Question

Let’s say I have a variable’s name stored in another variable:

myvar=123
varname=myvar

Now, I’d like to get 123 by just using $varname variable.
Is there a direct way for that? I found no such bash builtin for lookup by name, so came up with this:

function var { v="\$$1"; eval "echo "$v; }

so

var $varname  # gives 123

Which doesn’t look too bad in the end, but I’m wondering if I missed something more obvious.

Answer 1

From the man page of bash:

${!varname}

If the first character of parameter is an exclamation point, a level of
variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable;
this variable is then expanded and that value is used in the rest of
the substitution, rather than the value of parameter itself. This is
known as indirect expansion.