Attempting to replace two character string with /dev/random hex

Question

Example string

CA DA 00 17 11 38 88 C5 03 

Desired output

 AB 3C 6C 8F DA 88 24 78 6C

Commands attempted

 $ tr -dc 0-9A-F < /dev/urandom filename  ## prints too many chars

awk '{gsub(length($1)==2,{printf "%02")}}' filename  ## syntax doesn't work, unsure how to add hex

 $ sed 's/[a-z0-9]\{2\}//g' filename  ## only replaces digits, unsure how to add hex as a replacement

I ended up using vim to do a partial conversion for some level of randomization.

 :s/\d\d/AA/g

Can anyone provide a working solution?

It would be nice to see solutions (and explanations) leveraging tr/awk/sed for knowledge sharing purposes.

Thanks.

Answer 1

To replace each field with a random 2-digit hex number with awk is just:

$ awk -v seed="$RANDOM" 'BEGIN{srand(seed)} {for (i=1; i<=NF; i++) $i=sprintf("%02X",rand()*256)} 1' file
C7 A1 02 1A 4A 94 95 A0 1E

$ awk -v seed="$RANDOM" 'BEGIN{srand(seed)} {for (i=1; i<=NF; i++) $i=sprintf("%02X",rand()*256)} 1' file
1C 50 A9 D3 8B B0 24 9C 14

Hopefully it's very obvious what it's doing.

Answer 2

Here is an idea on how to get a random hex (mac address?)

awk -v seed=$RANDOM '
    BEGIN{
        srand(seed);
        split("0 1 2 3 4 5 6 7 8 9 A B C D E F",hex," ");
        for (i=1; i<=6; i++) 
            printf "%s%s ",hex[int(rand()*16)+1],hex[int(rand()*16)+1];
        print ""
    }'
D8 D9 BA 00 6A C6