根据GLSL中向量的特定分量进行最小-最大的最快方法? [英] Fastest way to do min-max based on specific component of vectors in GLSL?

问题描述

我需要在GLSL代码中多次调用这种函数.

I need to call this kind of function many many times in my GLSL code.

vec2 minx(vec2 a, vec2 b) {
    if (a.x < b.x) return a;
    else return b;
}

我担心分支过多.有没有办法避免if-else构造?

I'm concerned of the excessive branching. Is there a way to do this avoiding if-else constructs?

推荐答案

我建议使用GLSL函数

I suggest to use the GLSL functions mix and step.

mix 插入2个值之间根据浮点内插值a在[0.0，1.0]范围内.如果a等于0.0，则返回第一个值；如果a等于1.0，则返回第二个值.

mix interpolates between 2 values according to a floating point interpolation value a in the range [0.0, 1.0]. If the a is equal 0.0 then the 1st value is returned and if the a is equal 1.0 then the 2nd value is returned.

step 测试值是否小于边缘值.如果小于此值，则返回0.0，否则返回1.0.

step tests whether a value is less than an edge value. If it is less then 0.0 is returned, else 1.0 is returned.

如果将这两个功能结合在一起，则代码将如下所示:

If you combine the 2 functions your code will look like this:

vec2 minx(vec2 a, vec2 b)
{
    return mix( a, b, step( b.x, a.x ) );
}

请注意， step 恰好是0.0或恰好是1.0，这会导致 返回第一个值或返回第二个值.

Note, the result of step is either exactly 0.0 or exactly 1.0, this causes that mix either returns the 1st value or returns the 2nd value.

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