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Re: triviality regarding $# counts

On Sat, Apr 13, 2024, at 11:26 PM, Ray Andrews wrote:
> On 2024-04-13 17:30, Lawrence Velázquez wrote:
>> A different output method should make this clearer: 
> Yes, thanks.  Up till now I've basically ignored $* completely.

In a sense, you've been using it all along.  Given an array "arr",
$arr is the same as $arr[*] (by default), which is why double-quoting
it produces a single word.

	% arr=(a '' b 'c   d' '' e)
	% printf '<%s>' $arr; echo
	<a><b><c   d><e>
	% printf '<%s>' $arr[*]; echo
	<a><b><c   d><e>
	% printf '<%s>' $arr[@]; echo
	<a><b><c   d><e>
	% printf '<%s>' ${(@)arr}; echo
	<a><b><c   d><e>
	% printf '<%s>' "$arr"; echo
	<a  b c   d  e>
	% printf '<%s>' "$arr[*]"; echo
	<a  b c   d  e>
	% printf '<%s>' "$arr[@]"; echo
	<a><><b><c   d><><e>
	% printf '<%s>' "${(@)arr}"; echo
	<a><><b><c   d><><e>

-- 
vq
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