Convert decimal to Base-4 in bash - bash

Question

I have been using a pretty basic, and for the most part straight forward, method to converting base-10 numbers {1..256} to base-4 or quaternary numbers. I have been using simple division $(($NUM/4)) to get the main result in order to get the remainders $(($NUM%4)) and then printing the remainders in reverse to arrive at the result. I use the following bash script to do this:
#!/bin/bash
NUM="$1"
main() {
local EXP1=$(($NUM/4))
local REM1=$(($NUM%4))
local EXP2=$(($EXP1/4))
local REM2=$(($EXP1%4))
local EXP3=$(($EXP2/4))
local REM3=$(($EXP2%4))
local EXP4=$(($EXP3/4))
local REM4=$(($EXP3%4))
echo "
$EXP1 remainder $REM1
$EXP2 remainder $REM2
$EXP3 remainder $REM3
$EXP4 remainder $REM4
Answer: $REM4$REM3$REM2$REM1
"
}
main
This script works fine for numbers 0-255 or 1-256. But beyond this(these) ranges, results become mixed and often repeated or inaccurate. This isn't so much of a problem as I don't intend to convert numbers beyond 256 or less than 0 (negative numbers [yet]).
My question is: "Is there a more simplified method to do this, possibly using expr or bc?

Answer 1

Base 4 conversion in bash
int2b4() {
local val out num ret=\\n;
for ((val=$1;val;val/=4)){
out=$((val%4))$out;
}
printf ${2+-v} $2 %s${ret[${2+1}]} $out
}
Invoked with only 1 argument, this will convert to base 4 and print the result followed by a newline. If a second argument is present, a variable of this name will be populated, no printing.
int2b4 135
2013
int2b4 12345678
233012011032
int2b4 5432 var
echo $var
1110320
Detailled explanation:
The main part is (could be written):
out=""
for (( val=$1 ; val > 0 ; val = val / 4 )) ;do
out="$((val%4))$out"
done
We're conversion loop could be easily understood (i hope)
local ensure out val num to be local empty variables and initialise locally ret='\n'
printf line use some bashisms
${2+-v} is emppty if $2 is empty and represent -v if not.
${ret[${2+1}]} become respectively ${ret[]} ( or ${ret[0]} ) and ${ret[1]}
So this line become
printf "%s\n" $out
if no second argument ($2) and
printf -v var "%s" $out
if second argument is var (Note that no newline will be appended to a populated variable, but added for terminal printing).
Conversion back to decimal:
There is a bashism letting you compute with arbitrary base, under bash:
echo $((4#$var))
5432
echo $((4#1110320))
5432
In a script:
for integer in {1234..1248};do
int2b4 $integer quaternary
backint=$((4#$quaternary))
echo $integer $quaternary $backint
done
1234 103102 1234
1235 103103 1235
1236 103110 1236
1237 103111 1237
1238 103112 1238
1239 103113 1239
1240 103120 1240
1241 103121 1241
1242 103122 1242
1243 103123 1243
1244 103130 1244
1245 103131 1245
1246 103132 1246
1247 103133 1247
1248 103200 1248

Answer 2

Create a look-up table taking advantage of brace expansion
$ echo {a..c}
a b c
$ echo {a..c}{r..s}
ar as br bs cr cs
$ echo {0..3}{0..3}
00 01 02 03 10 11 12 13 20 21 22 23 30 31 32 33
and so, for 0-255 in decimal to base-4
$ base4=({0..3}{0..3}{0..3}{0..3})
$ echo "${base4[34]}"
0202
$ echo "${base4[255]}"
3333