Bash. Hex to ascii. Is possible without xxd or perl? - hex

Question

I'm developing a script on which I have a hex string 31323334353637383930313233 and I want to transform it into ASCII. Desired output is 1234567890123.
I already have it working using:
echo "31323334353637383930313233" | xxd -r -p
or
echo "31323334353637383930313233" | perl -pe 's/(..)/chr(hex($1))/ge'
But the point is try to use the minimum possible requirements for the script. I want it working in suse, fedora, debian, ubuntu, arch, etc... It seems the xxd command is included in vim package. I'm wondering if there is a way to achieve this using only awk or any internal Linux tool which is going to be present by default in all Linux systems.

Answer 1

Found this script here:
#!/bin/bash
function hex2string () {
I=0
while [ $I -lt ${#1} ];
do
echo -en "\x"${1:$I:2}
let "I += 2"
done
}
hex2string "31323334353637383930313233"
echo
You may change the line hex2string "31323334353637383930313233" so that it takes the hex value from parameters, that is:
#!/bin/bash
function hex2string () {
I=0
while [ $I -lt ${#1} ];
do
echo -en "\x"${1:$I:2}
let "I += 2"
done
}
hex2string "$1"
echo
So when executed as:
./hexstring.sh 31323334353637383930313233
It will provide the desired ascii output.
NOTE: Can't test if it works in all Linux systems.

Answer 2

Using gawk, from HEX to ASCII
$ gawk '{
gsub(/../,"0x& ");
for(i=1;i<=NF;i++)
printf("%c", strtonum($i));
print ""
}' <<<"31323334353637383930313233"
1234567890123
Using any awk
$ cat hex2asc_anyawk.awk
BEGIN{
split("0 1 2 3 4 5 6 7 8 9 A B C D E F", d, / /)
for(i in d)Decimal[d[i]]=i-1
}
function hex2dec(hex, h,i,j,dec)
{
hex = toupper(hex);
i = length(hex);
while(i)
{
dec += Decimal[substr(hex,i,1)] * 16 ^ j++
i--
}
return dec;
}
{
gsub(/../,"& ");
for(i=1;i<=NF;i++)
printf("%d",hex2dec($i));
print ""
}
Execution
$ awk -f hex2asc_anyawk.awk <<<"31323334353637383930313233"
1234567890123
Explanation
Steps :
Get the decimal equivalent of hex from table.
Multiply every digit with 16 power of digit location.
Sum all the multipliers.
Example :
BEGIN{
# Here we created decimal conversion array, like above table
split("0 1 2 3 4 5 6 7 8 9 A B C D E F", d, / /)
for(i in d)Decimal[d[i]]=i-1
}
function hex2dec(hex, h,i,j,dec)
{
hex = toupper(hex); # uppercase conversion if any A,B,C,D,E,F
i = length(hex); # length of hex string
while(i)
{
# dec var where sum is stored
# substr(hex,i,1) gives 1 char from RHS
# multiply by 16 power of digit location
dec += Decimal[substr(hex,i,1)] * 16 ^ j++
i-- # decrement by 1
}
return dec;
}
{
# it modifies record
# suppose if given string is 31323334353637383930313233
# after gsub it becomes 31 32 33 34 35 36 37 38 39 30 31 32 33
# thus re-evaluate the fields
gsub(/../,"& ");
# loop through fields , NF gives no of fields
for(i=1;i<=NF;i++)
# convert from hex to decimal
# and print equivalent ASCII value
printf("%c",hex2dec($i));
# print newline char
print ""
}
Meaning of dec += Decimal[substr(hex,i,1)] * 16 ^ j++
dec += Decimal[substr(hex,i,1)] * 16 ^ j++
^ ^ ^
| | |
| | 2.Multiply every digit with 16 power of digit location.
| |
| 1.Gives decimal equivalent of hex
|
|
3. Sum all the multipliers

Answer 3

here's a special cheating trick for u - due to ingenuity of how they originally mapped decimal digits to bytes, their hex are all x3[0-9],
so therefore, if u already know they would decode out to digits and nothing else, here's a fast shortcut :
echo "31323334353637383930313233" |
mawk 'gsub("..","_&") + gsub("_3",_)^_'
1234567890123
if it's already URL-percent-encoded, then it's even simpler :
echo '%31%32%33%34%35%36%37%38%39%30%31%32%33' |
mawk 'gsub("%3",_)^_'
or
gawk ++NF FS='%3' OFS=
1234567890123
This specialized approach can handle hex of absolutely any arbitrary size, even for awks that don't have built-in support for bigints
TL;DR : don't "do math" when none is needed

Answer 4

Alternate (g)awk solution:
echo "31323334353637383930313233" | awk 'RT{printf "%c", strtonum("0x"RT)}' RS='[0-9]{2}'