为什么不bash的标志-e退出时，子shell不成？ [英] Why doesn't bash flag -e exit when a subshell fails?

问题描述

我有点困惑在这里。在这里我的目标是有一个非零退出code中的bash脚本退出时，任何在脚本中的命令失败。使用-e标志，我以为用子shell时，这将是情况下，即使。下面是一个简化的例子：

I'm a bit confused here. My goal here is to have the bash script exit with a non-zero exit code when any of the commands within the script fails. Using the -e flag, I assumed this would be the case, even when using subshells. Below is a simplified example:

#!/bin/bash -e

(false)

echo $?
echo "Line reached!"

下面是输出时跑了：

[$]>Tests/Exec/continuous-integration.sh 
1
Line reached!

Bash的版本：3.2.25在CentOS

Bash version: 3.2.25 on CentOS

推荐答案

看来好像这是关系到你的庆典的版本。在我访问时，bash版本3.1.17和3.2.39出现此行为机，庆典4.1.5没有。

It appears as though this is related to your version of bash. On machines that I have access to, bash version 3.1.17 and 3.2.39 exhibit this behaviour, bash 4.1.5 does not.

虽然丑了一点，在这两个版本的有效的解决方案可能是这样的：

Although a bit ugly, a solution that works in both versions could be something like this:

#!/bin/bash -e

(false) || exit $?

echo $?
echo "Line reached!"

有其中涉及与设置-e 选项错误bash的源更新日志的一些注意事项。

There are some notes in the bash source changelog which related to bugs with the set -e option.

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