为什么golang RGBA.RGBA（）方法使用| 和<<？

Question

In the golang color package, there is a method to get r,g,b,a values from an RGBA object:

func (c RGBA) RGBA() (r, g, b, a uint32) {
    r = uint32(c.R)
    r |= r << 8
    g = uint32(c.G)
    g |= g << 8
    b = uint32(c.B)
    b |= b << 8
    a = uint32(c.A)
    a |= a << 8
    return
}

If I were to implement this simple function, I would just write this

func (c RGBA) RGBA() (r, g, b, a uint32) {
    r = uint32(c.R)
    g = uint32(c.G)
    b = uint32(c.B)
    a = uint32(c.A)
    return
}

What's the reason r |= r << 8 is used?

Answer 1

From the the excellent "The Go image package" blogpost:

[...] the channels have a 16-bit effective range: 100% red is represented by RGBA returning an r of 65535, not 255, so that converting from CMYK or YCbCr is not as lossy. Third, the type returned is uint32, even though the maximum value is 65535, to guarantee that multiplying two values together won't overflow.

and

Note that the R field of an RGBA is an 8-bit alpha-premultiplied color in the range [0, 255]. RGBA satisfies the Color interface by multiplying that value by 0x101 to generate a 16-bit alpha-premultiplied color in the range [0, 65535]

So if we look at the bit representation of a color with the value c.R = 10101010 then this operation

r = uint32(c.R)
r |= r << 8

effectively copies the first byte to the second byte.

   00000000000000000000000010101010 (r)
 | 00000000000000001010101000000000 (r << 8)
--------------------------------------
   00000000000000001010101010101010 (r |= r << 8)

This is equivalent to a multiplication with the factor 0x101 and distributes all 256 possible values evenly across the range [0, 65535].