(z) word splitting on one word

(z) word splitting on one word

[Julian Prein]

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Hi!

I am trying to get the last word of LBUFFER.

What I did previously was:

    ${${(z)LBUFFER}[-1]}

This worked flawlessly until I noticed today that it misbehaves if LBUFFER
contains only one word. In this case it seems like `[-1]` acts on a scalar, as
it expands to the last character instead of LBUFFER as I would expect.

Is this expected behaviour?

I now work around it by prepending LBUFFER with a dummy word so that it always
has at least two words (I know that it is not empty). But this feels very hacky.

Do you know of a better way?

Julian

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Re: (z) word splitting on one word

[Mikael Magnusson]

On 9/22/22, Julian Prein <druckdev@protonmail.com> wrote:
> Hi!
>
> I am trying to get the last word of LBUFFER.
>
> What I did previously was:
>
>     ${${(z)LBUFFER}[-1]}
>
> This worked flawlessly until I noticed today that it misbehaves if LBUFFER
> contains only one word. In this case it seems like `[-1]` acts on a scalar,
> as
> it expands to the last character instead of LBUFFER as I would expect.
>
> Is this expected behaviour?
>
> I now work around it by prepending LBUFFER with a dummy word so that it
> always
> has at least two words (I know that it is not empty). But this feels very
> hacky.
>
> Do you know of a better way?

${${(Az)LBUFFER}[-1]}

-- 
Mikael Magnusson

Re: (z) word splitting on one word

[Julian Prein]

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> $\{$\{(Az)LBUFFER\}\[-1\]\}

Perfect, thank you very much!

Julian

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