Re: The ~ parameter expansion flag: bug or misunderstanding

[Clint Hepner <clint.hepner@xxxxxxxxx>]

Thanks. One thing I still can't quite wrap my mind around is a scenario
where

    ${(j.|.)~array}

and

    ${~j.|.)array}

would behave differently. I assume it would involve some additional
modifications made to $array; I tried something like

    array=(foo bar)
    [[ "?" = ${(~j.|.)array:s/foo/?/} ]]   # match
    [[ "?" = ${(j.|.)~array:s/foo/?/} ]]   # match

but that was not sufficient.

On Wed, Sep 3, 2014 at 11:44 AM, Bart Schaefer <schaefer@xxxxxxxxxxxxxxxx>
wrote:

> On Sep 3, 10:58am, Clint Hepner wrote:
> }
> } My reading of the ~ flag leads me to believe that you can replace the
> } literal string with a pattern, so that
> }
> } % print -l ${(~s.[12].):-a1b2c}
> } a
> } b
> } c
>
> No, that's not what the flag means.  It means that after the expansion
> of the parameter has been completed, any pattern characters inserted
> into the result are active (i.e., not considered quoted).  It doesn't
> otherwise change the operation of the other flags themselves.
>
> This doesn't have any effect for the (s) flag because (s) removes
> characters from the result, rather than inserting them.
>
> The purpose of the (~) flag is really to work with the (j) flag, so
> that you can create a pattern from an array without having any other
> pattern characters in the values of the array also become active.
> It also works with (l) and (r), though the uses for treating padding
> as pattern characters are pretty obscure.
>
> To use a pattern for processing the value during expansion, you need
> to use the trailing // notation, e.g.
>
>     % print -l ${=${:-a1b2c}//[12]/ }
>
> or
>
>     % print -l ${(s:1:)${:-a1b2c}//[12]/1}
>