA颜色 [英] Alpha Blending 2 RGBA colors in C

问题描述

可能重复：

如何快速进行alpha混合？

什么alpha混合2种RGBA（整数）颜色的最快方法是什么？

What is the fastest way to alpha blend 2 RGBA (Integer) colors?

请注意，要混合的目标颜色始终是不透明的，只有第二种颜色可以具有不同的颜色

As a note, the target color where to blend is always opaque, only the second color can have different levels of transparency.

我正在尝试找到C语言中最快的方法，同时考虑到混合所产生的最终颜色必须最终不透明且完全不透明（alpha = 0xff）

I am trying to find the fastest way in C, taking into account that the final resulting color from the blend must end up with no transparency, fully opaque (alpha = 0xff)

推荐答案

int blend(unsigned char result[4], unsigned char fg[4], unsigned char bg[4])
{
    unsigned int alpha = fg[3] + 1;
    unsigned int inv_alpha = 256 - fg[3];
    result[0] = (unsigned char)((alpha * fg[0] + inv_alpha * bg[0]) >> 8);
    result[1] = (unsigned char)((alpha * fg[1] + inv_alpha * bg[1]) >> 8);
    result[2] = (unsigned char)((alpha * fg[2] + inv_alpha * bg[2]) >> 8);
    result[3] = 0xff;
}

我不知道它有多快，但是全都是整数。它通过将alpha（和inv_alpha）转换为8.8个定点表示而起作用。不必担心alpha的最小值为1的情况。在这种情况下，fg [3]为0，表示前景是透明的。混合将为1 * fg + 256 * bg，这意味着fg的所有位都将移出结果。

I don't know how fast it is, but it's all integer. It works by turning alpha (and inv_alpha) into 8.8 fixed-point representations. Don't worry about the fact that alpha's min value is 1. In that case, fg[3] was 0, meaning the foreground is transparent. The blends will be 1*fg + 256*bg, which means that all the bits of fg will be shifted out of the result.

您可以非常快地完成操作，实际上，如果将RGBA打包为64位整数。然后，您可以使用单个表达式并行计算所有三种结果颜色。

You could do it very fast, indeed, if you packed your RGBAs in 64 bit integers. You could then compute all three result colors in parallel with a single expression.

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