shell script函数如何传递数组参数 - scz

admin 2022-11-17 AM 79℃ 0条

参看《20.26 $0 ${1+"$@"}什么意思》

给BASH函数传递数组参数的幺蛾子真多,测试整理了几种方案,其他方案大概率是这几种方案的变种。不得不说,BASH变得跟Perl一样神经病。

!/bin/bash

For bash 5.1.16

./arraytest.sh

Passing arrays as parameters in bash - [2009-06-30]

https://stackoverflow.com/questions/1063347/passing-arrays-as-parameters-in-bash

How to pass an array as function argument - [2015-09-15]

https://askubuntu.com/questions/60218/how-to-add-a-directory-to-the-path

How to pass an array as function argument but with other extra parameters - [2022-02-14]

https://unix.stackexchange.com/questions/690603/how-to-pass-an-array-as-function-argument-but-with-other-extra-parameters

function test_0 ()
{

#
# 甚至可以不显式引用"${@}"
#
# for e in "${@}"
#
for e
do
    echo "${e}"
done

}

function test_1 ()
{

while [ "${#}" -gt 0 ]
do
    echo "${1}"
    shift 1
done

}

function test_2 ()
{

#
# help declare
#
# 若需要indexed array,必须指定-a
#
local -a arr=("${@}")
#
# "${#arr[@]}"
#
local -i len="${#}"
for ((i = 0; i < len; i++))
do
    echo "${arr[$i]}"
done

}

传引用,要求 bash 4.3+

function test_3 ()
{

while [ "${#}" -gt 0 ]
do
    local -n ref="${1}"
    for e in "${ref[@]}"
    do
        echo "${e}"
    done
    shift 1
done

}

可传任意多个数组参数进来

function test_4 ()
{

while [ "${#}" -gt 0 ]
do
    #
    # The ! in ${!1} expands the arg 1 variable
    #
    # ${!parameter} is called indirect reference or sometimes double
    # referenced, this means that instead of using $1's value, we use
    # the value of the expanded value of $1
    #
    for e in "${!1}"
    do
        echo "${e}"
    done
    shift
done

}

function test_main ()
{

local arr_0=("1 one" "2 two" "3 three")
local arr_1=("4 four")
local arr_2=("5 five... 5" "6 six___ 6")

#
# 数组会被展开再传参
#
echo test_0
test_0 "${arr_0[@]}" "${arr_1[@]}" "${arr_2[@]}"
echo

echo test_1
test_1 "${arr_0[@]}" "${arr_1[@]}" "${arr_2[@]}"
echo

echo test_2
test_2 "${arr_0[@]}" "${arr_1[@]}" "${arr_2[@]}"
echo

#
# 传引用
#
echo test_3
test_3 arr_0 arr_1 arr_2
echo

#
# array are passed as names and are expanded in the function. Thus no
# $ is needed when given as parameters.
#
echo test_4
test_4 arr_0[@] arr_1[@] arr_2[@]
echo

}

test_main

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