# How to calculate hexadecimal xor (^) from shell?

Unfortunately bc and calc don't support xor.

## 4 Answers

With any POSIX shell:

``````\$ printf '%#x\n' "\$((0x11 ^ 0x22))"
0x33
``````
• `%#x` works in C too, nice! – Steven Penny Jun 27 '16 at 22:54

Like this:

``````echo \$(( 0xA ^ 0xF ))
``````

Or if you want the answer in hex:

``````printf '0x%X\n' \$(( 0xA ^ 0xF ))
``````

On a side note, `calc(1)` does support `xor` as a function:

``````\$ calc
base(16)
0xa
xor(0x22, 0x33)
0x11
``````

gdb has powerful expression calculator:

``````gdb -q -ex 'print/x 0xA ^ 0xF' -ex q
``````

A shell function:

``````calc_gdb() { gdb -q -ex "print/x \$*" -ex q;}
calc_gdb 0xA ^ 0xF

\$1 = 0x5
``````
• Amusing, but it seems like a sledgehammer to crack a nut! – abligh Jun 28 '16 at 8:16
• Useful for the more general case of evaluating expressions, but not for XOR – Peter Cordes Jun 28 '16 at 8:57

It is possible to do that in bc:

``````echo 'xor(10,15)' | bc -l logic.bc
``````

Or in hex:

``````echo 'obase=16;ibase=16; xor(AA,FF)' | bc -l logic.bc
``````

Using the logic file from here.

Just do `wget http://phodd.net/gnu-bc/code/logic.bc` to get it.

• Wow, that site is the `bc(1)` geek's heaven. :) Thank you for the link. – Satō Katsura Jun 28 '16 at 17:24
• My version of "bc" doesn't seem to have the XOR function and just says: `Runtime error (func=(main), adr=51): Function xor not defined.` – slacy Oct 5 '16 at 21:13
• @slacy did you get the `logic.bc` file referenced there? That is what defines `xor` it seems – Eric Renouf Oct 11 '16 at 17:53