这是一篇旧时习作，Shell脚本练得多了，想练练移位计算，就有了下面的脚本。

#!/bin/bash
IPPattern="(\<([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])\>\.){3}\<([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])\>"
maskPattern="[1-9]|[12][0-9]|3[01]"while :
doecho -e -n "Please input a legal IP address [such as 192.168.1.1/16]: "read userInputIPMask=`echo $userInput | egrep "$IPPattern/$maskPattern"`  #Such as 192.168.1.1/24if [ -z "$IPMask" ] ;thenecho "Please input the right format. [such as 192.168.1.1/1-31] "continuefiIPAddr=`echo $IPMask | cut -d/ -f1`IPType[1]=`echo $IPAddr | cut -d. -f1`IPType[2]=`echo $IPAddr | cut -d. -f2`IPType[3]=`echo $IPAddr | cut -d. -f3`IPType[4]=`echo $IPAddr | cut -d. -f4`mask=`echo $IPMask | cut -d/ -f2`echo "IP address is ${IPType[1]}*${IPType[2]}*${IPType[3]}*${IPType[4]} , Mask is $mask ."((IPHex[1]=IPType[1]<<24))((IPHex[2]=IPType[2]<<16))((IPHex[3]=IPType[3]<<8))((IPHex[4]=IPType[4]))#192 c0 ; 1 1 1((iph=${IPHex[1]}+${IPHex[2]}+${IPHex[3]}+${IPHex[4]}))#echo $iph#0xffffffff#declare -i strMask1=4294967295declare -i strMask1=0xffffffff#echo $strMask1((strMask1=strMask1<<(32-mask) & 0xffffffff))#echo $strMask1((strMask2=~strMask1))#echo $strMask2((networkAddr=iph & strMask1))((bcastAddr= (iph | strMask2) &  0xffffffff))#echo $networkAddr | awk '{printf "%x\n",$0}'#echo $bcastAddr | awk '{printf "%x\n",$0}'((IPHex[1]=networkAddr>>24 & 0x000000ff))((IPHex[2]=networkAddr>>16 & 0x000000ff))((IPHex[3]=networkAddr>>8 & 0x000000ff))((IPHex[4]=networkAddr & 0x000000ff))echo -e "Network Address   : ${IPHex[1]}.${IPHex[2]}.${IPHex[3]}.${IPHex[4]}"((IPHex[1]=bcastAddr>>24 & 0x000000ff))((IPHex[2]=bcastAddr>>16 & 0x000000ff))((IPHex[3]=bcastAddr>>8 & 0x000000ff))((IPHex[4]=bcastAddr & 0x000000ff))echo -e "Broadcast Address : ${IPHex[1]}.${IPHex[2]}.${IPHex[3]}.${IPHex[4]}"
done