引用数组中所有元素时${arr[*]}和${arr[@]}是有细微区别的

Example:

#!/bin/sh
function showarr(){
        arr=$1
        for b in ${arr[*]};do
                echo $b
        done
        return 0
}
regions=('aa pp' 'bb' 'cc')
showarr $regions
exit 0

详解shell数组${arr[*]}和${arr[@]}区别

$regions其实只引用了数组的第一个元素

#!/bin/sh
function showarr(){
        arr=$1
        for b in ${arr[*]};do
                echo $b
        done
        return 0
}
regions=('aa pp' 'bb' 'cc')
showarr ${regions[*]}
echo ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
showarr ${regions[@]}
exit 0

详解shell数组${arr[*]}和${arr[@]}区别

引用了数组全部元素,但是showarr函数中arr变量只获取了第一个参数的值 "aa"

#!/bin/sh
function showarr(){
        arr=$1
        for b in ${arr[*]};do
                echo $b
        done
        return 0
}
regions=('aa pp' 'bb' 'cc')
showarr "${regions[*]}"
echo ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
showarr "${regions[@]}"
exit

详解shell数组${arr[*]}和${arr[@]}区别

此种情况跟 $* $@比较类似,${regions[*]}把参数打散,作为一个字符串整体传递,原有参数结构被破坏

${regions[@]}保持了原参数结构,因此$1其实为 'aa pp'

#!/bin/sh
function showarr(){
        arr=$1
        for b in ${arr[*]};do
                echo $b
        done
        echo "\$1: $1"
        echo "\$2: $2"
        echo "\$3: $3"
        return 0
}
regions=('aa pp' 'bb' 'cc')
showarr "${regions[*]}"
echo ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
showarr "${regions[@]}"
exit 0

详解shell数组${arr[*]}和${arr[@]}区别

改进后的脚本执行情况验证了我们的猜想

结论: $@ $* ${arr[@]} ${arr[*]} 类似,加不加"", 使用@还是*根据实际情况选择

不加""的时候,@ 和 * 完全一样,加""时,@可以保持原有参数结构,*将原有参数结构打乱

shell对 "$@"会做特殊处理,"$*"可以认为是普通字符串

#!/bin/sh
for p in "$*";do
        echo $p
done
echo ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
for p in "$@";do
        echo $p
done

详解shell数组${arr[*]}和${arr[@]}区别

"$@" 已经不是简单的字符串了

原文地址:https://www.cnblogs.com/dissipate/p/13724003.html

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