复制带有空元素的 Bash 数组 [英] Copy a Bash array with empty elements

问题描述

我在 bash(4.2.25 版)复制带有空元素的数组时遇到问题.当我将数组复制到另一个变量中时，它不会随之复制任何空元素.

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.

#!/bin/bash

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

IFS=$'
'

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

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

当我这样做时，输出如下:

When I do this, here is the output:

--- array (3) ---
one

three

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

原始数组包含所有三个元素，包括空元素，但副本没有.我在这里做错了什么?

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

推荐答案

你有一个引用问题，你应该使用 @，而不是 *.使用:

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

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

来自 bash(1) 手册页:

From the bash(1) man page:

可以使用 ${name[subscript]} 引用数组的任何元素.需要大括号以避免与路径名扩展发生冲突.如果subscript 是 @ 或 *，单词扩展到 name 的所有成员.这些仅当单词出现在双引号中时，下标才会有所不同.如果该单词被双引号括起来，${name[*]} 扩展为单个单词IFS 的第一个字符分隔的每个数组成员的值特殊变量，而 ${name[@]} 将 name 的每个元素扩展为一个单独的单词.

Any element of an array may be referenced using ${name[subscript]}. The braces are required to avoid conflicts with pathname expansion. If subscript is @ 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 IFS special variable, and ${name[@]} expands each element of name to a separate word.

更改后的示例输出:

--- array (3) ---
one

three

--- copy (3) ---
one

three

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