有没有写一个中止程序的执行，所以不管怎么叫bash函数的方法吗？ [英] Is there a way to write a bash function which aborts the whole execution, no matter how it is called?

问题描述

我是用我的bash功能退出1语句来终止整个脚本，它工作得很好：

I was using "exit 1" statement in my bash functions to terminate the whole script and it worked fine:

function func()
{
   echo "Goodbye"
   exit 1
}
echo "Function call will abort"
func
echo "This will never be printed"

但后来我意识到，当所谓的喜欢它不会做的工作：

But then I realized that it doesn't do the work when called like:

res=$(func)

据我所知，我创建了一个子shell和退出1，将取消子shell，而不是主之一....

I understand that I created a subshell and "exit 1" aborts that subshell and not the primary one....

但有没有写从而中止程序的执行，所以不管怎么称呼？我只需要得到真正的返回值（由函数呼应）功能的方法。

But is there a way to write a function which aborts the whole execution, no matter how it is called? I just need to get the real return value (echoed by the function).

推荐答案

在什么能的做的，就是注册了 TERM顶级壳信号退出，然后发送一个 TERM 顶端级别外壳：

What you could do, is register the top level shell for the TERM signal to exit, and then send a TERM to the top level shell:

#!/bin/bash
trap "exit 1" TERM
export TOP_PID=$$

function func()
{
   echo "Goodbye"
   kill -s TERM $TOP_PID
}

echo "Function call will abort"
echo $(func)
echo "This will never be printed"

那么，你的函数发送 TERM 信号回至顶层壳，其被捕获并使用提供的命令，在这种情况下，退出1。

So, your function sends a TERM signal back to the top level shell, which is caught and handled using the provided command, in this case, "exit 1".

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