#!/bin/bash
count=$(echo $1 | wc -c)
let "count --"
i=0
n=$count
result=0
while [ $i -lt $count ]
do
c=$(echo $1|cut -b $n)
if [ $c -gt 1 ]; then #对每一位判断是否大于1,即可判断是否为一个二进制数,换成其他进制也是一样
echo "$* is not a valuable bin number."
exit 1
else
let result+=c*2**i #转化步骤
fi
let i++
let n--
done
echo "$1(b)=$result(d).";