How to substitute quoted, multi-word strings as arguments?

Question

I\'m trying to substitute a string variable, containing multiple quoted words, as a parameter to a command.

Thus, given the following example script (Note the -x in

Answer 1

I don't think it is doing what you think it is doing.

[~]$ myArg="\"hello\" \"world\""
[~]$ echo "string is:" $myArg
string is: "hello" "world"

I see no extra quotes of any kind- echo gets three argument strings.

[~]$ cargs(){ echo $#; }
[~]$ cargs "string is:" $myArg
3

Bash will expand the variable first, so

cargs "string is:" $myArg

becomes (though without the literal backslashes- this is why string escaping is a PITA)

cargs "string is:" "\"hello\"" "\"world\""

And the args array is:

0x00:string is:0
0x0B:"hello"0
0x13:"world"0
0x1B:0

Now, if you add the *, or glob path expansion in one of those, Bash will at this point expand it, unless you escape it, or use single quotes in your literal command.

[~]$ cargs "string is:" $myArg *
19
[~]$ cargs "string is:" $myArg "\*"
4
[~]$ cargs "string is:" $myArg '*'
4