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How to compare hex values using AWK? – Debian

Consider using the following program that filters hex number in the range 0x10 – 0xa0.

echo -e "0x10n0xffn0x16n0x80n0x50" | awk '$1 >= 0x10 && $1 <= 0xa0 { print $0 }'

The program works fine, if I replace all numbers using decimal equivalents, but fails with hex numbers.

The version of awk that I’m using doesn’t support --non-decimal-data – it’s the default version that comes with Debian 10 Buster.

How can I make the comparison work with hex numbers?

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2

Answers


  1. 0

    awk has strtonum function for this conversion from hex to decimal:

    echo -e "0x10n0xffn0x16n0x80n0x50" |
awk '{n = strtonum($1)} n >= 0x10 && n <= 0xa0'

0x10
0x16
0x80
0x50

    In absence of gnu awk, you may use this perl one liner also:

    echo -e "0x10n0xffn0x16n0x80n0x50" |
perl -lane 'print if hex($F[0]) >= 0x10 && hex($F[0]) <= 0xa0'
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  2. 0

    one trick is doing string comparison instead. Since "a">"9", this will work

    $ echo -e "0x10n0xffn0x16n0x80n0x50" | 
  awk '$1 >= "0x10" && $1 <= "0xa0"'


0x10
0x16
0x80
0x50
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