Bash array variables: [@] or [*]?

Question

bash-3.2$ echo astr | sed 'hah' | sed 's/s/z/'
sed: 1: "hah": extra characters at the end of h command
bash-3.2$ echo ${PIPESTATUS[*]}
0 1 0
bash-3.2$ echo astr | sed 'hah' | sed 's/s/z/'
sed: 1: "hah": extra characters at the end of h command
bash-3.2$ PIPERET=("${PIPESTATUS[*]}")
bash-3.2$ echo ${PIPERET[*]}
0 1 0
bash-3.2$

This indicates that [*] works fine. But this tut mentions to use [@] instead.

Are both equally valid?

Answer 1

The difference matters mainly when the array elements contain spaces etc. and especially multiple spaces, and is only manifest when the expressions are enclosed in double quotes:

$ x=( '   a  b  c   ' 'd  e  f' )
$ printf "[%s]\n" "${x[*]}"
[   a  b  c    d  e  f]
$ printf "[%s]\n" "${x[@]}"
[   a  b  c   ]
[d  e  f]
$ printf "[%s]\n" ${x[@]}
[a]
[b]
[c]
[d]
[e]
[f]
$ printf "[%s]\n" ${x[*]}
[a]
[b]
[c]
[d]
[e]
[f]
$

Outside double quotes, there's no difference. Inside double quotes, * means 'a single string' and @ means 'array elements individually'.

It is closely analogous to the way $* and $@ (and "$*" and "$@") work.

See the bash manual on:

• arrays
• special parameters

Answer 2

To quote man bash

If subscript is @ or *, the word expands to all members of name. These subscripts differ only when the word appears within double quotes. If the word is double-quoted, ${name[*]} expands to a single word with the value of each array member separated by the first character of the IFS special variable, and ${name[@]} expands each element of name to a sep- arate word. When there are no array members, ${name[@]} expands to nothing. If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word.