IP Address Converter

Question

There is IP address: 66.102.13.19, and from this address as that received this address

http://1113984275

But how? And how I can make this

Answer 1

Binary shifting is always faster than multiplying or dividing.
Using a binary AND is faster than mod.

ip2dec(){ # Convert an IPv4 IP number to its decimal equivalent.
          declare -i a b c d;
          IFS=. read a b c d <<<"$1";
          echo "$(((a<<24)+(b<<16)+(c<<8)+d))";
        }
dec2ip(){ # Convert an IPv4 decimal IP value to an IPv4 IP.
          declare -i a=$((~(-1<<8))) b=$1; 
          set -- "$((b>>24&a))" "$((b>>16&a))" "$((b>>8&a))" "$((b&a))";
          local IFS=.;
          echo "$*";
        }

ip=66.102.13.19
a=$(ip2dec "$ip")
b=$(dec2ip "$a")
echo "$ip DecIP=$a IPv4=$b "

Note: This system will fail with values like 0008.
Bash thinks it is an octal number.

To solve that, each value will have to be cleaned with something like:

IsDecInt()(     # Is the value given on $1 a decimal integer (with sign)?
                declare n=$1; set --
                if [[ $n =~ ^([+-]?)((0)|0*([0-9]*))$ ]]; then
                    set -- "${BASH_REMATCH[@]:1}"
                else
                    exit 1
                fi
                echo "$1$3$4";
          )

  a=$(IsDecInt "$a")||{ Echo "Value $a is not an integer" >&2; exit 1; }

For (very) old shells: bash since 2.04, or dash or any (reasonable) shell:

ip2dec(){ # Convert an IPv4 IP number to its decimal equivalent.
          local a b c d;
          IFS=. read a b c d <<-_EOF_
$1
_EOF_
          echo "$(((a<<24)+(b<<16)+(c<<8)+d))";
        }

dec2ip(){   # Convert an IPv4 decimal IP value to an IPv4 IP.
            local a=$((~(-1<<8))) b=$1; 
            set -- "$((b>>24&a))" "$((b>>16&a))" "$((b>>8&a))" "$((b&a))";
            local IFS=.;
            echo "$*";
        }

ip=$1
a=$(ip2dec "$ip")
b=$(dec2ip "$a")
echo "$ip DecIP=$a IPv4=$b "