How to perform bitwise operations on hexadecimal numbers in bash?

Of course you can do bitwise operations (inside an arithmetic expansion):

$ echo "$((0x12345678 << 1))"
610839792

Or:

$ echo "$(( 16#12345678 << 1 ))"
610839792

The value could be set in a variable as well:

$ var=0x12345678         # or var=16#12345678
$ echo "$(( var << 1 ))"
610839792

And you can do OR, AND, XOR and/or NOT:

$ echo "$(( 0x123456 | 0x876543 ))"
9925975

And to get the result in hex as well:

$ printf '%X\n' "$(( 0x12345678 | 0xDEADBEEF ))"     # Bitwise OR
DEBDFEFF

$ printf '%X\n' "$(( 0x12345678 & 0xDEADBEEF ))"     # Bitwise AND
12241668

$ printf '%X\n' "$(( 0x12345678 ^ 0xDEADBEEF ))"     # Bitwise XOR
CC99E897

$ printf '%X\n' "$(( ~ 0x2C8B ))"                    # Bitwise NOT
FFFFFFFFFFFFD374

The only detail with a bitwise not (~) is that it flips all available bits. If the number representation use 64 bits, the result will have 64 bits. All leading zero bits will be flipped to ones.

To limit such conversion, just use an AND:

$ printf '%X\n' "$(( ( ~ 0x2C8B ) & 0xFFFF ))"
D374

Note that a bitwise NOT ~ is not a logical NOT !. A logical NOT turns input into 0 or 1 only, not any other number.

$ printf '%X\n' "$(( ! 0xdead ))" "$(( ! 0 ))"
0
1

You can easily bitshift such numbers in an arithmetic context:

$ hex="0x12345678"
$ result=$((hex << 1))
$ printf "Result in hex notation: 0x%x\n" "$result"
0x2468acf0

Yes.

Arithmetic expressions support base 16 numbers and all the usual C operators.

Example:

$ hex="0xff"
$ echo $(( hex >> 1 ))
127