2 
# Create array
arrayLong=(one two three four)
for element in "${arrayLong[@]}"
do
    echo "$element"
done
echo "${#arrayLong[@]}"

Output:

one
two
three
four
4

Then:

# Make new array with only first half of values
arrayShort=("${arrayLong[@]:0:2}")
for element in "${arrayShort[@]}"
do
    echo "$element"
done
echo "${#arrayShort[@]}"

The output of this is

one two
1

Why is my short array not actually an array? It's just one element. How can I split my array when the results are full on arrays?

My bash version is GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin16)

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10
  • 1
    It works as it should for me. Try not quoting arrayLong though, like: arrayShort=(${arrayLong[@]:0:2}). Can you also post the output of: echo "${IFS@Q}" Commented Sep 22, 2017 at 23:02
  • 3
    What's your bash version? Commented Sep 22, 2017 at 23:17
  • 1
    GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin16) Commented Sep 22, 2017 at 23:47
  • 1
    @teradyl I'm running the same version (under macOS 10.12.5), and I get "2" for the length of arrayShort. Try declare -p arrayShort and see what that says (I get declare -a arrayShort='([0]="one" [1]="two")'). Commented Sep 23, 2017 at 3:33
  • 2
    Based on his output I think it's safe to say that his arrayShort is declare -a arrayShort='([0]="one two")'. My suspicion is that his IFS does not include a space. Commented Sep 23, 2017 at 10:28

1 Answer 1

Reset to default
3

I was unaware that IFS does not reset automatically after using it. Earlier in the code, I had set the IFS=$'\n' without storing the original value. Here's what I should have done:

# set Internal Field Separator to new line only to split files
oIFS="$IFS"
IFS=$'\n'
array=(${all_files})
# Return IFS to initial value
IFS="$oIFS"

To double check what the IFS is at a given moment, try printf "%q\n" "$IFS". The default value should be $' \t\n'

Thanks @MiniMax & @Jesse_b for helping with this.

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