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设计Bash脚本以删除恶意软件(Designing a Bash Script to Remove Malware)

 我在大学的技术台工作。 人们总是在他们的macbook上加入大量恶意软件（我以为mac不能得病毒？）。 通常的过程是删除不良应用程序并对系统/库文件进行排序，以查找相关的plist或其他不良内容。 这个过程可能需要很长时间，所以我想我会尝试创建一个自动删除文件的脚本。 我对bash并不是很熟悉，但它会像这样。  
 问题是定义一个任意变量，不知道如何做到这一点。  
#!/bin/bash
TR = malware_quary
sudo find / -name malware_quary |
while read filename
do
    if(malware_quary = "*mackeeper*")
        read -p "Are you sure you want to remove " +malware_quary+"? " -n 1 -r
        echo
        if [[ ! $REPLY =~ ^[Yy]$ ]]
        then
            cat malware_quary < ~/Desktop/log.txt
        rm malware_quary
        fi
done
 
 我将继续添加各种其他恶意软件（即管道，genieo等）的条件，直到我构建了一个扩展的日志文件，我可以在其中运行所有结果。 

I work at a tech desk for my university. People are always coming in with globs of malware on their macbooks (I thought mac's couldn't get virus'?). The usual process is removing the bad applications and sorting through the system/library files to find associated plists or additional bad stuff. This process can take a hell of a long time so I thought I would try and create an automated script for removing files. I'm not super familiar with bash, but it would go something like this.  
The issue is defining an arbitrary variable, not sure how to do this. 
#!/bin/bash
TR = malware_quary
sudo find / -name malware_quary |
while read filename
do
    if(malware_quary = "*mackeeper*")
        read -p "Are you sure you want to remove " +malware_quary+"? " -n 1 -r
        echo
        if [[ ! $REPLY =~ ^[Yy]$ ]]
        then
            cat malware_quary < ~/Desktop/log.txt
        rm malware_quary
        fi
done
 
I will continue to add if conditions for various other malware (ie conduit, genieo, etc..) until I have built an extensive log file where I can just run all the results against it. 

原文：https://stackoverflow.com/questions/26852666

更新时间：2023-11-14 15:11

最满意答案

 
  
   这条生产线究竟如何运作？ 为什么参数解包允许该函数为可变参数模板中的每个参数执行？  
  
 
 参数包扩展是一些涉及参数包的模式，后跟...  
 所以expr(T)...是一个以expr(T)为模式的包扩展，它扩展为expr(T0), expr(T1), expr(T2), ..., expr(TN)参数包中的Ti 。  
 包扩展只能在某些上下文中使用，例如参数列表或初始化列表，因此在这种情况下，每个子表达式的结果用于形成数组int _[]的初始化列表。 该数组未使用且仅存在，因此其初始化程序可用作进行包扩展的上下文。 每个子表达式都是这样的形式(SubscribeToType<Ti>(blah, blah), 0)这意味着函数调用的结果被丢弃，表达式的值为0.这是一个允许包装的一点点扩展以生成包含N个整数的braced-init-list，因为这是初始化数组所需要的。  
 
  
   我非常肯定这条线可以用lambda代替。 你怎么能用lambda表达式替换该行？  
  
 
 你为什么想要？  
 它可以，但你需要在lambda中进行非常类似的包扩展，所以它不会简化任何事情。 

 
  
  How, exactly, does the line work? How come parameter unpacking is allowing that function to execute for each parameter in the variadic template? 
  
 
A parameter pack expansion is some pattern involving a parameter pack followed by ... 
So expr(T)... is a pack expansion with expr(T) as its pattern, and it expands to expr(T0), expr(T1), expr(T2), ..., expr(TN) for each Ti in the parameter pack. 
A pack expansion can only be used in certain contexts, such as argument lists, or initializer lists, so in this case the results of each sub-expression is being used to form an initializer list for the array int _[]. The array is unused and only exists so that its initializer can be used as the context in which to do the pack expansion. Each sub-expression is of the form (SubscribeToType<Ti>(blah, blah), 0) which means the result of the function call is discarded and the expression has the value 0. This is a bit of a kluge to allow the pack expansion to produce a braced-init-list containing N integers, because that's what's needed to initialize the array. 
 
  
  I am very certain this line could be replaced by a lambda. How could you replace the line with a lambda expression? 
  
 
Why would you want to? 
It could, but you'd need a very similar pack expansion in the lambda, so it wouldn't simplify anything.

设计Bash脚本以删除恶意软件(Designing a Bash Script to Remove Malware)

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