参见英文答案 >
Why is bash errexit not behaving as expected in function calls?3个答案set -e(或以#!/ bin / sh -e开头的脚本)对于在出现问题时自动轰炸非常有用。它使我不必错误地检查可能失败的每个命令。
- #!/bin/sh -e
- echo "the following command could fail:"
- false
- echo "this is after the command that fails"
输出符合预期:
- the following command could fail:
现在我想把它包装成一个函数:
- #!/bin/sh -e
- my_function() {
- echo "the following command could fail:"
- false
- echo "this is after the command that fails"
- }
- if ! my_function; then
- echo "dealing with the problem"
- fi
- echo "run this all the time regardless of the success of my_function"
预期产量:
- the following command could fail:
- dealing with the problem
- run this all the time regardless of the success of my_function
实际产量:
- the following output could fail:
- this is after the command that fails
- run this all the time regardless of the success of my_function
(即函数忽略set -e)
这可能是预期的行为。我的问题是:如何在shell函数中获得效果并使用set -e?我发现the same question在Stack Overflow之外被问到但没有合适的答案。
从set -e的文档:
When this option is on,if a simple command fails for any of the
reasons listed in Consequences of
Shell Errors or returns an exit status
value >0,and is not part of the
compound list following a while,
until,or if keyword,and is not a
part of an AND or OR list,and is not
a pipeline preceded by the ! reserved
word,then the shell shall immediately
exit.
在您的情况下,false是前面的管道的一部分!和if的一部分。因此解决方案是重写代码,使其不重写。
换句话说,这里的功能没什么特别之处。尝试:
- set -e
- ! { false; echo hi; }