(Bash) 从函数内编辑数组的内容

Question

我试图用空格将数组的所有项目填充到 20 个字符，但似乎无法让我的循环正常工作。它似乎通过数组项正确递增，但不会更改项。我哪里出错了？

#!/bin/bash

testArray=( "bish" "bash" "bosh")

padLine () {
array=( "${@}" )
testLength=20
counter=0

##loop begins here##
for i in "${array[@]}";
do
size=${#array[$counter]}
testLength=20

#echo ""
#echo "size: " $size
#echo "Tlength: " $testLength
#echo "count: " ${array[$counter]}
#echo ""

if [ $size -lt $testLength ]
then 
    offset=$( expr $testLength - $size )

    #echo "Offset: " $offset

    case $offset in
        0)
            l0=""
            ;;
        1)
            l1=" "
            array[$counter]=${array[$counter]/%/$l1};;
        2)
            l2="  "
            array[$counter]="${array[$counter]/%/$l2}";;
        3)
            l3="   "
            array[$counter]=${array[$counter]/%/$l3};;
        4)
            l4="    "
            array[$counter]="${array[$counter]/%/$l4}";;
        5)
            l5="     "
            array[$counter]="${array[$counter]/%/$l5}";;
        6)
            l6="      "
            array[$counter]=${array[$counter]/%/$l6};;
        7)
            l7="       "
            array[$counter]=${array[$counter]/%/$l7};;
        8)
            l8="        "
            array[$counter]=${array[$counter]/%/$l8};;
        9)
            l9="         "
            array[$counter]=${array[$counter]/%/$l9};;
        10)
            l10="          "
            array[$counter]=${array[$counter]/%/$l10};;
        11)
            l11="           "
            array[$counter]=${array[$counter]/%/$l11};;
        12)
            l12="            "
            array[$counter]=${array[$counter]/%/$l12};;
        13)
            l13="             "
            array[$counter]=${array[$counter]/%/$l13};;
        14)
            l14="              "
            array[$counter]=${array[$counter]/%/$l14};;
        15)
            l15="               "
            array[$counter]=${array[$counter]/%/$l15};;
        16)
            l16="                "
            array[$counter]=${array[$counter]/%/$l16};;
        17)
            l17="                 "
            array[$counter]=${array[$counter]/%/$l17};;
        18)
            l18="                  "
            array[$counter]=${array[$counter]/%/$l18};;
        19)
            l19="                   "
            array[$counter]=${array[$counter]/%/$l19};;

        *)
    esac
fi
counter=$( expr $counter + 1 )  
done
}

padLine "${testArray[@]}"


echo -e "${testArray[0]}" 
echo -e "${testArray[1]}"
echo -e "${testArray[2]}"

预期输出：

bish                #lines end here, padded to 20 chars
bash                #                
bosh                #

实际输出：

bish# no padding
bash
bosh

Answer 1

仅用于输出：

array=( bish bash bosh )
printf '%-20s#\n' "${array[@]}"

这会产生

bish                #
bash                #
bosh                #

...#出现在第 21 列的地方。

要创建一个新数组（并打印它）：

array=( bish bash bosh )

for elem in "${array[@]}"; do
    padarr+=( "$( printf '%-20s#' "$elem" )" )
done

printf '%s\n' "${padarr[@]}"

使用/bin/sh，只需打印：

set -- bish bash bosh
printf '%-20s#\n' "$@"

使用/bin/sh，$@就地修改：

set -- bish bash bosh
i=0
while [ "$i" -lt "$#" ]; do
    set -- "$@" "$( printf '%-20s#' "$1" )"
    shift
    i=$(( i + 1 ))
done

printf '%s\n' "$@"

该printf格式化字符串%-20s储量20个字符左对齐字符串。

作为bash(4.3+) 函数：

pad_array () {
    local padlen=$1
    local -n localarray=$2

    local -a tmp
    local elem

    for elem in "${localarray[@]}"; do
        tmp+=( "$( printf '%-*s#' "$padlen" "$elem" )" )
    done

    localarray=( "${tmp[@]}" )
}

myarray=( bish bash bosh )
pad_array 20 myarray

printf '%s\n' "${myarray[@]}"

pad_array此处的功能还允许您选择填充量。

该数组按其名称传递，并由名称引用变量中的函数接收。这意味着无论何时在函数中访问名称引用，都会实际使用命名变量。