bash 使用 unset 与将变量设置为空
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Using unset vs. setting a variable to empty
提问by helpermethod
I'm currently writing a bash testing framework, where in a test function, both standard bash tests ([[) as well as predefined matchers can be used. Matchers are wrappers to '[[' and besides returning a return code, set some meaningful message saying what was expected.
我目前正在编写一个 bash 测试框架，其中在测试函数中，可以使用标准的 bash 测试 ( [[) 以及预定义的匹配器。匹配器是 '[[' 的包装器，除了返回返回码之外，还设置一些有意义的消息来说明预期的内容。
Example:
例子：
string_equals() {
if [[ ! = ]]; then
error_message="Expected '' to be ''."
return 1
fi
}
So, when a matcher is used, and it fails, only then an error_message is set.
因此，当使用匹配器并且失败时，仅设置 error_message 。
Now, at some point later, I test whether the tests succeeded. If it succeeded, I print the expectation in green, if it failed in red.
现在，在稍后的某个时刻，我测试测试是否成功。如果成功，我用绿色打印期望值，如果失败用红色打印。
Furthermore, there may be an error_message set, so I test if a message exists, print it, and then unset it (because the following test may not set an error_message):
此外，可能有一个 error_message 设置，所以我测试一个消息是否存在，打印它，然后取消设置它（因为下面的测试可能没有设置一个error_message）：
if [[ $error_message ]]; then
printf '%s\n' "$error_message"
unset -v error_message
fi
Now my question is, if it is better to unset the variable, or to just set it to '', like
现在我的问题是，是否最好取消设置变量，或者只是将其设置为 ''，例如
error_message=''
Which one is better? Does it actually make a difference? Or maybe should I have an additional flag indicating that the message was set?
哪一个更好？它真的有区别吗？或者也许我应该有一个额外的标志来表明消息已设置？
回答by cdarke
Mostly you don't see a difference, unless you are using set -u:
大多数情况下，您看不到区别，除非您使用set -u：
/home/user1> var=""
/home/user1> echo $var
/home/user1> set -u
/home/user1> echo $var
/home/user1> unset var
/home/user1> echo $var
-bash: var: unbound variable
So really, it depends on how you are going to test the variable.
所以实际上，这取决于您将如何测试变量。
I will add that my preferred way of testing if it is set is:
我将补充说，我的首选测试方式是：
[[ -n $var ]] # True if the length of $var is non-zero
or
或者
[[ -z $var ]] # True if zero length
回答by Steven Penny
As has been said, using unset is different with arrays as well
如前所述，使用 unset 也与数组不同
$ foo=(4 5 6)
$ foo[2]=
$ echo ${#foo[*]}
3
$ unset foo[2]
$ echo ${#foo[*]}
2
回答by PdC
So, by unset'ting the array index 2, you essentially remove that element in the array and decrement the array size (?).
因此，通过取消设置数组索引 2，您实际上删除了数组中的该元素并减小数组大小 (?)。
I made my own test..
我做了我自己的测试..
foo=(5 6 8)
echo ${#foo[*]}
unset foo
echo ${#foo[*]}
Which results in..
这导致..
3
0
So just to clarify that unset'ting the entire array will in fact remove it entirely.
所以只是为了澄清取消整个数组实际上会完全删除它。
回答by Jonathan H
Based on the comments above, here is a simple test:
根据上面的评论，这里是一个简单的测试：
isunset() { [[ "${!1}" != 'x' ]] && [[ "${!1-x}" == 'x' ]] && echo 1; }
isset() { [ -z "$(isunset "")" ] && echo 1; }
Example:
例子：
$ unset foo; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's unset
$ foo= ; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's set
$ foo=bar ; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's set
