Converting Hex To Decimal In Awk Or Sed
Answer :
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.
This seems to work:
awk -F, '{ p1 = substr($4, 1, 6);
p2 = ("0x" substr($4, 7, 4)) + 0;
p3 = ("0x" substr($4, 11, 4)) + 0;
printf "%s,%s,%s,%s,%d,%d\n", $1, $2, $3, p1, p2, p3;
}'
For your sample input data, it produces:
123711184642,02,3583090366663629,639f02,292,14292
123715942138,01,3538710295145500,639f02,45014,50755
123711616258,02,3548370476972758,639f02,72,22322
The string concatenation of '0x' plus the 4-digit hex followed by adding 0 forces awk
to treat the numbers as hexadecimals.
You can simplify this to:
awk -F, '{ p1 = substr($4, 1, 6);
p2 = "0x" substr($4, 7, 4);
p3 = "0x" substr($4, 11, 4);
printf "%s,%s,%s,%s,%d,%d\n", $1, $2, $3, p1, p2, p3;
}'
The strings prefixed with 0x are forced to integer when presented to printf()
and the %d
format.
The code above works beautifully with the native awk
on MacOS X 10.6.5 (version 20070501); sadly, it does not work with GNU gawk
3.1.7. That, it seems, is permitted behaviour according to POSIX (see the comments below). However, gawk
has a non-standard function strtonum
that can be used to bludgeon it into performing correctly - pity that bludgeoning is necessary.
gawk -F, '{ p1 = substr($4, 1, 6);
p2 = "0x" substr($4, 7, 4);
p3 = "0x" substr($4, 11, 4);
printf "%s,%s,%s,%s,%d,%d\n", $1, $2, $3, p1, strtonum(p2), strtonum(p3);
}'
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.
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