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Re: Re: How might I return tied scalar that does not have to be dereferenced as $$scalar?

by rr (Sexton)
on Sep 23, 2002 at 17:01 UTC ( #200164=note: print w/replies, xml ) Need Help??


in reply to Re: How might I return tied scalar that does not have to be dereferenced as $$scalar?
in thread How might I return tied scalar that does not have to be dereferenced as $$scalar?

Hello Aldo

I guess that what I want to do is not possible without creating new basic variable types. This project is for an OSS database abstraction layer that I wrote that automatically scavenges the db for meta information, creates classes to represent rows in tables. Each class then uses a collection of the Datum instances. This Datum class does type checking based on it's SQL type in the DB, and formats things correctly. You can provide additional type checking by configuration file (like dbname.table.colum Datum's must be all CAPS).

My purpose in posting was to find a way to make coding with these objects faster by getting rid of using methods to update/get the values of the Datum instances. Using the overload+tie kinks above give me most of what I need. Too bad you cannot overload the assignment operator.

Again, thanks to all all for your input.

rr

  • Comment on Re: Re: How might I return tied scalar that does not have to be dereferenced as $$scalar?

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