Remove last word in bash variable

Question

I have something like that:

...
args=$*
echo $args
...

result is

unusable1 unusable2 useful useful ... useful unusable3

I need remove all "unusable" args. They always at first, second and last position. After some investigation i find ${*:3} bash syntax. It help remove first two.

...
args=${*:3}
echo $args
...

result is

useful useful ... useful unusable3

But I can't find how to remove last word using same nice syntax.

Answer 1

You can use a function/script like this to print all but last arguments:

func() {
   echo "${@:1:$#-1}";
}

func aa bb cc dd ee
aa bb cc dd

func foo bar baz hello how are you
foo bar baz hello how are