Bash expansion hexadecimal

Question

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}

Note that `echo {0-9A-F}` works in zsh with the `BRACE_CCL` option. — TonioElGringo, Feb 20 '17 at 16:26
Note that in base 10 you could also say `{00..99}` to get the same output. — Ruslan, Jan 09 '20 at 21:16

score 23 · Accepted Answer · answered Feb 19 '17 at 19:15

23

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

answered Feb 19 '17 at 19:15

chepner

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This is a nicer trick than manually write each digit in 0-9A-F! – dhag Feb 19 '17 at 20:40

Kusalananda · Answer 2 · 2020-01-09T21:57:56.900

15

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.

edited Jan 09 '20 at 21:57

answered Feb 19 '17 at 15:25

Kusalananda

320,670
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score 7 · Answer 3 · answered Feb 19 '17 at 15:32

7

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.

answered Feb 19 '17 at 15:32

Satō Katsura

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Bash expansion hexadecimal

3 Answers3