Bash: split line into multiple lines

Question

I have a list of words in lines:

aaaa bbbb ccc dddd
eee fff ggg hhh
iii jjj kkk

I want each word in a separate line:

aaaa
bbbb
ccc
dddd
eee
fff
ggg
hhh
iii
jjj
kkk

How to do that in bash with least number of characters? Without awk preferably.

Answer 1

With pure bash:

while IFS=" " read -r -a line
do
    printf "%s\n" "${line[@]}"
done < file

See:

$ while IFS=" " read -r -a line; do printf "%s\n" "${line[@]}"; done < file
aaaa
bbbb
ccc
dddd
eee
fff
ggg
hhh
iii
jjj
kkk

With xargs:

xargs -n 1 < file

With awk:

awk '{for(i=1;i<=NF;i++) print $i}' file

or

awk -v OFS="\n" '$1=$1' file

With GNU sed:

sed 's/ /\n/g' file

With OS sed:

sed $'s/ /\\\n/g' file

With cut:

cut -d' ' --output-delimiter=$'\n' -f1- file

With grep:

grep -o '[^ ]\+' file

or

grep -Po '[^\s]+' file