Question

我正在学习bash。现在我想给一个函数一个空数组。但是，它不起作用。请参考以下内容，

function display_empty_array_func() {
  local -a aug1=$1
  local -a aug2=${2:-no input}
  echo "array aug1 = ${aug1[@]}"
  echo "array aug2 = ${aug2[@]}"
}

declare -a empty_array=()

display_empty_array_func "${empty_array[@]}" "1"

输出如下，

array aug1 = 1  # I expected it is empty
array aug2 = no input  # I expected it is 1

根据我的理解，引用变量允许我们给出一个空变量，喜欢以下，

function display_empty_variable_func() {  
  local aug1=$1
  local aug2=${2:-no input}
  echo "variable aug1 = ${aug1}"
  echo "variable aug2 = ${aug2}"
}

display_empty_variable_func "" "1"

其产出如下

variable aug1 =    # it is empty as expected
variable aug2 = 1  # it is 1 as expected

我不知道传递空数组有什么问题。知道机制或解决方案的人。请让我知道。非常感谢你。

Answer 1

enter image description here如果位置参数为空，则shell脚本＆amp; shell函数不会考虑该值，而是将下一个非空值作为其值（位置参数值）。如果我们想要取空值，我们必须将该值放在引号中。

Ex: I'm having small script like 
cat test_1.sh
#!/bin/bash

echo "First Parameter is :"$1;

echo "Second Parameter is :"$2;

case -1
If i executed this script as 
sh test_1.sh

First Parameter is :

Second Parameter is :

Above two lines are empty because i have not given positional parameter values to script.

case-2
If i executed this script as 
sh test_1.sh 1 2

First Parameter is :1

Second Parameter is :2

case-2
If i executed this script as 
sh test_1.sh  2 **# here i given two spaces in between .sh and 2 and i thought 1st param is space and second param is 2 but**

First Parameter is :2

Second Parameter is :

The output looks like above. My first statement will apply here.

case-4
If i executed this script as 
sh test_1.sh " " 2

First Parameter is :

Second Parameter is :2

Here i kept space in quotes. Now i'm able to access the space value.

**This will be help full for you. But for your requirement please use below code.** 

In this you have **quote** the space value(""${empty_array[@]}"") which is coming from **an empty array**("${empty_array[@]}"). So you have to use additional quote to an empty array.

function display_empty_array_func() {
  local -a aug1=$1
  local -a aug2=${2:-no input}
  echo "array aug1 = ${aug1[@]}"
  echo "array aug2 = ${aug2[@]}"
}

declare -a empty_array=()

display_empty_array_func ""${empty_array[@]}"" "1"

The output is:
array aug1 =
array aug2 = 1

如何在bash中提供一个空数组？

1 个答案: