Re: triviality regarding $# counts

[Ray Andrews <rayandrews@eastlink.ca>]

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On 2024-04-14 07:06, Mark J. Reed wrote:
>
> What? No.  Why do you keep bringing (f) into this? That flag has 
> nothing to do with copying; in fact it intentionally changes things: 
> anywhere there used to be a single string containing a newline,you get 
> two strings instead.
% hhh=( "${(@f)iii}" )
% typeset -p hhh; typeset -p iii
typeset -a hhh=( 'abc\ndef\nghi' )
typeset iii='abc\ndef\nghi'

... newlines still there.  One element.


Besides it worked as I showed.
>
> But if you drop the f, you do get your Xerox copy:
>
>     % fff=( "${(@)ddd}" )
>
... so far so good with tests.  The (f) seems not to have removed 
anything -- I was sure it did, yesterday :( -- but that was then. But 
yours seems to work and it's simpler, more intuitive.  The (f) does not 
seem to be missed, either.  And to think that just yesterday I thought I 
had this all sorted.

> Would it be nicer if you could just do *fff=$ddd* and not have to 
> include the parens and quotes and @? Sure. You could even make a case 
> that it /should /work that way, since we're in Zshland where *$ddd* 
> expands to the whole array instead of a single element. But that's not 
> the way assignment works. Though, as I said, if the array has no empty 
> elements, you can get away with just *fff=($ddd)*.
'IF' ... what I'm wanting is some universal copy that has no gotchas.

BTW, just philosophically speaking, the parens are an interesting 
question.  Whereas I'd say that a copy is a copy is a copy:

aaa=$bbb

... so whatever bbb may happen to be, aaa will become the same.  But if 
aaa is a scalar, should it be promoted 'silently' ?  Seems to me the 
parens are explicit that aaa will become an array so I like them.  One 
can never be too clear as to what's happening.

cut


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