How to parse arguments with space inside bash function

Question

In bash I am calling function:

myFunction "1 2"
myFunction()
{
        echo "$1"
        echo "$2"
}

this prints me "1 2" and empty row. How I can parse arguments so it will print on one row 1 and on another 2 ? I can not call myFunction "1" "2" because arguments are stored in some other variable

Answer 1

Sounds like you want to split the first argument your function receives on space characters. For that, you could use the split+glob operator after having configured the split part to use space as the delimiter and disabled the glob part:

myfunction() {
  local -       # make changes to options local to the function.
                # needs bash 4.4 or newer.
  local IFS=' ' # split on space only
  set -o noglob # disable glob part
  set -- $1     # split+glob invoked on $1 by leaving that $1 unquoted,
                # result stored in $1, $2... using set --
  printf '%s\n' "$1"
  printf '%s\n' "$2"
}
myfunction "1 2"

Answer 2

That's the same as splitting any variable on spaces. Use word splitting or read:

With word splitting:

var="foo bar"
set -f              # disable globbing
IFS=' '             # make sure IFS contains (just) a space
printf "%s\n" $var  

With read, for a standard shell (if you know there's only two pieces to split into):

var="foo bar"
IFS=' ' read a b <<EOF
$var
EOF
printf "%s\n" "$a" "$b" 

The same with a here-string (Bash/ksh/zsh):

var="foo bar"
IFS=' ' read a b <<< "$var"
printf "%s\n" "$a" "$b" 

With read -a in Bash, or read -A in ksh/zsh, you can split the string to an arbitrary number of pieces and put them in an array:

var="foo bar"
IFS=' ' read -a arr <<< "$var"     # Bash
printf "%s\n" "${arr[@]}"

In all of the above, you can use $1 in place of $var as usual.

The variants with read also assume the string doesn't contain multiple lines.

However, in Bash you can also split a multi-line string to an array using any whitespace as separator:

IFS=$' \t\n' read -d '' -a arr <<< "$var"

---

Of course if you have the string in a variable outside the function, and run myFunction $var, the variable will be split to multiple arguments before the function runs.

Answer 3

Try like this,

myFunction()
{
        echo "$1"
        echo "$2"
}

myFunction 1 2