Packing float into vec4 - how does this code work?

Question

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?

Nicol Bolas Nicol Bolas · Accepted Answer · 2012-03-27T03:14:16

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.

Packing float into vec4 - how does this code work?

3 Answers