Bash: all of array except last element

Question

Bash has a neat way of giving all elements in an array except the first:

"${a[@]:1}"

To get all except the last I have found:

"${a[@]:0:$((${#a[@]}-1))}"

But, man, that is ugly.

Is there an elegant alternative?

You can hack with b=("${a[@]}"); unset "b[-1]"; do_whatever_with "${b[@]}" — anishsane
– anishsane, Commented Jul 6, 2017 at 4:35

heemayl · Accepted Answer · 2017-07-06 04:39:33Z

36

I am not sure how much improvement it would be, but you can drop the arithmetic operator ($(())) and starting index (0 here):

${a[@]::${#a[@]}-1}

So:

$ foo=( 1 2 3 )

$ echo "${foo[@]::${#foo[@]}-1}"
1 2

As you can see, the improvement is purely syntactical; the idea remains the same.

answered Jul 6, 2017 at 4:39

heemayl

42.5k10 gold badges86 silver badges87 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Note for zsh users: the simplification works if you keep the starting index.