Prevent shell from escaping single quotes - bash

Question

Script:
#!/bin/sh -x
ARGS=""
CMD="./run_this_prog"
. . .
ARGS="-first_args '-A select[val]' "
. . .
$CMD $ARGS
I want the commandline to be expanded like this when I run this shell script:
./run_this_prog -first_args '-A select[val]'
Instead what shell does (note the added '\' before each single quote):
+ ARGS=
+ CMD='./run_this_prog'
+ ARGS='-first_args '\''-A select[val]'\'' '
and what it ran on commandline (escaped every special char - Not what I want):
./run_this_prog -first_args \'\-A select\[val\]\'
I tried escaping single quotes like :
ARGS="-first_args \'-A select[val]\' "
But that resulted in (added '\' after each backslash):
+ ARGS=
+ CMD='./run_this_prog'
+ ARGS='-first_args \'\''-A select[val]\'\'' '
I did my googling but couldn't find anything relevant. What am I missing here?
I am using sh-3.2 on rel6 centOS.

Answer 1

Once a quote is inside a string, it will not work the way you want: Inside a string quotes are not syntactic elements, they are just literal characters. This is one reason why bash offers arrays.
Replace:
#!/bin/sh -x
...
ARGS="-first_args '-A select[val]' "
$CMD $ARGS
With:
#!/bin/bash -x
...
ARGS=(-first_args '-A select[val]')
"$CMD" "${ARGS[#]}"
For a much more detailed discussion of this issue, see: "I'm trying to put a command in a variable, but the complex cases always fail!"