A minor syntax question

A minor syntax question

[Jesper Nygårds]

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I am writing a function that's basically a wrapper to an invocation of
'find'. I have the need to transform some arguments to my function into an
expression I pass on to 'find'. Here's a simplified example of what I want
to do:

xff() {
    zparseopts -D -E t+:=types || return 1
    typearg="("${"$(print -- '-o -type '${^types:#-t})"#-o }")"
    print $typearg
}

What I want to do is to construct a nested expression for the variable
number of file types that can be given. They should be contained within
parenthesis, and joined by "-o". So, for example:

xff -t f   ->  "(-type f)"
xff -t f -t d   -> "(-type f -o -type d)"

As you can see, my function strips away the '-t' elements, then expands the
parameter list with '-o -type ' before each element, and finally strips
away the leading '-o'. It works as intended, but I feel it is more
complicated than required. In particular, I couldn't find a way to make the
'${^...}' parameter expansion trigger without the embedded print statement.
At the same time, I'm rather happy with having found a one-liner expressing
what I want.

So my question is: is there a more elegant way of solving this which is
still compact?

Re: A minor syntax question

[Oliver Kiddle]

Jesper Nygårds wrote:
>     typearg="("${"$(print -- '-o -type '${^types:#-t})"#-o }")"

> away the leading '-o'. It works as intended, but I feel it is more
> complicated than required. In particular, I couldn't find a way to make the
> '${^...}' parameter expansion trigger without the embedded print statement.

Joining the array to form a string should avoid the need for a print,
allowing the #-o to apply to the string as a whole. E.g:

typearg=${${(j. .):-'-o -type '${^types:#-t}}#-o }

Shorter versions should be possible such as the following:

typearg="${${=types//-t/-o -type}[2,-1]}"

I'd be inclined to keep typearg as an array, however.

Oliver

Re: A minor syntax question

[Jesper Nygårds]

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Both your examples work great, Oliver. Thank you.

Inspired by your second example I also found a solution which more closely
follows the way I have been thinking about the problem, and that I find
quite readable:

typearg="("${(j: -o :)${types:#-t}/#/-type }")"

On Tue, Jun 7, 2016 at 1:54 PM, Oliver Kiddle <okiddle@yahoo.co.uk> wrote:

> Jesper Nygårds wrote:
> >     typearg="("${"$(print -- '-o -type '${^types:#-t})"#-o }")"
>
> > away the leading '-o'. It works as intended, but I feel it is more
> > complicated than required. In particular, I couldn't find a way to make
> the
> > '${^...}' parameter expansion trigger without the embedded print
> statement.
>
> Joining the array to form a string should avoid the need for a print,
> allowing the #-o to apply to the string as a whole. E.g:
>
> typearg=${${(j. .):-'-o -type '${^types:#-t}}#-o }
>
> Shorter versions should be possible such as the following:
>
> typearg="${${=types//-t/-o -type}[2,-1]}"
>
> I'd be inclined to keep typearg as an array, however.
>
> Oliver
>