将float打包到vec4中-此代码如何工作?

如果按照建议进行操作，仅将浮点值写入帧缓冲区的红色通道，则只能获得8位精度.使用这种方法，您可以为所有32位工作.I am trying to study shadow mapping in WebGL. I see same piece of shader code copied in various libraries and examples that achieve this. However nowhere did I find the explanation of how it works.The idea is to save a depth value (a single float) into the color buffer (vec4). There is a pack function that saves float to vec4 and unpack function that retrieves the float from vec4.vec4 pack_depth(const in float depth){ const vec4 bit_shift = vec4(256.0*256.0*256.0, 256.0*256.0, 256.0, 1.0); const vec4 bit_mask = vec4(0.0, 1.0/256.0, 1.0/256.0, 1.0/256.0); vec4 res = fract(depth * bit_shift); res -= res.xxyz * bit_mask; return res;}float unpack_depth(const in vec4 rgba_depth){ const vec4 bit_shift = vec4(1.0/(256.0*256.0*256.0), 1.0/(256.0*256.0), 1.0/256.0, 1.0); float depth = dot(rgba_depth, bit_shift); return depth;}I would have imagined that packing a float into vec4 should be a trivial problem, just copy it into one of the 4 slots of vec4 and leave others unused. That's why the bit shifting logic in above code is puzzling to me.Can anyone shed some light? 解决方案 It's not storing a GLSL float in a GLSL vec4. What it's doing is storing a value in a vec4 which, when written to an RGBA8 framebuffer (32-bit value) can be read as a vec4 and then reconstituted into the same float that was given previously.If you did what you suggest, just writing the floating-point value to the red channel of the framebuffer, you'd only get 8 bits of accuracy. With this method, you get all 32-bits working for you. 这篇关于将float打包到vec4中-此代码如何工作?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！