Is there a way to find the length of the array *(files names) in zsh without using a for loop to increment some variable?
I naively tried echo ${#*[@]} but it didn't work. (bash syntax are welcome as well)
2 Answers
${#*[@]} would be the length of the $* array also known as $@ or $argv, which is the array of positional parameters (in the case of a script or function, that's the arguments the script or function received). Though you'd rather use $# for that.
* alone is just a glob pattern. In list context, that's expanded to the list of files in the current directory that match that pattern. As * is a pattern that matches any string, it would expand to all file names in the current directory (except for the hidden ones).
Now you need to find a list context for that * to be expanded, and then somehow count the number of resulting arguments. One way could be to use an anonymous function:
() {echo There are $# non hidden files in the current directory} *(N) Instead of *, I used *(N) which is * but with the N (for nullglob) globbing qualifier which makes it so that if the * pattern doesn't match any file, instead of reporting an error, it expands to nothing at all.
The expansion of *(N) is then passed to that anonymous function. Within that anonymous function, that list of file is available in the $@/$argv array, and we get the length of that array with $# (same as $#argv, $#@, $#* or even the awkward ksh syntax like ${#argv[@]}).
answered Feb 9, 2019 at 20:01
- Would you actually call
$*an array at all? A "string" would be a better word for it, IMHO. Maybe it's different inzsh?– Kusalananda♦CommentedFeb 10, 2019 at 7:58 - @Kusalananda how could it be a string if the elements it's made of are words/strings themselves? When
$*is used inside double quotes, the words it's made of ($1,$2, ...) will be joined by the first char ofIFSwithout its elements being split onIFSor spaces before that:(set -- 'a/b' 'a b'; IFS=:/; echo "<$*>")(this latter digression is to illustrate that$*is not somehow a string split an then rejoined in this particular syntax).– user313992CommentedFeb 10, 2019 at 10:05 - @mosvy I know he mentions
$*and$@unquoted, but"$*"is at least clearly a string whereas"$@"is not a single string. Calling$*an array and saying another name for it is$@seems to confuse the purpose of these two variables.– Kusalananda♦CommentedFeb 10, 2019 at 10:11 - @Kusalananda, it's the array of positional parameters.
zshandyashhave real arrays and arrays whose indice start at 1, so they can represent that array as a normal array variable. That array goes by the*,@andargvnames inzsh. Except that$@is special inside double quotes on list contexts like it was in the Bourne shell.$*inside double quotes expands to the concatenation of the elements like$argvor$argv[*](or even$*[*]) do, but that doesn't make it less that$argvis an array variable, not a scalar variable.CommentedFeb 17, 2019 at 18:02
files=(*) printf 'There are %d files\n' "${#files[@]}" or
set -- * printf 'There are %d files\n' "$#" You have to name the array first (as I did above with files) or use the built-in array $@ by populating it with the wildcard, as I did in the second example. In the former, the "length" (number of files) of the array is done with the ${#arrayname[@]} syntax. The number of elements in the built-in array is in $#.
answered Feb 9, 2019 at 1:43
- So the
*acts as a regex expression? I thought that it was a special array... although surprise me that there is no such array...CommentedFeb 9, 2019 at 1:59 - wait, in zsh
*[0] or *[1]gives me some file name... now I'm puzzled.CommentedFeb 9, 2019 at 2:02 - The array is $@... $* equals a single string with each argument separated by $IFS (usually space). ZSH and Bash doesn't handle usage of an "index" in space-delimited strings the same way.– svin83CommentedFeb 9, 2019 at 3:19
- Cristiano, the
*[0]in zsh would expand to filenames that start with anything (or nothing) followed by any of the following characters:0. The square brackets designates a set of characters in that position. You likely have files that end with a zero in your current directory. To tell zsh to give you back the first element of a*wildcard, you'd use parenthesis for array syntax:print -l *([1])-- noting that zsh arrays start at 1 unless you set$KSH_ARRAYSCommentedFeb 9, 2019 at 17:14
*suggest you want to use shell globing mechanism here)?*is not an array in the way you are using it, it is a shell glob. Arrays have nothing to do with your question unless you create an array as Jeff did in his answer. Your question is "How do I find how many files are in the current directory"echo *[0]in zsh prints the 1st file name...