How to get the variable or function address in memory?

[hash]

FOr example in C code:
addr_ptr = argv[1];
I get the address:
./getaddrvar HOME
HOME is stored at address 0xbffffeef
So how can i do this in perl? I`ve read something about:
$b = \$a;
But the hexadecimal`s value has only 7 digits,
on the oher hand using the C code above i got 8! Where is the problem? Thank you all for helping me.

[Joost]

There really is no reason to get at the address from perl. In fact both your examples are completely useless, but I assume that's because they're examples :-) If you want to interface with C code at this low level, you're better off using XS, or Inline::C.

Note that a perl scalar variable is a much more complex data structure than a C string, but both XS and Inline::C can make the conversion (almost) transparant.

updated: corrected typo

[Mr. Muskrat]

Generally with perl, just having the reference is enough. However, if you really think you need it then you can stringify the reference (sounds like what you have done already) to find out what type of reference it is (not the best way to do so, see ref) and where in perl's memory it resides.

[ihb]

An even better way than using ref() is to use &reftype and &refaddr in Scalar::Util.

ihb

[zentara]

I'm not a devel expert, but here is some code posted awhile back.

#!/usr/bin/perl
use Devel::Peek;

my $str = 'the key is:';
my $key = 'hello';
Dump($key);
decrypt( $str, $key );
$key = 'gone!';
# $key = 'x'x100000;   # a string that won't fit will change PV 
Dump($key);
decrypt( $str, $key );

sub decrypt { warn "\nGot @_\n\n" }

[rcaputo]

The memory address is the numeric version of the reference. Some sample code follows. As an added bonus, it illustrates the use of the mysterious "P" unpack() template.

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