Re: question about variabies/references (ignore my previous botched entry)

[chromatic]

in reply to question about variabies/references (ignore my previous botched entry)

Here's a funny thing. $scott and $main::scott refer to the same variable, assuming you haven't declared that you're in another package. Another piece of the puzzle is that putting curly braces around anything that looks like a variable causes Perl to interpolate the value of that variable. Doing: $main::{$test} = "skot2"; first causes interpolation -- so that the interpreter now has: $main::scott = "skot2"; From there, I assume plaid is correct in that the interpreter decides that you really meant to do a typeglob assignment, and takes "skot2" to be a symbolic reference (that is, the name of another variable). Since it hasn't been declared before, $scot2 (and $main::scot2) are autovivified. Next, $main::scott is aliased to $scot2. The $$test line is even more tricky. You might also write it ${$test} = "surprise!"; to be consistent with my explanation. If $test were a normal reference, this would dereference it. Since it contains the name of a variable ($skot2 has just been created automatically), Perl considers it a symbolic reference, looks up $main::skot2 in the symbol table, and assigns the value of "surprise!" to it.

Because $main::scott is aliased to $skot2, accessing it gets you the same value ("surprise!"). Because you're in package main, you can leave off the $main:: portion, and accessing $scott also gets you your "surprise!".

These are three or four things that will eventually surprise any new Perl programmer. Using strict and -w will warn you when these things happen (except for autovivification in hashes, which is worth another page of explanation). Use them liberally!

[btrott]

You (chromatic) wrote:

>   Another piece of the puzzle is that putting curly
>   braces around anything that looks like a variable
>   causes Perl to interpolate the value of that
>   variable. Doing: 
>   
>   $main::{$test} = "skot2";
>   
>   first causes interpolation -- so that the
>   interpreter now has:
>   
>   $main::scott = "skot2";

No, I don't think this is true. When you do

$main::{$test} = "skot2";

you're manipulating the symbol table of package main; this is a very different thing than modifying a scalar variable ($main::scott). $main::scott and $main::{"scott"} are two different things.

[chromatic]

You're right, the post-interpolated line should read: *main::scott = "scot2"; Here's what I've been using to test my assumptions:

#!/usr/bin/perl

$test1 = "value of test1";
$test2 = "test1";
$test3 = "Why are you here?";

# refers to *main::test1
print "=>$main::{$test2}<=\n";

print "Original \$test1: ...$test1...\n";

$main::{$test2} = "test3";

# $main::{"test1"} = "test1";
print "\$test1 = >>", $test1, "<<\n";
print "Main package: ", $main::test1, "\n";

As you can see, $main::{$test2} evaluates to a typeglob. Saying that $main::{$test2} = "test3" changes the scalar is technically incorrect. (Though it is what my previous answer would lead one to believe!)

Dunno where that u came from. Maybe it's a mu?

[btrott]

chromatic wrote:


> As you can see, $main::{$test2} evaluates to a typeglob.

Right; because a typeglob *is* an entry in the symbol table.

> Saying that $main::{$test2} =
> "test3" changes the scalar is technically incorrect.

Sorry, did I say that?