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But it looks like I can't replace that piece of data by modifying the original array.

Of course you can, but you must not destroy the internal reference of the array slot [0] to the scalar originally stored in it.  If you do so, you have no access to that scalar any longer via $a[0].

Change @a = ('a','b','c') to  pop @a; @a[0..2] = qw(a b c); and things will work fine.

In the former case, 3 new scalars are being placed into the array slots, while in the latter case, the old scalars are being modified.

#!/usr/bin/perl -w use strict; print "--------Trying with arrays-------\n"; my @a = (0, 1, 2, 3); my $ar = \@a; # Array reference my $air = \$a[0]; # Reference to member of an array print "Define array >@a<\nCreate a reference to \$a[0]\n"; print "array ref: >@$ar< \n"; print "array index ref: >$$air<\n"; pop @a; @a[0..2] = qw(a b c); print "Change array to >@a<\n"; print "array ref: >@$ar< \n"; print "array index ref: >$$air<\n"; # <--- I'd like to get "a" here __END__ --------Trying with arrays------- Define array >0 1 2 3< Create a reference to $a[0] array ref: >0 1 2 3< array index ref: >0< Change array to >a b c< array ref: >a b c< array index ref: >a<

Your misunderstanding is that you think Perl works like C, i.e. that references directly point to the memory locations where the actual data is stored.  This is not how Perl works.


In reply to Re^3: Update references by Eliya
in thread Update references by Anonymous Monk

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