Some experiments in manipulating bytes for an LED Driver. Javascript because it's fast and easy.

Javascript Binary operations

/* 
I wrote this to do some experimentation on
bit manipulation. I'm faster in Javascript than I am in C
and this let me work out the logic for shifting
the bits into the right positions.

The goal was to turn an integer into an on/off state
for the NXP9532D - an LED driver.
(https://www.nxp.com/docs/en/data-sheet/PCA9532.pdf - see page 8)

If I have a number, 5 in binary: 1001
that turns into a 1-byte value with two bits for each number.
A 1 becomes a 10 (setting the value to PWM0 rate)
So a 1001
to LS0 becomes 1000 0010

A 22: 0001 0110
becomes 

LS1: 0000 0010
LS0: 0010 1000

*/


//This is just a simple method for displaying numbers in binary 
let b = (num) => {
    return num.toString(2).padStart(8,0).slice(0,4) + " " + num.toString(2).padStart(8,0).slice(4,8)
}

var v = 22;                 // 0001 0110
var n1 = v & 15;            // lower nibble:  15 = 0000 1111
var n2 = (v & 240) >> 4;    // upper nibble: 240 = 1111 0000 - then shifted right

var lower = 0;
var upper = 0;

/* 
// This is a programmatic method of stepping through the 
// bits, and shifting out  

var c = 4

while(c > 0) {
    // console.log(b(8 & n1));
    
    if((8 & n1)) {  // 8 AND'd with the nibble tells us if there's a 1 in the 8 position
        lower = ((1 << (c * 2 - 1) ) | lower); //OR the         
        //c is the position of the bit.
        console.log(`c is ${c}. New position is: ${b(1 << (c * 2 - 1))}. lower is now ${b(lower)}`);
    }
    
    n1 = n1 << 1 //shift right by 1.
    c--;
} */

/* 
Example:
If 4 bit is true
Shift a bit into 32 bit
Combine with the existing lower
*/

// This isn't as programatically attractive, but is more efficient.
if((8 & n1)) lower = ((1 << 7) | lower);
if((4 & n1)) lower = ((1 << 5) | lower);
if((2 & n1)) lower = ((1 << 3) | lower);
if((1 & n1)) lower = ((1 << 1) | lower);

console.log(b(lower));
//0010 1000

if((8 & n2)) upper = ((1 << 7) | upper);
if((4 & n2)) upper = ((1 << 5) | upper);
if((2 & n2)) upper = ((1 << 3) | upper);
if((1 & n2)) upper = ((1 << 1) | upper);

console.log(b(upper));