bash - zsh 与 bash : how do parenthesis alter variable assignment behavior?

Question

我对如何在不同的现有 shell 中处理变量赋值和括号有一些麻烦和误解。

目前令我困惑的是:

总是使用下面的命令

./script.sh a b c d

运行以下代码时

#!/bin/zsh

bar=$@

for foo in $bar
do
    echo $foo
done

输出是

a b c d

与

#!/bin/zsh

bar=($@)

for foo in $bar
do
    echo $foo
done

是(我最初想要的)

a
b
c
d

但使用 bash 或 sh

#!/bin/bash

bar=$@

for foo in $bar
do
    echo $foo
done

给予

a
b
c
d

和

#!/bin/bash

bar=($@)

for foo in $bar
do
    echo $foo
done

只是

a

那里发生了什么？

Answer 1

联合行动

对于涉及的两个 shell，给出的示例将假定一个显式设置的 argv 列表:

# this sets $1 to "first entry", $2 to "second entry", etc
$ set -- "first entry" "second entry" "third entry"

在这两个 shell 中，declare -p 可用于以明确的形式发出变量名称的值，尽管它们表示该形式的方式可能有所不同。

在庆典中

bash 中的扩展规则通常与 ksh 兼容，并且在适用的情况下与 POSIX sh 语义兼容。与这些 shell 兼容需要不带引号的扩展执行字符串拆分和 glob 扩展(例如，将 * 替换为当前目录中的文件列表)。

在变量赋值中使用括号使其成为数组。比较这三个作业:

# this sets arr_str="first entry second entry third entry"
$ arr_str=$@
$ declare -p arr_str
declare -- arr="first entry second entry third entry"

# this sets arr=( first entry second entry third entry )
$ arr=( $@ )
declare -a arr='([0]="first" [1]="entry" [2]="second" [3]="entry" [4]="third" [5]="entry")'

# this sets arr=( "first entry" "second entry" "third entry" )
$ arr=( "$@" )
$ declare -p arr
declare -a arr='([0]="first entry" [1]="second entry" [2]="third entry")'

同样，在展开时，引号和印记很重要:

# quoted expansion, first item only
$ printf '%s\n' "$arr"
first entry

# unquoted expansion, first item only: that item is string-split into two separate args
$ printf '%s\n' $arr
first
entry

# unquoted expansion, all items: each word expanded into its own argument
$ printf '%s\n' ${arr[@]}
first
entry
second
entry
third
entry

# quoted expansion, all items: original arguments all preserved
$ printf '%s\n' "${arr[@]}"
first entry
second entry
third entry

在 zsh 中

zsh 做了很多神奇的事情来尝试按照用户的意思去做，而不是与历史 shell(ksh、POSIX sh 等)兼容。然而，即使在那里，做错事也会产生与你想要的不同的结果:

# Assigning an array to a string still flattens it in zsh
$ arr_str=$@
$ declare -p arr_str
typeset arr_str='first entry second entry third entry'

# ...but quotes aren't needed to keep arguments together on array assignments.
$ arr=( $@ )
$ declare -p arr
typeset -a arr
arr=('first entry' 'second entry' 'third entry')

# in zsh, expanding an array always expands to all entries
$ printf '%s\n' $arr
first entry
second entry
third entry

# ...and unquoted string expansion doesn't do string-splitting by default:
$ printf '%s\n' $arr_str
first entry second entry third entry