Re^4: Referencing in Function not working correctly.

"Please explain why you are doing these things."
I thought it would be safer to copy it in the functions. So, what your saying is if I returned @bonly (as below), perl will not copy it will just give me pointer:
@bonly = deDuppedArray(\@currentListing, \@previousListing, \@isec_plu
+s_bonly );
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My original problem is @bonly changes after the function call and before function ended? Do you know why?
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Re^5: Referencing in Function not working correctly. by jwkrahn (Abbot) on Apr 02, 2010 at 19:31 UTC
I thought it would be safer to copy it in the functions. Yes it would be safer. So, what your saying is if I returned @bonly (as below), perl will not copy it will just give me pointer: `@bonly = deDuppedArray(\@currentListing, \@previousListing, \@isec_plus_bonly );` No, the list returned from `deDuppedArray()` will have to be copied to assign it to the array `@bonly`. My original problem is @bonly changes after the function call and before function ended? Do you know why? From the code you posted, there is nothing in the `deDuppedArray` subroutine that modifies `@bonly`, directly or indirectly, so no, I don't know why. What sort of changes are you seeing?	[reply] [d/l] [select]

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Re^5: Referencing in Function not working correctly.
by jwkrahn (Abbot) on Apr 02, 2010 at 19:31 UTC

I thought it would be safer to copy it in the functions.

Yes it would be safer.

So, what your saying is if I returned @bonly (as below), perl will not copy it will just give me pointer:
@bonly = deDuppedArray(\@currentListing, \@previousListing, \@isec_plus_bonly );

No, the list returned from deDuppedArray() will have to be copied to assign it to the array @bonly.

My original problem is @bonly changes after the function call and before function ended? Do you know why?

From the code you posted, there is nothing in the deDuppedArray subroutine that modifies @bonly, directly or indirectly, so no, I don't know why. What sort of changes are you seeing?

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