A confusion about ${array[*]} versus ${array[@]} in the context of a bash completion

Question

I'm taking a stab at writing a bash completion for the first time, and I'm a bit confused about about the two ways of dereferencing bash arrays (${array[@]} and ${array[*]}).

Here's the relevant chunk of code (it works, by the way, but I would like to understand it better):

_switch() {     local cur perls     local ROOT=${PERLBREW_ROOT:-$HOME/perl5/perlbrew}     COMPREPLY=()     cur=${COMP_WORDS[COMP_CWORD]}     perls=($ROOT/perls/perl-*)     # remove all but the final part of the name     perls=(${perls[*]##*/})      COMPREPLY=( $( compgen -W "${perls[*]} /usr/bin/perl" -- ${cur} ) ) } 

bash's documentation says:

Any element of an array may be referenced using ${name[subscript]}. The braces are required to avoid conflicts with the shell's filename expansion operators. If the subscript is ‘@’ or ‘*’, the word expands to all members of the array 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 variable, and ${name[@]} expands each element of name to a separate word.

Now I think I understand that compgen -W expects a string containing a wordlist of possible alternatives, but in this context I don't understand what "${name[@]} expands each element of name to a separate word" means.

Long story short: ${array[*]} works; ${array[@]} doesn't. I would like to know why, and I would like to understand better what exactly ${array[@]} expands into.

Answer 1

(This is an expansion of my comment on Kaleb Pederson's answer -- see that answer for a more general treatment of [@] vs [*].)

When bash (or any similar shell) parses a command line, it splits it into a series of "words" (which I will call "shell-words" to avoid confusion later). Generally, shell-words are separated by spaces (or other whitespace), but spaces can be included in a shell-word by escaping or quoting them. The difference between [@] and [*]-expanded arrays in double-quotes is that "${myarray[@]}" leads to each element of the array being treated as a separate shell-word, while "${myarray[*]}" results in a single shell-word with all of the elements of the array separated by spaces (or whatever the first character of IFS is).

Usually, the [@] behavior is what you want. Suppose we have perls=(perl-one perl-two) and use ls "${perls[*]}" -- that's equivalent to ls "perl-one perl-two", which will look for single file named perl-one perl-two, which is probably not what you wanted. ls "${perls[@]}" is equivalent to ls "perl-one" "perl-two", which is much more likely to do something useful.

Providing a list of completion words (which I will call comp-words to avoid confusion with shell-words) to compgen is different; the -W option takes a list of comp-words, but it must be in the form of a single shell-word with the comp-words separated by spaces. Note that command options that take arguments always (at least as far as I know) take a single shell-word -- otherwise there'd be no way to tell when the arguments to the option end, and the regular command arguments (/other option flags) begin.

In more detail:

perls=(perl-one perl-two) compgen -W "${perls[*]} /usr/bin/perl" -- ${cur} 

is equivalent to:

compgen -W "perl-one perl-two /usr/bin/perl" -- ${cur} 

...which does what you want. On the other hand,

perls=(perl-one perl-two) compgen -W "${perls[@]} /usr/bin/perl" -- ${cur} 

is equivalent to:

compgen -W "perl-one" "perl-two /usr/bin/perl" -- ${cur} 

...which is complete nonsense: "perl-one" is the only comp-word attached to the -W flag, and the first real argument -- which compgen will take as the string to be completed -- is "perl-two /usr/bin/perl". I'd expect compgen to complain that it's been given extra arguments ("--" and whatever's in $cur), but apparently it just ignores them.

Answer 2

Your title asks about ${array[@]} versus ${array[*]} (both within {}) but then you ask about $array[*] versus $array[@] (both without {}) which is a bit confusing. I'll answer both (within {}):

When you quote an array variable and use @ as a subscript, each element of the array is expanded to its full content regardless of whitespace (actually, one of $IFS) that may be present within that content. When you use the asterisk (*) as the subscript (regardless of whether it's quoted or not) it may expand to new content created by breaking up each array element's content at $IFS.

Here's the example script:

#!/bin/sh  myarray[0]="one" myarray[1]="two" myarray[3]="three four"  echo "with quotes around myarray[*]" for x in "${myarray[*]}"; do         echo "ARG[*]: '$x'" done  echo "with quotes around myarray[@]" for x in "${myarray[@]}"; do         echo "ARG[@]: '$x'" done  echo "without quotes around myarray[*]" for x in ${myarray[*]}; do         echo "ARG[*]: '$x'" done  echo "without quotes around myarray[@]" for x in ${myarray[@]}; do         echo "ARG[@]: '$x'" done 

And here's it's output:

with quotes around myarray[*] ARG[*]: 'one two three four' with quotes around myarray[@] ARG[@]: 'one' ARG[@]: 'two' ARG[@]: 'three four' without quotes around myarray[*] ARG[*]: 'one' ARG[*]: 'two' ARG[*]: 'three' ARG[*]: 'four' without quotes around myarray[@] ARG[@]: 'one' ARG[@]: 'two' ARG[@]: 'three' ARG[@]: 'four' 

I personally usually want "${myarray[@]}". Now, to answer the second part of your question, ${array[@]} versus $array[@].

Quoting the bash docs, which you quoted:

The braces are required to avoid conflicts with the shell's filename expansion operators.

$ myarray= $ myarray[0]="one" $ myarray[1]="two" $ echo ${myarray[@]} one two 

But, when you do $myarray[@], the dollar sign is tightly bound to myarray so it is evaluated before the [@]. For example:

$ ls $myarray[@] ls: cannot access one[@]: No such file or directory 

But, as noted in the documentation, the brackets are for filename expansion, so let's try this:

$ touch one@ $ ls $myarray[@] one@ 

Now we can see that the filename expansion happened after the $myarray exapansion.

And one more note, $myarray without a subscript expands to the first value of the array:

$ myarray[0]="one four" $ echo $myarray[5] one four[5]