bash shell实现二进制与十进制数的互转

二进制转十进制

#!/bin/bash
#b2d.sh: convert binary number to decimal number
#Usage: ./b2d.sh number
BAD_ARGS=65
WRONG_ARGS=66
ARGS=1                                          #参数数目
if [ $# -ne $ARGS ]
then
    echo "Usage: `basename $0` binary_number"
    exit $BAD_ARGS
fi
case $1 in
    [01]*)                                      #判断是否为二进制数
        count=$(echo $1 | wc -c)
        let "count--"                           #求出所输入的二进制数的位数                        
        i=0
        n=$count
        result=0
        while [ $i -lt $count ]
        do
            c=$(echo $1 | cut -b$n)             #从最低位开始依次得到该位上的数，如1101，将依次得到1、0、1、1
            let "result+=c*2**i"                #依次累加转换为十进制数
            let "i++"
            let "n--"
        done
        echo "The decimal number of $1 is $result."
        exit 0
        ;;
    *)
        echo "Please run this script with a binary number"
        exit $WRONG_ARGS
        ;;
esac

十进制转二进制

#!/bin/bash
#d2b.sh: convert a decimal number to a binary number
#Usage: ./d2b.sh decimal_number
BAD_ARGS=65
WRONG_ARGS=66
ARGS=1                                          #参数数目
if [ $# -ne $ARGS ]
then
    echo "Usage: `basename $0` decimal_number"
    exit $BAD_ARGS
fi
function is_positive_int()                      #用于判断输入是否为正整数，是返回1，否返回0
{
    if [ $# -lt 1 ]
    then
        return 0
    fi
    if [[ $1 =~ ^[1-9][0-9]*$ ]]
    then
        return 1
    fi
    if [[ $1 =~ ^0$ ]]
    then
        return 1
    fi
    return 0
}
is_positive_int $1                              #调用该函数进行判断
if [ $? -ne 1 ]                                 #不为1，则不是十进制正整数，提示并退出                   
then
    echo "Please run this script with a decimal number."
    exit $WRONG_ARGS
else
    #echo "$1 is a decimal number."
    num=$1
    while [ $num -gt 0 ]                        #辗转相除法
    do
        let r=num%2
        result=$r$result
        let num=num/2
    done
    echo "The binary number of $1 is $result."
    exit 0
fi