bash 完成上下文中 ${array[*]} 与 ${array[@]} 的混淆

A confusion about ${array[*]} versus ${array[@]} in the context of a bash completion

提问by Telemachus

I'm taking a stab at writing a bash completion for the first time, and I'm a bit confused about about the two ways of dereferencing bash arrays (${array[@]}and ${array[*]}).

我第一次尝试编写 bash 完成，我对取消引用 bash 数组（${array[@]}和${array[*]}）的两种方法有点困惑。

Here's the relevant chunk of code (it works, by the way, but I would like to understand it better):

这是相关的代码块（顺便说一下，它可以工作，但我想更好地理解它）：

_switch()
{
    local cur perls
    local ROOT=${PERLBREW_ROOT:-$HOME/perl5/perlbrew}
    COMPREPLY=()
    cur=${COMP_WORDS[COMP_CWORD]}
    perls=($ROOT/perls/perl-*)
    # remove all but the final part of the name
    perls=(${perls[*]##*/})

    COMPREPLY=( $( compgen -W "${perls[*]} /usr/bin/perl" -- ${cur} ) )
}

bash's documentation says:

bash 的文档说：

Any element of an array may be referenced using ${name[subscript]}. The braces are required to avoid conflicts with the shell's filename expansion operators. If the subscript is ‘@' or ‘*', the word expands to all members of the array name. These subscripts differ only when the word appears within double quotes. If the word is double-quoted, ${name[*]} expands to a single word with the value of each array member separated by the first character of the IFS variable, and ${name[@]} expands each element of name to a separate word.

可以使用 ${name[subscript]} 引用数组的任何元素。需要大括号以避免与 shell 的文件名扩展运算符发生冲突。如果下标是 '@' 或 '*'，则单词扩展为数组名称的所有成员。这些下标仅在单词出现在双引号内时才不同。如果单词是双引号，则 ${name[*]} 扩展为单个单词，每个数组成员的值由 IFS 变量的第一个字符分隔，${name[@]} 扩展 name 的每个元素到一个单独的词。

Now I think I understand that compgen -Wexpects a string containing a wordlist of possible alternatives, but in this context I don't understand what "${name[@]} expands each element of name to a separate word" means.

现在我想我明白compgen -W需要一个包含可能替代词列表的字符串，但在这种情况下，我不明白“${name[@]} 将 name 的每个元素扩展为一个单独的词”是什么意思。

Long story short: ${array[*]}works; ${array[@]}doesn't. I would like to know why, and I would like to understand better what exactly ${array[@]}expands into.

长话短说：${array[*]}作品；${array[@]}没有。我想知道为什么，我想更好地理解到底${array[@]}扩展到什么。

回答by Gordon Davisson

(This is an expansion of my comment on Kaleb Pederson's answer -- see that answer for a more general treatment of [@]vs [*].)

（这是我对 Kaleb Pederson 答案的评论的扩展——请参阅该答案以了解对[@]vs的更一般处理[*]。）

When bash (or any similar shell) parses a command line, it splits it into a series of "words" (which I will call "shell-words" to avoid confusion later). Generally, shell-words are separated by spaces (or other whitespace), but spaces can be included in a shell-word by escaping or quoting them. The difference between [@]and [*]-expanded arrays in double-quotes is that "${myarray[@]}"leads to each element of the array being treated as a separate shell-word, while "${myarray[*]}"results in a single shell-word with all of the elements of the array separated by spaces (or whatever the first character of IFSis).

当 bash（或任何类似的 shell）解析一个命令行时，它会将其拆分为一系列“单词”（我将其称为“shell-words”以避免以后混淆）。通常，shell-words 由空格（或其他空格）分隔，但空格可以通过转义或引用它们来包含在 shell-word 中。双引号中的[@]和[*]-expanded 数组之间的区别在于，导致数组的"${myarray[@]}"每个元素都被视为一个单独的 shell-word，而"${myarray[*]}"导致一个单独的 shell-word，数组的所有元素用空格（或无论 的第一个字符IFS是什么）。

Usually, the [@]behavior is what you want. Suppose we have perls=(perl-one perl-two)and use ls "${perls[*]}"-- that's equivalent to ls "perl-one perl-two", which will look for single file named perl-one perl-two, which is probably not what you wanted. ls "${perls[@]}"is equivalent to ls "perl-one" "perl-two", which is much more likely to do something useful.

通常，[@]行为就是你想要的。假设我们有perls=(perl-one perl-two)并使用ls "${perls[*]}"- 这相当于ls "perl-one perl-two"，它将查找名为 的单个文件perl-one perl-two，这可能不是您想要的。 ls "${perls[@]}"相当于ls "perl-one" "perl-two"，这更有可能做一些有用的事情。

Providing a list of completion words (which I will call comp-words to avoid confusion with shell-words) to compgenis different; the -Woption takes a list of comp-words, but it must be in the form of a single shell-word with the comp-words separated by spaces. Note that command options that take arguments always (at least as far as I know) take a single shell-word -- otherwise there'd be no way to tell when the arguments to the option end, and the regular command arguments (/other option flags) begin.

提供完成词列表（我将其称为 comp-words 以避免与 shell-words 混淆）compgen是不同的；该-W选项需要一个合成词列表，但它必须是单个 shell-word 的形式，合成词用空格分隔。请注意，带参数的命令选项始终（至少据我所知）采用单个 shell 字——否则将无法判断选项的参数何时结束，以及常规命令参数（/other选项标志）开始。

In more detail:

更详细地：

perls=(perl-one perl-two)
compgen -W "${perls[*]} /usr/bin/perl" -- ${cur}

is equivalent to:

相当于：

compgen -W "perl-one perl-two /usr/bin/perl" -- ${cur}

...which does what you want. On the other hand,

...它可以满足您的需求。另一方面，

perls=(perl-one perl-two)
compgen -W "${perls[@]} /usr/bin/perl" -- ${cur}

is equivalent to:

相当于：

compgen -W "perl-one" "perl-two /usr/bin/perl" -- ${cur}

...which is complete nonsense: "perl-one" is the only comp-word attached to the -W flag, and the first real argument -- which compgen will take as the string to be completed -- is "perl-two /usr/bin/perl". I'd expect compgen to complain that it's been given extra arguments ("--" and whatever's in $cur), but apparently it just ignores them.

...这完全是胡说八道：“perl-one”是唯一附加到 -W 标志的 comp-word，第一个真正的参数——compgen 将作为要完成的字符串——是“perl-two /usr/bin/perl”。我希望 compgen 会抱怨它被赋予了额外的参数（“--”以及 $cur 中的任何内容），但显然它只是忽略了它们。

回答by Kaleb Pederson

Your title asks about ${array[@]}versus ${array[*]}but then you ask about $array[*]versus $array[@]which is a bit confusing. I'll answer both:

您的标题询问了${array[@]}vs，${array[*]}但随后您询问了$array[*]vs $array[@]，这有点令人困惑。我会回答两个：

When you quote an array variable and use @as a subscript, each element of the array is expanded to its full content regardless of whitespace (actually, one of $IFS) that may be present within that content. When you use the asterisk (*) as the subscript (regardless of whether it's quoted or not) it may expand to new content created by breaking up each array element's content at $IFS.

当您引用一个数组变量并@用作下标时，数组的每个元素都会扩展到其完整内容，而不管该内容$IFS中可能存在的空格（实际上是 之一）。当您使用星号 ( *) 作为下标时（无论它是否被引用），它可能会扩展为通过在 处分解每个数组元素的内容而创建的新内容$IFS。

Here's the example script:

这是示例脚本：

#!/bin/sh

myarray[0]="one"
myarray[1]="two"
myarray[3]="three four"

echo "with quotes around myarray[*]"
for x in "${myarray[*]}"; do
        echo "ARG[*]: '$x'"
done

echo "with quotes around myarray[@]"
for x in "${myarray[@]}"; do
        echo "ARG[@]: '$x'"
done

echo "without quotes around myarray[*]"
for x in ${myarray[*]}; do
        echo "ARG[*]: '$x'"
done

echo "without quotes around myarray[@]"
for x in ${myarray[@]}; do
        echo "ARG[@]: '$x'"
done

And here's it's output:

这是它的输出：

with quotes around myarray[*]
ARG[*]: 'one two three four'
with quotes around myarray[@]
ARG[@]: 'one'
ARG[@]: 'two'
ARG[@]: 'three four'
without quotes around myarray[*]
ARG[*]: 'one'
ARG[*]: 'two'
ARG[*]: 'three'
ARG[*]: 'four'
without quotes around myarray[@]
ARG[@]: 'one'
ARG[@]: 'two'
ARG[@]: 'three'
ARG[@]: 'four'

I personally usually want "${myarray[@]}". Now, to answer the second part of your question, ${array[@]}versus $array[@].

我个人通常想要"${myarray[@]}"。现在，回答您问题的第二部分，${array[@]}而不是$array[@].

Quoting the bash docs, which you quoted:

引用您引用的 bash 文档：

The braces are required to avoid conflicts with the shell's filename expansion operators.

需要大括号以避免与 shell 的文件名扩展运算符发生冲突。

$ myarray=
$ myarray[0]="one"
$ myarray[1]="two"
$ echo ${myarray[@]}
one two

But, when you do $myarray[@], the dollar sign is tightly bound to myarrayso it is evaluated before the [@]. For example:

但是，当您这样做时$myarray[@]，美元符号myarray与[@]. 例如：

$ ls $myarray[@]
ls: cannot access one[@]: No such file or directory

But, as noted in the documentation, the brackets are for filename expansion, so let's try this:

但是，正如文档中所指出的，括号用于文件名扩展，所以让我们试试这个：

$ touch one@
$ ls $myarray[@]
one@

Now we can see that the filename expansion happened after the $myarrayexapansion.

现在我们可以看到文件名扩展发生在扩展之后$myarray。

And one more note, $myarraywithout a subscript expands to the first value of the array:

还有一点要注意，$myarray没有下标会扩展到数组的第一个值：

$ myarray[0]="one four"
$ echo $myarray[5]
one four[5]