Bash shell，尝试创建和评估掩码
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Bash shell, trying to create and evaluate a mask
提问by Neuromante
I'm trying to create a mask and use the bitwise operator "&" to compare to another variable and see the output. Let there be code:
我正在尝试创建一个掩码并使用按位运算符“&”与另一个变量进行比较并查看输出。让有代码：
mask=00000
mesk=00010
mosk=$mask&$mesk
echo $mosk
echo meec
I'm trying to expand this functionality to be able to have more characters (different error/success codes), but those lines just don't work: Executing the script will print an empty line, then "meec".
我正在尝试扩展此功能以包含更多字符（不同的错误/成功代码），但这些行不起作用：执行脚本将打印一个空行，然后是“meec”。
I came from an object oriented programming background, and although I've read through several documents on this subject, it seems there's something I'm missing. Any help would be appreciated.
我来自面向对象的编程背景，虽然我已经阅读了关于这个主题的几个文档，但似乎我遗漏了一些东西。任何帮助，将不胜感激。
Edit: For some reason, turns out the code doesn't work, it says "command 00010 not found" >_>
编辑：由于某种原因，代码不起作用，它说“找不到命令 00010”>_>
采纳答案by Some programmer dude
It's because usually the &character in the shell is the modifier to put a command in the background.
这是因为通常&shell 中的字符是将命令置于后台的修饰符。
You have to use Arithmetic Expansionof Bash (for example) for it to work:
您必须使用Bash 的算术扩展（例如）才能使其工作：
mosk=$(($mask & $mesk))
