如何在shell函数中获得效果并使用“set -e”？ [重复]

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。

如何在函数内获得相应的内容？

例如，我有以下脚本在出错时立即退出并出现错误退出状态：

#!/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。

set -e
! { false; echo hi; }