如何在bash中防止嵌套函数失败被忽略

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英文:

How to prevent nested function failures being missed in bash

问题

以下是代码的中文翻译：

  1. #!/usr/bin/env bash
  3. set -Eeuo pipefail
  4. shopt -s huponexit
  5. shopt -s inherit_errexit
  7. function child_function {
  8. 	return 1
  9. }
  10. function parent_function {
  11. 	child_function
  12. 	echo "父进程注意到子进程的退出码为 $?"
  13. }
  14. function grandparent_function {
  15. 	local ec
  16. 	ec=0 && parent_function || ec="$?"
  17. 	echo "祖父进程注意到父进程的退出码为 $ec"
  18. }
  20. grandparent_function

令人惊讶的是，上述代码的输出结果为：

  1. 父进程注意到子进程的退出码为 1
  2. 祖父进程注意到父进程的退出码为 0

将代码修改为：

  1. #!/usr/bin/env bash
  3. set -Eeuo pipefail
  4. shopt -s huponexit
  5. shopt -s inherit_errexit
  7. function child_function {
  8. 	return 1
  9. }
  10. function parent_function {
  11. 	child_function || return "$?"
  12. 	echo "父进程注意到子进程的退出码为 $?"
  13. }
  14. function grandparent_function {
  15. 	local ec
  16. 	ec=0 && parent_function || ec="$?"
  17. 	echo "祖父进程注意到父进程的退出码为 $ec"
  18. }
  20. grandparent_function

将返回预期结果：

  1. 祖父进程注意到父进程的退出码为 1

是否需要设置其他选项来修复这个问题？还是这是bash的一个错误？

我认为这是一个错误的原因是，将代码修改为不使用 || ec="$?" 将会遵守 -e/errexit 选项（如果命令以非零状态退出，则立即退出），并且每个失败的函数都会立即退出：

  1. #!/usr/bin/env bash
  3. set -Eeuo pipefail
  4. shopt -s huponexit
  5. shopt -s inherit_errexit
  7. function child_function {
  8. 	return 1
  9. }
  10. function parent_function {
  11. 	child_function
  12. 	echo "父进程注意到子进程的退出码为 $?"
  13. }
  14. function grandparent_function {
  15. 	parent_function
  16. 	echo "祖父进程注意到父进程的退出码为 $?"
  17. }
  19. grandparent_function

不会输出任何内容，并返回退出码 1。

英文:

The following code:

  1. #!/usr/bin/env bash
  3. set -Eeuo pipefail
  4. shopt -s huponexit
  5. shopt -s inherit_errexit
  7. function child_function {
  8. 	return 1
  9. }
  10. function parent_function {
  11. 	child_function
  12. 	echo "parent noticed child exit code as $?"
  13. }
  14. function grandparent_function {
  15. 	local ec
  16. 	ec=0 && parent_function || ec="$?"
  17. 	echo "grandparent noticed parent exit code as $ec"
  18. }
  20. grandparent_function

Will surprisingly output:

  1. parent noticed child exit code as 1
  2. grandparent noticed parent exit code as 0

Changing the code to:

  1. #!/usr/bin/env bash
  3. set -Eeuo pipefail
  4. shopt -s huponexit
  5. shopt -s inherit_errexit
  7. function child_function {
  8. 	return 1
  9. }
  10. function parent_function {
  11. 	child_function || return "$?"
  12. 	echo "parent noticed child exit code as $?"
  13. }
  14. function grandparent_function {
  15. 	local ec
  16. 	ec=0 && parent_function || ec="$?"
  17. 	echo "grandparent noticed parent exit code as $ec"
  18. }
  20. grandparent_function

Returns the expected result of:

  1. grandparent noticed parent exit code as 1

Is there an additional setting that I need to set to fix this? Or is this a bug in bash?

The reason I believe it is a bug, is that changing the code to not use the || ec="$?" will respect the -e/errexit option (Exit immediately if a command exits with a non-zero status) and have each failed function exit immediately:

  1. #!/usr/bin/env bash
  3. set -Eeuo pipefail
  4. shopt -s huponexit
  5. shopt -s inherit_errexit
  7. function child_function {
  8. 	return 1
  9. }
  10. function parent_function {
  11. 	child_function
  12. 	echo "parent noticed child exit code as $?"
  13. }
  14. function grandparent_function {
  15. 	parent_function
  16. 	echo "grandparent noticed parent exit code as $?"
  17. }
  19. grandparent_function

Outputs nothing and returns exit code 1

答案1

得分: 1

感谢这些评论者：

还有这些支持资源：

我使用子shell技术为问题的代码示例创建了这个解决方法：

  1. #!/usr/bin/env bash
  3. set -e
  5. function child_function {
  6. 	return 1
  7. }
  8. function parent_function {
  9. 	child_function
  10. 	echo "parent noticed child exit code as $?"
  11. }
  12. function grandparent_function {
  13. 	local ec=0
  14. 	set +e; (set -e; parent_function); ec="$?"; set -e
  15. 	echo "grandparent noticed parent exit code as $ec"
  16. }
  18. grandparent_function

然而，子shell技术会阻止副作用，因为修改不会逃离子shell。

我花了几天时间研究了这个问题，并创建了一个包含各种技术的gist，其中包括一个可以处理副作用的方法：

https://gist.github.com/balupton/21ded5cefc26dc20833e6ed606209e1b

英文:

Thanks to the commenters:

And these supporting resources:

I created this workaround for the question's code sample using the subshell technique:

  1. #!/usr/bin/env bash
  3. set -e
  5. function child_function {
  6. 	return 1
  7. }
  8. function parent_function {
  9. 	child_function
  10. 	echo "parent noticed child exit code as $?"
  11. }
  12. function grandparent_function {
  13. 	local ec=0
  14. 	set +e; (set -e; parent_function); ec="$?"; set -e
  15. 	echo "grandparent noticed parent exit code as $ec"
  16. }
  18. grandparent_function

However, the subshell technique prevents side effects, as modifications do not escape the subshell.

I spent a few more days on this issue, and created a gist with various techniques, including one that works with side effects:

https://gist.github.com/balupton/21ded5cefc26dc20833e6ed606209e1b

huangapple
  • 本文由 发表于 2023年8月8日 23:40:33
  • 转载请务必保留本文链接：https://go.coder-hub.com/76861137.html
  • bash
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