Bit mask in Bash

Question

Is something like the following code possible within a shell script?

var1=0xA (0b1010)
if ( (var1 & 0x3) == 0x2 ){
    ...perform action...
}

Just to make my intentions 100% clear my desired action would be to check bits of var1 at 0x3 (0b0011) and make sure that it equals 0x2 (0b0010)

 0b1010
&0b0011
_______
 0b0010 == 0x2 (0b0010)

Generally non-completely-trivial calculations are performed through an external command, such as `bc` or `dc` or `expr`. — Matteo Italia, Apr 26 '18 at 19:14

chepner · Accepted Answer · 2019-08-02T16:10:30.357

7

Bit manipulation is supported in POSIX arithmetic expressions:

if [ $(( var1 & 0x3 )) -eq $(( 0x2 )) ]; then

However, it's a bit simpler use an arithmetic statement in bash:

if (( (var1 & 0x3) == 0x2 )); then

edited Aug 02 '19 at 16:10

answered Apr 26 '18 at 19:32

chepner

0x3 == 0x2 will always be false in the "simpler" command. Braces around (var1 & 0x3) required. – Ivan Kovtun Aug 02 '19 at 16:07
Thanks; I forgot (or didn't consider) that `&` had lower precedence than `==`. – chepner Aug 02 '19 at 16:10

score 4 · Answer 2 · answered Apr 26 '18 at 19:32

4

Yes, this is entirely possible:

#!/bin/bash

var1=0xA # (0b1010)
if (( (var1 & 0x3) == 0x2 ))
then
  echo "Match"
fi

answered Apr 26 '18 at 19:32

that other guy

2 Answers2