I would like to know if there is a way of using bash expansion to view all possibilities of combination for a number of digits in hexadecimal. I can expand in binaries
In base 2:
echo {0..1}{0..1}{0..1}
Which gives back:
000 001 010 011 100 101 110 111
In base 10:
echo {0..9}{0..9}
Which gives back:
00 01 02...99
But in hexadecimal:
echo {0..F}
Just repeat:
{0..F}
You can; you just need to break the range {0..F} into two separate ranges {0..9} and {A..F}:
$ printf '%s\n' {{0..9},{A..F}}{{0..9},{A..F}}
00
01
...
FE
EF
Using printf:
$ printf '%.2x\n' {0..255}
The format string %.2x says to format the output as a zero-filled, two-digit, lower-case, hexadecimal number (%02x would have done the same).
If you want upper-case, use %.2X.
Bash only understands base 10 integer ranges or ranges between ASCII characters in brace expansions of intervals.
It's possible but it isn't nice:
echo {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}{0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}
As far as I can tell bash has no notion of hex ranges.
echo {0-9A-F}works in zsh with theBRACE_CCLoption.{00..99}to get the same output.