Print array elements on separate lines in Bash?

Question

Try doing this :

$ printf '%s\n' "${my_array[@]}"

The difference between $@ and $*:

• Unquoted, the results are unspecified. In Bash, both expand to separate args and then wordsplit and globbed.

• Quoted, "$@" expands each element as a separate argument, while "$*" expands to the args merged into one argument: "$1c$2c..." (where c is the first char of IFS).

You almost always want "$@". Same goes for "${arr[@]}".

Always quote them!

Just quote the argument to echo:

( IFS=$'\n'; echo "${my_array[*]}" )

the sub shell helps restoring the IFS after use

Using for:

for each in "${alpha[@]}"
do
  echo "$each"
done

Using history; note this will fail if your values contain !:

history -p "${alpha[@]}"

Using basename; note this will fail if your values contain /:

basename -a "${alpha[@]}"

Using shuf; note that results might not come out in order:

shuf -e "${alpha[@]}"

Another useful variant is pipe to tr:

echo "${my_array[@]}" | tr ' ' '\n'

This looks simple and compact

I tried the answers here in a giant for...if loop, but didn't get any joy - so I did it like this, maybe messy but did the job:

 # EXP_LIST2 is iterated    
 # imagine a for loop
     EXP_LIST="List item"    
     EXP_LIST2="$EXP_LIST2 \n $EXP_LIST"
 done 
 echo -e $EXP_LIST2

although that added a space to the list, which is fine - I wanted it indented a bit. Also presume the "\n" could be printed in the original $EP_LIST.