Bash and IFS: string split to an array with specific separator add an extra empty element

About split a string into an array we have two scenarios:

if the string has empty spaces how a separator, according with the following post:

• Reading a space-delimited string into an array in Bash

So If I use:

string="Hello Unix World"
array1=($string)
echo ${array1[@]}
echo "size: '${#array1[@]}'"

read -a array2 <<< $string
echo ${array2[@]}
echo "size: '${#array2[@]}'"

The output is:

Hello Unix World
size: '3'
Hello Unix World
size: '3'

Both approaches works as expected.

Now, if the string has something different than an empty space how a separator, according with the following post:

• How to split a string into an array in Bash?

So If I use:

path="/home/human/scripts"
IFS='/' read -r -a array <<< "$path"

echo "Approach 1"
echo ${array[@]}
echo "size: '${#array[@]}'"

echo "Approach 2"
for i in "${array[@]}"; do
   echo "$i"
done

echo "Approach 3"
for (( i=0; i < ${#array[@]}; ++i )); do
    echo "$i: ${array[$i]}"
done

It prints:

Approach 1
home human scripts    <--- apparently all is ok, but see the line just below!
size: '4'
Approach 2
                      <--- an empty element
home
human
scripts
Approach 3
0:                    <--- confirmed, the empty element
1: home
2: human
3: scripts

Why appears that empty element? How fix the command to avoid this situation?

>Solution :

Your string split into 4 parts: an empty one, and the three words.

path="/home/human/scripts"
IFS='/' read -r -a array <<< "$path"
declare -p array

Output:

declare -a array=([0]="" [1]="home" [2]="human" [3]="scripts")

There are many ways to fix it. One is to delete the empty values. Another is to exclude the beginning slash before splitting.

for i in "${!array[@]}"; do
    [[ ${array[i]} ]] || unset 'array[i]'
done

Or

IFS='/' read -r -a array <<< "${path#/}"

The first one is adaptable to path forms were slashes are repeated not only in the beginning.