Bash scripting printf'ing an escape sequence contained in a variable

Question

Bash code:

yellow="\e[1;33m"
chosen_colour="${yellow}"
declare -i score=300
printf '%s %d\n' "${chosen_colour}" "${score}"

Result:

\e[1;33m 300

Should be:

300 /* in yellow */

---

How can I interpolate a string value containing an ANSI escape sequence into a printf statement without using either of these syntaxes:

Avoid 1: (works, in fact, but it's wasteful when doing a lot of +=s)

s="${yellow}"
s+="${score}"
s+=...
s+=...
s+=...
s+=...
s+=...
s+=...
s+=...
s+=...

Avoid 2: (hard to do in my case for the sheer number of variables requiring this construct)

printf "${yellow}${score}${a1}${a2}${a3}${a4}${a5}${a6}${a7}${a8}........."

---

I want to be able to pass the values to be replaced, according to a predefined FORMAT string, using the arguments part of the printf call, as I naively do in the 1st example.

I could live with something like this:

printf \
  '%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s...' \
    "${a1}" \
    "${a2}" \
    "${a3}" \
    "${a4}" \
    "${a5}" \
    "${a6}" \
    ...

although ultimately, for my many variables, I would use a construct like this:

${!a*} # or similar

Answer 1

You do:

printf "^[%s foo" "${a1}" # that is ctrl+v, ESC, followed by %s

or:

printf "\033%s foo" "${a1}"  # 033 octal for ESC