Re: Re: Why was it neccessary to pass a DBI handler by reference?

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.

Comment on Re: Re: Why was it neccessary to pass a DBI handler by reference?

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Re: Why was it neccessary to pass a DBI handler by reference? by Abigail-II (Bishop) on Jan 30, 2004 at 08:42 UTC
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. No, this is Perl, not C. A reference is a reference, and not a pointer. There are pointers involved in the internals, but pointers are used for other scalar types as well. Abigail	[reply]