如何将数组作为参数传递给 Bash 中的函数 [英] How to pass array as an argument to a function in Bash

问题描述

众所周知，在 bash 编程中，传递参数的方式是 $1, ..., $N.但是，我发现将数组作为参数传递给接收多个参数的函数并不容易.下面是一个例子:

As we know, in bash programming the way to pass arguments is $1, ..., $N. However, I found it not easy to pass an array as an argument to a function which receives more than one argument. Here is one example:

f(){
 x=($1)
 y=$2

 for i in "${x[@]}"
 do
  echo $i
 done
 ....
}

a=("jfaldsj jflajds" "LAST")
b=NOEFLDJF

f "${a[@]}" $b
f "${a[*]}" $b

如上所述，函数f接收两个参数:第一个被分配给x，这是一个数组，第二个被分配给y.

As described, function freceives two arguments: the first is assigned to x which is an array, the second to y.

f 可以通过两种方式调用.第一种方式使用"${a[@]}"作为第一个参数，结果为:

f can be called in two ways. The first way use the "${a[@]}" as the first argument, and the result is:

jfaldsj 
jflajds

第二种方式使用"${a[*]}"作为第一个参数，结果为:

The second way use the "${a[*]}" as the first argument, and the result is:

jfaldsj 
jflajds 
LAST

结果都不如我所愿.那么，有没有人知道如何正确地在函数之间传递数组?

Neither result is as I wished. So, is there anyone having any idea about how to pass array between functions correctly?

推荐答案

不能传递数组，只能传递其元素(即扩展数组).

You cannot pass an array, you can only pass its elements (i.e. the expanded array).

#!/bin/bash
function f() {
    a=("$@")
    ((last_idx=${#a[@]} - 1))
    b=${a[last_idx]}
    unset a[last_idx]

    for i in "${a[@]}" ; do
        echo "$i"
    done
    echo "b: $b"
}

x=("one two" "LAST")
b='even more'

f "${x[@]}" "$b"
echo ===============
f "${x[*]}" "$b"

另一种可能性是按名称传递数组:

The other possibility would be to pass the array by name:

#!/bin/bash
function f() {
    name=$1[@]
    b=$2
    a=("${!name}")

    for i in "${a[@]}" ; do
        echo "$i"
    done
    echo "b: $b"
}

x=("one two" "LAST")
b='even more'

f x "$b"

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