如何在UNIX中将字符串转换为整数
问题描述：
我有d1="11"和d2="07".我想将d1和d2转换为整数并执行d1-d2.如何在UNIX中做到这一点?
I have d1="11" and d2="07". I want to convert d1 and d2 to integers and perform d1-d2. How do I do this in UNIX?
d1 - d2当前为我返回"11-07"作为结果.
d1 - d2 currently returns "11-07" as result for me.
答
标准解决方案:
expr $d1 - $d2
您也可以这样做:
echo $(( d1 - d2 ))
，但是请注意，这会将07视为八进制数字！ (因此07与7相同，但010与10不同).
but beware that this will treat 07 as an octal number! (so 07 is the same as 7, but 010 is different than 10).