Slicing an array that contains empty strings

Question

Consider the array foo, initialized like this:

$ foo=( a b '' d e f g ) 

foo contains 7 elements, one of which is an empty string.

Below are a few ways to print out the contents of foo, using the print built-in:

$ print -rl -- $foo a b d e f g $ print -rl -- "$foo" a b d e f g $ print -rl -- $foo[@] a b d e f g $ print -rl -- "$foo[@]" a b d e f g 

Note that only the form whose last token is "$foo[@]" interprets it as 7 separate arguments.

Now, suppose that I wanted to use print -rl -- ... to display only the first 5 elements of foo, one element per line?

This won't work:

$ print -rl -- "$foo[1,5]" a b d e 

Nor this:

$ print -rl -- $foo[1,5] a b d e 

I've tried other variants, but they all fail to produce the desired output, namely

a b d e 

What's the slice-equivalent of the full "$foo[@]"?

If no such equivalent exists, how do I create an array bar consisting of the first 5 elements of foo?

Answer 1

Though zsh doesn't do split+glob upon parameter expansion, it still does empty removal, so that's still one reason you want to quote variables there, so:

print -rl -- "$var[@]" 

Or

print -rl -- "${(@)var}" 

Those @ are to get Bourne "$@"-like behaviour.

For elements 1 to 5:

print -rl -- "${(@)var[1,5]}" 

The ksh-like variant will also work:

print -rl -- "${(@)var:0:5}"