RGB bit mask help

[MeghansUncle2]

I am working on understanding the structure and functioning of bitmath in Micro Python. Specifically, how a specific function works in an existing ILI9341 driver. Following is a code snippet from that driver.

def color565(r, g, b):
    """Return RGB565 color value.

    Args:
        r (int): Red value.
        g (int): Green value.
        b (int): Blue value.

    """
    return (r & 0xf8) << 8 | (g & 0xfc) << 3 | b >> 3

I understand bit manipulation, but this one has me stumped, so I will walk through my assumptions and ask where I made a left turn when I should have turned right.

Let's assume that our values are within the 565 range. 5-bits = 32 possible values 0 - 31, 6-bits = 64 possible values 0 - 64

So assuming r = 16, g = 30, and b = 7 (these are all arbitrary, I don't really care what values are used for this discussion.)
Further, assuming that int refers to a 32-bit value. I will explain what I am assuming below breaking down the r g and b values and the return statement:

r = 16 in a 32-bit field represented in binary = 0000 0000 0000 0000 0000 0000 0000 1101
g = 30 in a 32-bit field represented in binary = 0000 0000 0000 0000 0000 0000 0001 1110
b = 7 in a 32-bit field represented in binary = 0000 0000 0000 0000 0000 0000 0000 0111

So the first operation in the return statement is:

       (r & 0xf8)

to AND F8 with the r value doesn't seem correct. F8 looks like a bitmask for the 5 bits of the r value, but i don't understand how it's masking it using this statement

r = 0000 0000 0000 0000 0000 0000 0000 1101
F8 = 0000 0000 0000 0000 0000 0000 1111 1000

I would think for the mask to work correctly, the r value would have to shift left 3 bits first to align with the mask. Can someone please explain where I am failing understanding this.

[robert-hh]

r,g and b are initially 8 bit values, which are truncated and packed into the rgb565 word, wich is 16 bit, not 24. For this, the low orde bits of r,g,b are dropped. That's all

[robert-hh]

By dropping the low order bits, the range is kept, but the resolution drops.

[MeghansUncle2]

So the last 3 bits (least significant) are made 0 and the result is what is used. Thanks you for that. This makes perfect sense now!!!

I have not heard of dropping lsb color bits referenced as lowering of resolution before, it's a good way to think about it.