Re: is_reference() ?

At 22:30 21/3/2001, Björn Schotte wrote:

Hi, I recently had a discussion with Alexander Aulbach (submitter of bug reports with unset() and the inconsistencies with unset($foo) and unset($bar["bla"]) (the latter works, the former not))

Huh? Both work...?

and we agreed that there definitively would be need of a funtion called is_reference(varname) Sometimes we would like to know if $foo is a references variable or not (e.g. to ship around the unset() bug issue). What do you think?

This has been discussed before (as far as I recall). The engine may convert certain variables to be references on the fly, without the user requesting it. So, is_reference() won't mean much to end users. Zeev