Let's assume I have an array

> words=(foo bar baz)

Now I can join the elements

> echo ${(j., .)words}
foo, bar, baz

And I can append a string to the elements:

> echo ${^words}yeah
fooyeah baryeah bazyeah

With the following, I can append and join the elements:

> wordsyeah=(${^words}yeah)
> echo ${(j., .)wordsyeah}
fooyeah, baryeah, bazyeah

Is it possible to do print "fooyeah, baryeah, bazyeah" in a single expression, i.e. without using additional variables?

Bonus: Can I print that without using any variables at all?

As far as I can tell, this boils down to running parameter expansion on strings, but I wasn't able to find out how (or if) that is possible.

  • This is essentially a duplicate of this question on SO. As per my answer there you could achive this with echo ${(j:, :):-${^${=:-foo bar baz}}yeah}, which is actually harder to type and 12 characters longer than echo fooyeah, baryeah, bazyeah. So it mainly makes sense if variables are involved, for example echo ${(j:, :):-${^${words}}yeah} (with words being an array).
    – Adaephon
    Jul 8 '16 at 20:47
  • Wow, you're totally right. I couldn't find that question. Now I'm wondering why I posted this here, too. And I was writing the answer just as you commented. I'll let the mods decide what to do with this question now. Jul 8 '16 at 20:51

The answer is

> print ${(j., .)${:-${^words}yeah}}
fooyeah, baryeah, bazyeah


> print ${(j., .)${:-{foo,bar,baz}yeah}}
fooyeah, baryeah, bazyeah

without using variables at all

The critical part is ${name:-word}. It is explained in the manual.


If name is set, or in the second form is non-null, then substitute its value; otherwise substitute word. In the second form name may be omitted, in which case word is always substituted.

Thanks to phy1729 from #zsh for pointing this out to me :)

  • 1
    Can you please accept your own answer, so this question doesn't stay marked as "unanswered"? Thanks! Nov 8 '21 at 11:48

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