TinyChan

Thinking about byte manipulation stuff (I wrote this on my phone):

function printU8binary(bytes){
  let len=bytes.length,
      iPad=(""+(len-1)).length, //Math.max(1,Math.floor(Math.log10(len-1))+1),
      array=Array.from(bytes);
  console.log(array.map(x=>(x&0xFF).toString(16).toUpperCase().padStart(2,"0")).join(" "));
  console.log(array.map((x,i)=>{
    let byte=x&0xFF;
    return(""+i).padStart(iPad)+": "+ //index
          byte.toString(2).padStart(8)+ //binary
          ` (0x${byte.toString(16).padStart(2,"0")}; ${byte}; ${String.fromCharCode(byte)})`;
  }).join("\n"));
  console.log(array.map(x=>{ //visual representation
    let o="";
    for(let i=0;i<8;i++)o+=(x>>i)&1?"■":"□";
    return o;
  }).join("\n"));
  console.log(String.fromCharCode(...array));
}
function randomU8array(len){
  let array=new Uint8Array(len);
  for(let i=0;i<array.length;i++)
    array[i]=Math.floor(Math.random()*256);
  return array;
}

printU8binary(randomU8array(32)); //example

/*
Max 32-bit value is (2**32)-1 = 4294967295 = ~0>>>0
Power of two: 1<<n = 2**n
1s bitmask: (1<<n)-1 (problem: (1<<32)-1 = 0) → (2**n)-1
>>> should be used when shifting, >> adds 1s at the left/upper if it's signed.
*/
function extractBits(value,start,count){
  //Should start = 32-count-start?
  return((value>>>start)&((2**count)-1))>>>0;
}