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This Bash guide says:

If the index number is @ or *, all members of an array are referenced.

When I do this:

LIST=(1 2 3)
for i in "${LIST[@]}"; do
  echo "example.$i "

Gives: example.1 example.2 example.3 (desired result).

But when I use ${LIST[*]}, I get example.1 2 3 instead.


Edit: when using printf, @ and * actually do give the same results.

share|improve this question
It seems to work for me. I tried both @ and * and it seems to produce the same result both the times. What shell you are using? Run echo $SHELL and paste the output to your question. – Ramesh Jun 7 '14 at 14:26
My example was wrong, this actually happens only with echo, not with printf, I just noticed. – arjan Jun 7 '14 at 14:36
possible duplicate of What is the difference between $* and $@? – goldilocks Jun 7 '14 at 17:13
@goldilocks The other question is about $* and $@. Though, the answer would be similar and one question could be considered a subset of the other, they are different questions. – Stéphane Chazelas Jun 7 '14 at 19:31

1 Answer 1

up vote 4 down vote accepted

The difference is subtle; "$*" creates one argument, while "$@" will expand into separate arguments, so:

LIST=(1 2 3)
for i in "${LIST[@]}"; do
 echo "example.$i"

will deal with the list (print it) as multiple variables


LIST=(1 2 3)
for i in "${LIST[*]}"; do
 echo "example.$i"

will deal with the list as one variable.

share|improve this answer
I provided the wrong example, I just noticed it happens only with echo, not with printf. – arjan Jun 7 '14 at 14:38
@arjan ,see updates – Networker Jun 7 '14 at 14:39
Do you know where the difference between echo and printf comes from? Because with printf in the for loop, the * list reference is treated as multiple variables. – arjan Jun 7 '14 at 14:45
good to know, thanks @arjan – Networker Jun 7 '14 at 14:46

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