As bitwise is like converting a into binary then using operator add or xor or nor to get a different bit , can return as decimal or stay as bits. Or adding 1 zero behind a binary then data that is retreive is different so
A in sql after call from table.
I can add bitwise to get a different value
but the storing of A is still A no different in table data. Adding more column
looking for button F so
let x = 'a'.charCodeAt(0).toString(2) & 'f'.charCodeAt(0).toString(2);
x is 1099840  = char a & char f

But keyboard f will distinguish action button, solo f won't change to chopping to axing stone.
So i think the forest tree need to do converstion or walkman
doing converstion need basic  grab
char doing standing       A     &  F button  & resource a
                walking        97      &  F & a
                   jumping     61     & F &  a
                 squad down141    ??basically  not choping or axing so first 3 will do

It seems i ran into same number for a & f & a and a & f & b which both is 1099840 which cannot do for chopping and axing will be same picture

so this theory will be fail.

so this theory will be fail.

so i try << shifting

let x = 'a'.charCodeAt(0).toString(2) << 'f'.charCodeAt(0).toString(2) ;
let x2 = x << 'a'.charCodeAt(0).toString(2);

And it ended up with suprise
a  << f  get 842547200
a << f << a get 1685094400

a << f << b  -520093696
a <<f << c -1040187392
so i just need to auto remove - sign  Math.abs(x)

And it work all unique value to put xxx.png and will show different picture when chopping and axing stone

But when put to B character it show the same number for few 97 << f << a and 98 <<f<<a  and 61 <<f<<a  62 <<f<<a

still fail

so this theory will be fail.

so i try << shifting

let x = 'a'.charCodeAt(0).toString(2) ^ 'f'.charCodeAt(0).toString(2) ;
let x2 = x << 'a'.charCodeAt(0).toString(2);

And it ended up with suprise
a  ^ f    a  get 862
a  ^ f << a  get 862

a ^f  <<b  441344
a ^f  <<c   882688


so i just need to auto remove - sign  Math.abs(x)

And it work all unique value to put xxx.png and will show different picture when chopping and axing stone

But when put to B character it show the same number for few 97 ^ f << a and 98 ^f<<a  and 61 ^f<<a  62 ^f<<a

still fail

So everything fail so i just do the first if habitat.convert()
Tree will do this formula
tree name is 'a' convert to binaray with ^ f button as f is the keyboard

let x = 'a'.charCodeAt(0).toString(2) ^ 'f'.charCodeAt(0).toString(2) ;

it seems that decimal ^ f will get same number no result

 seems that hexa ^ f will get same number no result

Also octal ^ will get same number no result.

binaray ^ f  give unique

if habitat.convert()
rock will do another formula
 then my habitat need do switch more coding. Can only do straight if naming is a or b it bitwisse ^ with keyboard 'f' as compute is after press 'f'

So everything fail so i just do the first if habitat.convert()
Tree will do this formula
tree name is 'a' convert to binaray with ^ f button as f is the keyboard

let x = 'a'.charCodeAt(0).toString(2) ^ 'f'.charCodeAt(0).toString(2) ;


So after getting unique 431 i pass to walkman.doconvert again

inside will be also a but bitwise again ?
seems 431 bitwise a again will hit 10 which is d with key F. so fail
So no choice 431 + "a" , 431+161 all in string
431a, 431161, 43197
or if  recrouse a will be
a431   stand chopping png
61431  walk chopping png
141431 jump chopping png
97431 gravity chopping png

if recouse b +walkman will be
a420   mining png
61420  samsheing png
141420  jumping smash png
97420   gravity smash png

Which if person b call  resourse a will be 420a, 42061, 420141, 42097 again will fail

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