Re: triviality regarding $# counts

[Ray Andrews <rayandrews@xxxxxxxxxxx>]

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