The Weird and The Wonderful

A | B = ( A ^ B ) | ( A & B )

100 = 0x64
5 = 0x05
0x64 | 0x05 = 0x65

0x65 ? 0x05 = 0x64

0x65 ? 0x64 = 0x05

albert_redditt wrote:

I was told that there's no way to undo an OR as it loses value in the output..

1 | 1
1 | 0
0 | 1

A | B = C

6 | B = 7

( v1 ^ or_val ) + ( v1 & or_val )
=    (the left and right operands do not "intersect", their bitwise AND is zero, in that case A + B == A | B)
( v1 ^ or_val ) | ( v1 & or_val )
=    (OR distributes over AND)
(v1 ^ or_val | v1) & (v1 ^ or_val | or_val)
=    (A ^ B | B == A | B)
(or_val | v1) & (v1 | or_val)
=    (A AND A == A)
v1 | or_val

Rob Grainger wrote:

the input of a function (or operator) must be reconstructable from the input to that function

PIEBALDconsult wrote:

Only works with base-10 ?

function hexToDec(number) {
   if (number === 1) return 1;
   else if (number === 2) return 2;
    //...
   else if (number === A) return 10;
   else if (number === B) return 11;
   //...
}

It is also important to know if a number is odd, so additionally I created this other package @samuelmarina/is-odd

PIEBALDconsult wrote:

function isOdd(x) { return isEven(x-1); }