在带有函数的 bash 中使用 set -e / set +e
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Using set -e / set +e in bash with functions
提问by Michel Müller
I've been using a simple bash preamble like this in my scripts:
我一直在我的脚本中使用这样一个简单的 bash 序言：
#!/bin/bash
set -e
In conjunction with modularity / using functions this has bitten me today.
结合模块化/使用功能，这让我今天很头疼。
So, say I have a function somewhere like
所以，说我在某个地方有一个功能
foo() {
#edit: some error happens that make me want to exit the function and signal that to the caller
return 2
}
Ideally I'd like to be able to use multiple small files, include their functions in other files and then call these functions like
理想情况下，我希望能够使用多个小文件，将它们的函数包含在其他文件中，然后调用这些函数，例如
set +e
foo
rc=$?
set -e
. This works for exactly two layers of routines. But if foo is also calling subroutines like that, the last setting before the return will be set -e, which will make the script exit on the return - I cannot override this in the calling function. So, what I had to do is
. 这正好适用于两层例程。但是，如果 foo 也像这样调用子例程，则返回之前的最后一个设置将是set -e，这将使脚本在返回时退出 - 我无法在调用函数中覆盖它。所以，我必须做的是
foo() {
#calling bar() in a shielded way like above
#..
set +e
return 2
}
Which I find very counterintuitive (and also not what I want - what if in some contexts I'd like to use the function without shielding against failures, while in other contexts I want to handle the cleanup?) What's the best way to handle this? Btw. I'm doing this on OSX, I haven't tested whether this behaviour is different on Linux.
我觉得这非常违反直觉（也不是我想要的 - 如果在某些情况下我想使用该函数而不屏蔽故障，而在其他情况下我想处理清理呢？）处理这个问题的最佳方法是什么？ ? 顺便提一句。我在 OSX 上这样做，我还没有测试过这种行为在 Linux 上是否不同。
回答by Martin Tournoij
Shell functions don't really have "return values", just exit codes.
Shell 函数并没有真正的“返回值”，只有退出代码。
You could add && :to the caller, this makes the command "tested", and won't exit it:
您可以添加&& :到调用者，这会使命令“经过测试”，并且不会退出：
foo() {
echo 'x'
return 42
}
out=$(foo && :)
echo $out
The :is the "null command" (ie. it doesn't do anything). In this case it doesn't even get executed, since it only gets run if fooreturn 0 (which it doesn't).
这:是“空命令”（即它不做任何事情）。在这种情况下，它甚至不会被执行，因为它只有在foo返回 0 时才会运行（它没有）。
This outputs:
这输出：
x
It's arguably a bit ugly, but then again, all of shell scripting is arguably a bit ugly ;-)
可以说它有点难看，但话说回来，所有的 shell 脚本都可以说有点难看 ;-)
Quoting sh(1)from FreeBSD, which explains this better than bash's man page:
引用sh(1)FreeBSD，它比 bash 的手册页更好地解释了这一点：
-e errexit
Exit immediately if any untested command fails in non-interactive
mode. The exit status of a command is considered to be explicitly
tested if the command is part of the list used to control an if,
elif, while, or until; if the command is the left hand operand of
an “&&” or “||” operator; or if the command is a pipeline preceded
by the ! operator. If a shell function is executed and its exit
status is explicitly tested, all commands of the function are con‐
sidered to be tested as well.
