How to perform bitwise operations on hexadecimal numbers in bash?

Question

In my bash script I have a string containing a hexadecimal number, e.g. hex="0x12345678". Is it possible to treat it as a hex number and do bit shifting on it?

Answer 1

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