The file contains code like the below and need to convert each one into a decimal

00 00 00 04 17 03 06 01

So the output should come as

0 0 0 4 23 3 6 1

$ for i in 00 00 00 04 17 03 06 01 ; do echo "ibase=16; $i" | bc ; done

Without external programs

for i in 00 00 00 04 17 03 06 01 ; do
  printf "%d\n" "0x$i"
done

Hi Jayan

I have a file called hex.txt
the file contains numerous entries i, just gave 8 as an example , I need to pick up entire content from the file and convert to decimal and not just those 8 alone

Can u advise if that is possible

Please post a representative sample of hex.txt and the desired output..

hex.txt contains numerous entries like this

00 00 00 04 0A 02 05 01 00 00 00 00 0A 0A 32 C7 00 00 00 00 0A 0A 32 F3 0A 0A 2B 2D 00 00 00 00 00 00 00 08 80 01 06 6D 26 E4 00 02 00 00 00 00
06 6D 0A 80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

Try:

awk '{for(i=1;i<=NF;i++)$i=sprintf("%0d","0x"$i)}1' infile