Bash variable substitution in a function - regex

Question

I have a function read_command defined as:
function read_command {
local __newline __lines __input __newstr __tmp i;
exec 3< "$*";
__newline=$'\n';
__lines=();
while IFS= read <&3 -r __line && [ "$__line" != '####' ]; do
echo "$__line";
__lines+=("$__line");
done
while IFS= read <&3 -r __line && [ "$__line" != '####' ]; do
read -e -p "${__line#*:}$PS2" __input;
local ${__line%%:*}="$__input";
done
__command="";
for i in "${__lines[#]}"; do
__tmp=$(echo "${i}");
__command="${__command} ${__newline} ${__tmp}";
done
echo -e "$__command";
}
In the current directory there is a file named "test", with the following
content:
greet ${a:-"Bob"}
greet ${b:-"Jim"}
####
a: name of friend a
b: name of friend b
####
In the terminal, the command executed is
read_command test
With no input, I am expecting the output of the last statement to be:
greet Bob
greet Jim
But what I get is:
greet ${a:-"Bob"}
greet ${b:-"Jim"}
What is wrong here?
Edit: As suggested by David, adding eval works in some cases except the following one.
j=1;i="export DISPLAY=:0 && Xephyr :${j}&";k=$(eval echo "$i");
echo $k
export DISPLAY=:0
I am expecting k to be "export DISPLAY=:0 && Xephyr :1&", what's wrong here?
Edit: I tried with the following
k=$(eval "echo \"$i\"")
This is the link to the script I am working on.
https://gist.github.com/QiangF/565102ba3b6123942b9bf6b897c05f87

Answer 1

During the first while loop, in echo "$__line", you have __line='greet ${a:-"Bob"}'. When you try to print that, Bash won't be expanding ${a:-"Bob"} into Bob. (Even if you remove the quotes around $__line this won't happen.) To get that effect, you need to add eval, as in, e.g., eval echo "$__line". Unfortunately eval comes with its can of worms, you have to start worrying about interactions between quoting levels and such.