Given the IP and netmask, how can I calculate the network address using bash?

Question

In a bash script I have an IP address like 192.168.1.15 and a netmask like 255.255.0.0. I now want to calculate the start address of this network, that means using the &

Answer 1

Just adding an alternative if you have only network prefix available (no netmask):

IP=10.20.30.240
PREFIX=26
IFS=. read -r i1 i2 i3 i4 <<< $IP
IFS=. read -r xx m1 m2 m3 m4 <<< $(for a in $(seq 1 32); do if [ $(((a - 1) % 8)) -eq 0 ]; then echo -n .; fi; if [ $a -le $PREFIX ]; then echo -n 1; else echo -n 0; fi; done)
printf "%d.%d.%d.%d\n" "$((i1 & (2#$m1)))" "$((i2 & (2#$m2)))" "$((i3 & (2#$m3)))" "$((i4 & (2#$m4)))"

Answer 2

In addition to @Janci answer

IP=10.20.30.240
PREFIX=26
IFS=. read -r i1 i2 i3 i4 <<< $IP
D2B=({0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1})
binIP=${D2B[$i1]}${D2B[$i2]}${D2B[$i3]}${D2B[$i4]}
binIP0=${binIP::$PREFIX}$(printf '0%.0s' $(seq 1 $((32-$PREFIX))))
# binIP1=${binIP::$PREFIX}$(printf '0%.0s' $(seq 1 $((31-$PREFIX))))1
echo $((2#${binIP0::8})).$((2#${binIP0:8:8})).$((2#${binIP0:16:8})).$((2#${binIP0:24:8}))

Answer 3

For people who hit this while googling and need an answer that works in ash, the sh that's included in BusyBox and therefore on many routers, here's something for that case:

IP=10.20.30.240
MASK=255.255.252.0
IFS=. read -r i1 i2 i3 i4 << EOF
$IP
EOF
IFS=. read -r m1 m2 m3 m4 << EOF
$MASK
EOF
read masked << EOF
$(( $i1 & $m1 )).$(( $i2 & $m2 )).$(( $i3 & $m3 )).$(( $i4 & $m4 ))
EOF
echo $masked

And here's what to do if you only have the prefix length:

IP=10.20.30.240
PREFIX=22
IFS=. read -r i1 i2 i3 i4 << EOF
$IP
EOF
mask=$(( ((1<<32)-1) & (((1<<32)-1) << (32 - $PREFIX)) ))
read masked << EOF
$(( $i1 & ($mask>>24) )).$(( $i2 & ($mask>>16) )).$(( $i3 & ($mask>>8) )).$(( $i4 & $mask ))
EOF
echo $masked

Answer 4

Some Bash functions summarizing all other answers.

ip2int()
{
    local a b c d
    { IFS=. read a b c d; } <<< $1
    echo $(((((((a << 8) | b) << 8) | c) << 8) | d))
}

int2ip()
{
    local ui32=$1; shift
    local ip n
    for n in 1 2 3 4; do
        ip=$((ui32 & 0xff))${ip:+.}$ip
        ui32=$((ui32 >> 8))
    done
    echo $ip
}

netmask()
# Example: netmask 24 => 255.255.255.0
{
    local mask=$((0xffffffff << (32 - $1))); shift
    int2ip $mask
}


broadcast()
# Example: broadcast 192.0.2.0 24 => 192.0.2.255
{
    local addr=$(ip2int $1); shift
    local mask=$((0xffffffff << (32 -$1))); shift
    int2ip $((addr | ~mask))
}

network()
# Example: network 192.0.2.0 24 => 192.0.2.0
{
    local addr=$(ip2int $1); shift
    local mask=$((0xffffffff << (32 -$1))); shift
    int2ip $((addr & mask))
}

Answer 5

Great answer, though minor typo in answer above.

$ printf "%d.%d.%d.%d\n" "$((i1 & m1))" "$(($i2  <-- $i2 should be i2

If anyone knows how to calculate the broadcast address (XOR the network), then calculate the usable nodes between network and broadcast I'd be interested in those next steps. I have to find addresses in a list within a /23.

Answer 6

Use bitwise & (AND) operator:

$ IFS=. read -r i1 i2 i3 i4 <<< "192.168.1.15"
$ IFS=. read -r m1 m2 m3 m4 <<< "255.255.0.0"
$ printf "%d.%d.%d.%d\n" "$((i1 & m1))" "$((i2 & m2))" "$((i3 & m3))" "$((i4 & m4))"
192.168.0.0

Example with another IP and mask:

$ IFS=. read -r i1 i2 i3 i4 <<< "10.0.14.97"
$ IFS=. read -r m1 m2 m3 m4 <<< "255.255.255.248"
$ printf "%d.%d.%d.%d\n" "$((i1 & m1))" "$((i2 & m2))" "$((i3 & m3))" "$((i4 & m4))"
10.0.14.96