复制一个包含空元素的 Bash 数组

Copy a Bash array with empty elements

提问by Benjamin Leinweber

I'm having problems in bash (ver 4.2.25) copying arrays with empty elements. When I make a copy of an array into another variable, it does not copy any empty elements along with it.

我在 bash (ver 4.2.25) 中使用空元素复制数组时遇到问题。当我将一个数组复制到另一个变量中时，它不会同时复制任何空元素。

#!/bin/bash

array=( 'one' '' 'three' )
copy=( ${array[*]} )

IFS=$'\n'

echo "--- array (${#array[*]}) ---"
echo "${array[*]}"

echo
echo "--- copy (${#copy[*]}) ---"
echo "${copy[*]}"

When I do this, here is the output:

当我这样做时，这是输出：

--- array (3) ---
one

three

--- copy (2) ---
one
three

The original array has all three elements including the empty element, but the copy does not. What am I doing wrong here?

原始数组具有包括空元素在内的所有三个元素，但副本没有。我在这里做错了什么？

回答by Carl Norum

You have a quoting problem and you should be using @, not *. Use:

您有引用问题，您应该使用@，而不是*。用：

copy=( "${array[@]}" )

From the bash(1)man page:

从bash(1)手册页：

Any element of an array may be referenced using ${name[subscript]}. The braces are required to avoid conflicts with pathname expansion. If subscriptis @or *, the word expands to all members of 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 IFSspecial variable, and ${name[@]}expands each element of nameto a separate word.

可以使用 引用数组的任何元素${name[subscript]}。需要大括号以避免与路径名扩展发生冲突。如果 subscript是 @或*，则单词扩展到 的所有成员name。这些下标仅在单词出现在双引号内时才不同。如果单词是双引号，则${name[*]}扩展为单个单词，每个数组成员的值由IFS特殊变量的第一个字符分隔 ，并将 的${name[@]}每个元素扩展name为单独的单词。

Example output after that change:

更改后的示例输出：

--- array (3) ---
one

three

--- copy (3) ---
one

three

回答by Steven Penny

Starting with Bash 4.3, you can do this

从Bash 4.3开始，你可以这样做

$ alpha=(bravo charlie 'delta  3' '' foxtrot)

$ declare -n golf=alpha

$ echo "${golf[2]}"
delta  3