how to put all command arguments in one variable

Question

I want to execute a shell script that require 3 arguments.

The argument number 2 contains a string with space

I want to put all arguments in one variable like this:

Linux:~# kk="\"111\" \"222 222\" \"333\""
Linux:~# echo $kk
"111" "222 222" "333"

Now If I call a function:

func() {
  echo ---$1---
  echo ---$2---
  echo ---$3---
}

with the $kk variable in this way

func $kk

Then it will return

Linux:~# func $kk
---"111"---
---"222---
---222"---

And I was expecting to get this result

---111---
---222 222---
---333---

How to solve this issue without using eval ?

I know that the eval solve this issue but I do not want to use it (since it takes time if I execute many time a such call).

Answer 1

OP originally tagged question bash, then removed the tag. Please refer to Charles Duffy's great answer for a POSIX shell solution! the solution below uses arrays and is only suitable for bash.

---

You should not put your data in a string like this, but in an array. By the way, you're missing some quotes in your function:

func() {
    echo "---$1---"
    echo "---$2---"
    echo "---$3---"
}

or better yet (printf is preferable to echo):

func() {
    printf -- '---%s---\n' "$1"
    printf -- '---%s---\n' "$2"
    printf -- '---%s---\n' "$3"
}

$ kk=( "111" "222 222" "333" )
$ func "${kk[@]}"

will happily print:

---111---
---222 222---
---333---

You can even include newlines and such in your arguments:

$ kk=( $'one\none' "222 222" $' *   three\nthree' )
$ func "${kk[@]}"
---one
one---
---222 222---
--- *   three
three---

Answer 2

If features (like arrays) didn't make bash more expressive than POSIX sh, there would have been no reason to add them. :)

That said, you can work around it by overriding "$@" for your needs:

set -- 111 "222 222" 333
printf '%s\n' "$@"

...will print...

111
222 222
333

If you need to build the arguments list step by step, you can make several calls to set with "$@" as the first arguments (make sure you observe the quotes!):

set -- 111
set -- "$@" "222 222"
set -- "$@" 333
printf '%s\n' "$@"

...will print...

111
222 222
333