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

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";

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No, I don't think this is true. When you do 
$main::{$test} = "skot2";

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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.

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Symbol Table Madness
by chromatic (Archbishop) on Apr 05, 2000 at 21:57 UTC
    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";

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    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?

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

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

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

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      Sorry, did I say that?
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