in reply to Scalar variable value is name of a scalar variable
No. '$INFO' is a string containing a dollar sign followed by the letters I, N, F, and O. It is a constant. $INFO is a variable named $INFO whose value can be just about anything (although, given the context, I expect it's probably an integer).
While it is possible to do something like what you're talking about with soft references, that's bad ju-ju, so don't do it unless you enjoy ripping your hair out as you try to figure out how a variable which doesn't even appear in your code is getting modified.
The better approach in this case would be to simply remove the quotes from your assignments. e.g., Use $loglev = $DEBUG; instead of $loglev = '$DEBUG'; (There are other, even better, options which don't involve if/elsif/elsif/... chains, but this one is the quickest and simplest to implement, given your posted code as a starting point.)
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