Reading a space-delimited string into an array in Bash

Question

I have a variable which contains a space-delimited string:

line="1 1.50 string"

I want to split that string with space as a delimiter and store the result in an array, so that the following:

echo ${arr[0]}
echo ${arr[1]}
echo ${arr[2]}

outputs

1
1.50
string

Somewhere I found a solution which doesn't work:

arr=$(echo ${line})

If I run the echo statements above after this, I get:

1 1.50 string
[empty line]
[empty line]

I also tried

IFS=" "
arr=$(echo ${line})

with the same result. Can someone help, please?

Answer 1

In: arr=( $line ). The "split" comes associated with "glob".
Wildcards (*,? and []) will be expanded to matching filenames.

The correct solution is only slightly more complex:

IFS=' ' read -a arr <<< "$line"

No globbing problem; the split character is set in $IFS, variables quoted.

Answer 2

Try this:

arr=(`echo ${line}`);

Answer 3

If you need parameter expansion, then try:

eval "arr=($line)"

For example, take the following code.

line='a b "c d" "*" *'
eval "arr=($line)"
for s in "${arr[@]}"; do 
    echo "$s"
done

If the current directory contained the files a.txt, b.txt and c.txt, then executing the code would produce the following output.

a
b
c d
*
a.txt
b.txt
c.txt