Perl Pointer Question

[laurar]

Hi,

I was wondering if someone could figure out why my pointer assignment below won't work!
In my code, I try to create a new hash %newhash at the same location as %hashy using the following code:
%newhash=%$var;
However, if you run the following code you will notice that \%hashy and \%newhash do not contain the same address values.

This is wrecking my head so any help would be much appreciated!
Thanks in advance!

Note: To run it, you'll need a file named, ph_nos, containing data like the following:

sdfasd,sfads
sdfasdf,sdfasd
sdfasd,sdfads
I.e. 2 strings seperated by commas.

#!usr/bin/perl/
 
open(FILEREAD,"ph_nos");
@file_in=<FILEREAD>;
foreach $elem (@file_in)
   {
   @temp=split(",",$elem);
   %hashy = @temp;
   $var = \%hashy;
   print \%hashy;
   }
 
    %newhash=%$var;
   print $var;                       #should be same val as below but it isn't?????
   print \%newhash;

[nithinpes]

Hi,

I was wondering if someone could figure out why my pointer assignment below won't work!
In my code, I try to create a new hash %newhash at the same location as %hashy using the following code:
%newhash=%$var;
However, if you run the following code you will notice that \%hashy and \%newhash do not contain the same address values.

This is wrecking my head so any help would be much appreciated!
Thanks in advance!

Note: To run it, you'll need a file named, ph_nos, containing data like the following:
sdfasd,sfads
sdfasdf,sdfasd
sdfasd,sdfads
I.e. 2 strings seperated by commas.


#!usr/bin/perl/

open(FILEREAD,"ph_nos");
@file_in=<FILEREAD>;
foreach $elem (@file_in)
{
@temp=split(",",$elem);
%hashy = @temp;
$var = \%hashy;
print \%hashy;
}

%newhash=%$var;
print $var; #should be same val as below but it isn't?????
print \%newhash;

That is the expected result. In the line:

%newhash=%$var;
 

the memory location/address of the hash is not getting copied. Rather, you are copying the content of the hash. $var is being dereferred and the resulting hash is assigned to %newhash.
If you need to copy the reference address, you should try this way:

$newref = $var; #assigning reference
print "old reference:$var\n newreference:$newref\n";
 
## Dereference
 
print "The contents of the hash:\n";
print "$_ : $var->{$_}\n" foreach(keys %$var);
print "$_ : $newref->{$_}\n" foreach(keys %$newref);
 

[laurar]

Hi,
Thanks so much for your quick reply.
Although, im not so sure that im interpreting you correctly.
If I am, then this suggest that if i code the following:
print keys(%newhash)..
this should print the same as
print keys(%hashy)
.. but this is not the case.

Am I missing something?

Thanks again for your help.

[laurar]

Wait I think I've figured it out.
I didn't realise the hash would be destroyed when I went outside the scope of the for loop.
So this explains the query in my last post!

If I declare %newhash as a global variable and assign %newhash within my for loop as follows:
%newhash=%$var;
..will the hash structure exist at address &newhash outside the scope of the for loop or will it also be destroyed, like %hashy?

I know this is terrible programming but I just want to figure out how hash structured are created/destroyed/referenced in memory.

Thanks, once again!

[KevinADC]

In the code:

open(FILEREAD,"ph_nos");
@file_in=<FILEREAD>;
foreach $elem (@file_in)
{
@temp=split(",",$elem);
%hashy = @temp;
$var = \%hashy;
print \%hashy;
}
 

%hashy is overwritten each time by this line:

%hashy = @temp;

in the end %hashy will only equal whatever the last line of the file was. That might be contributing to some confusion on your part if you think %hashy should have all the lines of the file as key/value pairs.

SInce you are not using "strict" or even "my" all your variables are global, so there is no issue of variable scoping. Declaring a variable inside the foreach loop will not affect it outside the loop. The real problem is most likely what I described above.

[laurar]

Yup, that's sorted my confusion alright!
Thanks for all your help!