Sorry, my understanding was incorrect. The way I (incorrectly) understood it was that $y was a scalar representation of the reference \$x. (In the example: "my $y = \$x;") I then thought that when you dereferenced it, it converted it back into a reference.

Of course, I guess this is sortof how it works, in that \$x must figure out the address of x and return it so in a way it almost acts like a function. So \$x is basically a function with the parameter $x that returns the address of $x. (as a scalar of course) But then again \ isn't really a function, I imagine it interprets whatever follows it at runtime. However, I'm not about to start digging around the perl code to see exactly how it works.

In reply to Re: Re: Why was it neccessary to pass a DBI handler by reference? by MCS
in thread Why was it neccessary to pass a DBI handler by reference? by kudra

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