Sorting an array in lexicographical order. Pure bash - arrays

Question

I have two arrays with IPs. I want to combine them into a third array and apply an ascending ordering.
#!/bin/bash
#
dbip[0]=1.1.1.1
dbip[1]=1.1.1.2
dbip[2]=1.1.1.3
dbip[3]=1.1.1.4
dbip[4]=1.1.1.5
dbip[5]=1.1.1.10
dbip[6]=1.1.1.9
ngip[0]=1.1.1.5
ngip[1]=1.1.1.6
ngip[2]=1.1.1.7
ngip[3]=1.1.1.1
ngip[4]=1.1.1.11
#I am adding the dbip array into the final one
for (( i=0; i<${#dbip[#]}; i++ ))
do
allip[$i]=${dbip[$i]}
done
#Remembering the no. of elements in the final array
var=${#allip[#]}
echo "$var"
#Adding the ngip array into the final one
for (( i=0; i<${#ngip[#]}; i++ ))
do
allip[$var+$i]=${ngip[$i]}
done
#Printing the initial order of the elements in the array
echo "size= ${#allip[#]}"
for (( i=0; i<${#allip[#]}; i++ ))
do
echo "${allip[$i]}"
done
#Sorting the array in ascending order
for (( i=0; i<${#allip[#]}; i++ ))
do
for (( j=$i; j<${#allip[#]}; j++ ))
do
if [ allip[$i] \> allip[$j] ];
then
aux=${allip[$i]}
allip[$i]=${allip[$j]}
allip[$j]=$aux;
fi
done
done
echo "###############################"
#Printing the final form of the array
for (( i=0; i<${#allip[#]}; i++ ))
do
echo "${allip[$i]}"
done
Problem is that the output is not ordered in numerical or lexicographical way.
Output:
1.1.1.1
1.1.1.11
1.1.1.1
1.1.1.2
1.1.1.3
1.1.1.4
1.1.1.5
1.1.1.10
1.1.1.9
1.1.1.5
1.1.1.7
1.1.1.6
The output should be like this:
1.1.1.1
1.1.1.1
1.1.1.2
1.1.1.3
.......
1.1.1.10
1.1.1.11
or like this
1.1.1.1
1.1.1.1
1.1.1.10
1.1.1.11
so i can remove the duplicates later.
How can i do this in a pure programming way in bash. No pipes.
Note that IPs can be from different classes: 10.55.72.190, 10.55.70.1, 10.51.72.44, etc.

Answer 1

Here is "pure" bash:
dbip=( [0]=1.1.1.1 [1]=1.1.1.2 [2]=1.1.1.3 [3]=1.1.1.4 [4]=1.1.1.5 [5]=1.1.1.10 [6]=1.1.1.9 )
ngip=( [0]=1.1.1.5 [1]=1.1.1.6 [2]=1.1.1.7 [3]=1.1.1.1 [4]=1.1.1.11 )
allip=( $(printf "%s\n" "${dbip[#]}" "${ngip[#]}" | sort -V) )
printf "%s\n" "${allip[#]}"
1.1.1.1
1.1.1.1
1.1.1.2
1.1.1.3
1.1.1.4
1.1.1.5
1.1.1.5
1.1.1.6
1.1.1.7
1.1.1.9
1.1.1.10
1.1.1.11

Answer 2

A pure bash implementation (no call to any tool like sort).
The idea: convert IPs to integers, perform a bubble sort on these, convert back to x.x.x.x format.
#! /bin/bash
ips=(192.0.2.235 1.1.1.1 1.1.1.11 1.1.1.1 1.1.1.2 1.1.1.3 1.1.1.4
1.1.1.5 1.1.1.10 1.1.1.9 1.1.1.5 1.1.1.7 1.1.1.6)
# integer IPs
intips=()
echo Initial addresses:
for ((i=0; i<${#ips[#]}; ++i)); do
echo ${ips[$i]}
done
echo
# convert ip to int
ip2int() {
local -i intip
local -a ip
local IFS="."
ip=($1)
intip=$(( ${ip[0]}*2**24 + ${ip[1]}*2**16 + ${ip[2]}*2**8 + ${ip[3]} ))
echo $intip
}
# convert int to ip
int2ip() {
local -i a1 a2 a3 a4 intip=$1
a1=$(( intip / (2**24) )); ((intip -= a1*2**24))
a2=$(( intip / (2**16) )); ((intip -= a2*2**16))
a3=$(( intip / (2**8) )); ((intip -= a3*2**8))
a4=$(($intip))
echo "$a1.$a2.$a3.$a4"
}
# simple bubble sort (ascending) of an integer array
sortn_asc() {
local -a array=( "$#" )
local -i max i val1 val2
for (( max=$(( ${#array[#]} - 1 )); max > 0; max-- )); do
for (( i=0; i<max; i++ )); do
val1=${array[$i]}
val2=${array[$((i + 1))]}
# switch if necessary
if (( $val1 > $val2 )); then
local tmp=$val1
array[$i]=$val2
array[$(($i + 1))]=$tmp
fi
done
done
echo "${array[#]}"
}
# convert ips to integers
for ((i=0; i<${#ips[#]}; ++i)); do
intips[$i]=$(ip2int ${ips[$i]})
done
# sort
intips=( $(sortn_asc ${intips[#]}) )
# convert and echo sorted integers table
echo Sorted addresses:
for ((i=0; i<${#intips[#]}; ++i)); do
echo $(int2ip ${intips[$i]})
done
exit 0
The output:
br#lorien:~/tmp$ ./19398658.sh
Initial addresses:
192.0.2.235
1.1.1.1
1.1.1.11
1.1.1.1
1.1.1.2
1.1.1.3
1.1.1.4
1.1.1.5
1.1.1.10
1.1.1.9
1.1.1.5
1.1.1.7
1.1.1.6
Sorted addresses:
1.1.1.1
1.1.1.1
1.1.1.2
1.1.1.3
1.1.1.4
1.1.1.5
1.1.1.5
1.1.1.6
1.1.1.7
1.1.1.9
1.1.1.10
1.1.1.11
192.0.2.235