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; }
