Converting hex to decimal in awk or sed

By AWK

This answer concentrates on showing how to do the conversion by awk portably.

Using --non-decimal-data for gawk is not recommended according to GNU Awk User's Guide. And using strtonum() is not portable.

In the following examples the first word of each record is converted.

By user-defined function

The most portable way of doing conversion is by a user-defined awk function [reference]:

function parsehex(V,OUT)
{
    if(V ~ /^0x/)  V=substr(V,3);

    for(N=1; N<=length(V); N++)
        OUT=(OUT*16) + H[substr(V, N, 1)]

    return(OUT)
}

BEGIN { for(N=0; N<16; N++)
        {  H[sprintf("%x",N)]=N; H[sprintf("%X",N)]=N } }

{ print parsehex($1) }

By calling shell's printf

You could use this

awk '{cmd="printf %d 0x" $1; cmd | getline decimal; close(cmd); print decimal}'

but it is relatively slow. The following one is faster, if you have many newline-separated hexadecimal numbers to convert:

awk 'BEGIN{cmd="printf \"%d\n\""}{cmd=cmd " 0x" $1}END{while ((cmd | getline dec) > 0) { print dec }; close(cmd)}'

There might be a problem if very many arguments are added for the single printf command.

In Linux

In my experience the following works in Linux:

awk -Wposix '{printf("%d\n","0x" $1)}'

I tested it by gawk, mawk and original-awk in Ubuntu Linux 14.04. By original-awk the command displays a warning message, but you can hide it by redirection directive 2>/dev/null in shell. If you don't want to do that, you can strip the -Wposix in case of original-awk like this:

awk $(awk -Wversion >/dev/null 2>&1 && printf -- "-Wposix") '{printf("%d\n","0x" $1)}'

(In Bash 4 you could replace >/dev/null 2>&1 by &>/dev/null)

Note: The -Wposix trick probably doesn't work with nawk which is used in OS X and some BSD OS variants, though.

Here's a variation on Jonathan's answer:

awk $([[ $(awk --version) = GNU* ]] && echo --non-decimal-data) -F, '
    BEGIN {OFS = FS}
    {
        $6 = sprintf("%d", "0x" substr($4, 11, 4))
        $5 = sprintf("%d", "0x" substr($4,  7, 4))
        $4 = substr($4,  1, 6)
        print
    }'

I included a rather contorted way of adding the --non-decimal-data option if it's needed.

Edit

Just for the heck of it, here's the pure-Bash equivalent:

saveIFS=$IFS
IFS=,
while read -r -a line
do
    printf '%s,%s,%d,%d\n' "${line[*]:0:3}" "${line[3]:0:6}" "0x${line[3]:6:4}" "0x${line[3]:10:4}"
done
IFS=$saveIFS

The "${line[*]:0:3}" (quoted *) works similarly to AWK's OFS in that it causes Bash's IFS (here a comma) to be inserted between array elements on output. We can take further advantage of that feature by inserting array elements as follows which more closely parallels my AWK version above.

saveIFS=$IFS
IFS=,
while read -r -a line
do
    line[6]=$(printf '%d' "0x${line[3]:10:4}")
    line[5]=$(printf '%d' "0x${line[3]:6:4}")
    line[4]=$(printf '%s' "${line[3]:0:6}")
    printf '%s\n' "${line[*]}"
done
IFS=$saveIFS

Unfortunately, Bash doesn't allow printf -v (which is similar to sprintf()) to make assignments to array elements, so printf -v "line[6]" ... doesn't work.

Edit: As of Bash 4.1, printf -v can now make assignments to array elements. Example:

printf -v 'line[6]' '%d' "0x${line[3]:10:4}"

The quotes around the array reference are needed to prevent possible filename matching. If a file named "line6" existed in the current directory and the reference wasn't quoted, then a variable named line6 would be created (or updated) containing the printf output. Nothing else about the file, such as its contents, would come into play. Only the name - and only tangentially.