Where is the uninitialized value in the compare function? - perl

I tried to find the error myself, but don't see it. The following code produces warnings (same problem in Perl 5.18.2 and 5.32.1).
Use of uninitialized value in numeric comparison (<=>) at test.pl line 14.
while the comparison function of the sort is performed (and, as a consequence, the sort operation is not performed correctly). As far as I see, no value of the hash is initial, they all have defined numeric values.
use strict;
use warnings;
# Sort the occurrences of letters in a list of words, highest count first.
my #words = ('ACGT','CCGT','CATG'); # Just an example
my $a = join '',#words;
my $l = length $a;
my %count = ();
for (my $i = 0; $i < $l; $i++) {
my $x = substr( $a, $i, 1);
$count{$x}++;
}
for my $x (sort { $count{$b} <=> $count{$a} } keys %count) {
print "$x: $count{$x}\n";
}
Remark : it didn't help adding the hash element creation line before incrementing it with the $count{$x}++; statement - same result (expectedly, as undef counts like 0 for the increment operation):
...
$count{$x} = 0 unless defined $count{$x};
$count{$x}++;
...

Normally, the sort function uses two package variables called $a and $b to do the sorting. Specifically, sort sets the variables called $a and $b on the current package to be the current sort values. Those are not arguments to your { $count{$b} <=> $count{$a} } block; they're global variables in the current package.
Now, $b is fine. Since you never do anything else with it, Perl picks up the package variable just fine. But you declared a lexical (my) variable called $a earlier in your code, and that lexical is shadowing the package variable.
So sort is setting a variable called, effectively, $YourPackage::a, and your code is accessing a local variable called my $a, which is unrelated to the other one.
You can fix this by changing the my $a = join '',#words; variable to be called something else, and in fact you should probably do that. The names $a, $b, and $_ are used for things like this indiscriminately in Perl for historical reasons, so it's probably best to never have your own variables named any of those names.
But if you don't want to (or can't) change any of the rest of the code, you can expose the package variable with our.
for my $x (sort { our $a; $count{$b} <=> $count{$a} } keys %count) {
...
}

Related

Autovivification doesnt work even as lvalue

I have this simple code:
#!/usr/bin/perl
#inp = map 2**$_, 0..6;
#cc = grep {
my $num = $inp[$_];
my $sum; #---- HERE, I have to have the var declared first, before init.
$sum += $_ for (split //, $num);
print "$sum\n";
$sum % 2;
} 0..$#inp;
Here, the $sum will be used in for loop, However in this case:
#!/usr/bin/perl
#inp = map 2**$_, 0..6;
#cc = grep {
my $num = $inp[$_];
my $sum += $_ for (split //, $num); # HERE, Trying to autovificate - wont work
print "$sum\n";
$sum % 2;
} 0..$#inp;
But when I used var $sum at the same line with for loop - that means I am trying to declare and initiate at once - where should work the autovivifaction - As i would expect to autovivificate the $sum to zero (because used with math operator +=), but will not work, but why so? What are the rules for autovivification?
This is not autovivification. You have a syntax mistake. If you had use strict and use warnings turned on, it would be more obvious.
The post-fix for construct treats the left-hand side like a block. So there is a scope for the body of the loop. Therefore you are declaring your my $sum inside that loop body scope, and it's not visible outside.
If you turn on use warnings, you'll get Use of uninitialized value $sum in concatenation (.) or string at ... line 6, which is the print after.
You need to declare the variable first (and use strict and warnings!).
my has two effects:
At compile time, my declares the variable.
At run time, my allocates a new variable. More or less.
The first effect is what allows you to refer to the variable until the end of the enclosing block.
The second effect means $sum can't possibly hold the sum at the end of the loop since you call my to create a new variable each pass of the loop.
[ Warning: This section discusses Perl's guts. Feel free to jump ahead. ]
But why is it undef instead of containing the number from the last pass?
Well, that's cause my doesn't actually allocate a new variable when executed. It places an instruction on the stack to allocate a new one on scope exit!
The for statement modifier creates a lexical scope so that $_ can be properly restored when the statement is complete, so my $sum is replaced with a fresh variable at the end of each loop pass. (It's technically only being cleared rather than deallocated and reallocated thanks to an optimization.)
Your code could be written as follows:
#!/usr/bin/perl
use strict;
use warnings;
sub sum { my $acc; $acc += $_ for #_; $acc }
my #inp = map 2**$_, 0..6;
my #cc = grep { ( sum split // ) % 2 } #inp;
or even just
my #cc = grep { ( sum split //, 2**$_ ) % 2 } 0..6;
Always use use strict; use warnings;. Note that use warnings; would have made it more obvious that something was going wrong.
By the way, I don't know what you think autovivification means, but it's wrong.
Autovivification is the creation of a variable and a reference to it when deferencing an undefined value.
$ perl -e'
my $x;
CORE::say $x // "[undef]";
$x->[0] = 123;
CORE::say $x // "[undef]";
'
[undef]
ARRAY(0x35d7f56740)
Less formally, it could also refer to the creation of hash or array elements when using them as lvalues.
$ perl -e'
my $x;
CORE::say exists($h{x}) ? 1 : 0;
my $ref = \( $h{x} );
CORE::say exists($h{x}) ? 1 : 0;
'
0
1
There's no attempt to autovivify in your code.

Why isn't my sort working in Perl?

I have never used Perl, but I need to complete this exercise. My task is to sort an array in a few different ways. I've been provided with a test script. This script puts together the array and prints statements for each stage of it's sorting. I've named it foo.pl:
use strict;
use warnings;
use MyIxHash;
my %myhash;
my $t = tie(%myhash, "MyIxHash", 'a' => 1, 'abe' => 2, 'cat'=>'3');
$myhash{b} = 4;
$myhash{da} = 5;
$myhash{bob} = 6;
print join(", ", map { "$_ => $myhash{$_}" } keys %myhash) . " are the starting key => val pairs\n";
$t->SortByKey; # sort alphabetically
print join(", ", map { "$_ => $myhash{$_}" } keys %myhash) . " are the alphabetized key => val pairs\n";
$t->SortKeyByFunc(sub {my ($a, $b) = #_; return ($b cmp $a)}); # sort alphabetically in reverse order
print join(", ", map { "$_ => $myhash{$_}" } keys %myhash) . " are the reverse alphabetized key => val pairs\n";
$t->SortKeyByFunc(\&abcByLength); # use abcByLength to sort
print join(", ", map { "$_ => $myhash{$_}" } keys %myhash) . " are the abcByLength sorted key => val pairs\n";
print "Done\n\n";
sub abcByLength {
my ($a, $b) = #_;
if(length($a) == length($b)) { return $a cmp $b; }
else { return length($a) <=> length($b) }
}
Foo.pl uses a package called MyIxHash which I've created a module for called MyIxHash.pm. The script runs through the alphabetical sort: "SortByKey", which I've inherited via the "IxHash" package in my module. The last two sorts are the ones giving me issues. When the sub I've created: "SortKeyByFunc" is ran on the array, it passes in the array and a subroutine as arguments. I've attempted to take those arguments and associate them with variables.
The final sort is supposed to sort by string length, then alphabetically. A subroutine for this is provided at the bottom of foo.pl as "abcByLength". In the same way as the reverse alpha sort, this subroutine is being passed as a parameter to my SortKeyByFunc subroutine.
For both of these sorts, it seems the actual sorting work is done for me, and I just need to apply this subroutine to my array.
My main issue here seems to be that I don't know how, if possible, to take my subroutine argument and run my array through it as a parameter. I'm a running my method on my array incorrectly?
package MyIxHash;
#use strict;
use warnings;
use parent Tie::IxHash;
use Data::Dumper qw(Dumper);
sub SortKeyByFunc {
#my $class = shift;
my ($a, $b) = #_;
#this is a reference to the already alphabetaized array being passed in
my #letters = $_[0][1];
#this is a reference to the sub being passed in as a parameter
my $reverse = $_[1];
#this is my variable to contain my reverse sorted array
my #sorted = #letters->$reverse();
return #sorted;
}
1;
"My problem occurs where I try: my #sorted = #letters->$reverse(); I've also tried: my #sorted = sort {$reverse} #letters;"
You were really close; the correct syntax is:
my $reverse = sub { $b cmp $a };
# ...
my #sorted = sort $reverse #letters;
Also note that, for what are basically historical reasons, sort passes the arguments to the comparison function in the (slightly) magic globals $a and $b, not in #_, so you don't need to (and indeed shouldn't) do my ($a, $b) = #_; in your sortsubs (unless you declare them with a prototype; see perldoc -f sort for the gritty details).
Edit: If you're given a comparison function that for some reason does expect its arguments in #_, and you can't change the definition of that function, then your best bet is probably to wrap it in a closure like this:
my $fixed_sortsub = sub { $weird_sortsub->($a, $b) };
my #sorted = sort $fixed_sortsub #letters;
or simply:
my #sorted = sort { $weird_sortsub->($a, $b) } #letters;
Edit 2: Ah, I see the/a problem. When you write:
my #letters = $_[0][1];
what you end up with a is a single-element array containing whatever $_[0][1] is, which is presumably an array reference. You should either dereference it immediately, like this:
my #letters = #{ $_[0][1] };
or just keep is as a reference for now and dereference it when you use it:
my $letters = $_[0][1];
# ...
my #sorted = sort $whatever #$letters;
Edit 3: Once you do manage to sort the keys, then, as duskwuff notes in his original answer, you'll also need to call the Reorder() method from your parent class, Tie::IxHash to actually change the order of the keys. Also, the first line:
my ($a, $b) = #_;
is completely out of place in what's supposed to be an object method that takes a code reference (and, in fact, lexicalizing $a and $b is a bad idea anyway if you want to call sort later in the same code block). What it should read is something like:
my ($self, $sortfunc) = #_;
In fact, rather than enumerating all the things that seem to be wrong with your original code, it might be easier to just fix it:
package MyIxHash;
use strict;
use warnings;
use parent 'Tie::IxHash';
sub SortKeyByFunc {
my ($self, $sortfunc) = #_;
my #unsorted = $self->Keys();
my #sorted = sort { $sortfunc->($a, $b) } #unsorted;
$self->Reorder( #sorted );
}
1;
or simply:
sub SortKeyByFunc {
my ($self, $sortfunc) = #_;
$self->Reorder( sort { $sortfunc->($a, $b) } $self->Keys() );
}
(Ps. I now see why the comparison functions were specified as taking their arguments in #_ rather than in the globals $a and $b where sort normally puts them: it's because the comparison functions belong to a different package, and $a and $b are not magical enough to be the same in every package like, say, $_ and #_ are. I guess that could be worked around, but it would take some quite non-trivial trickery with caller.)
(Pps. Please do credit me and duskwuff / Stack Overflow when you hand in your exercise. And good luck with learning Perl — trust me, it'll be a useful skill to have.)
Your SortKeyByFunc method returns the results of sorting the array (#sorted), but it doesn't modify the array "in place". As a result, just calling $t->SortKeyByFunc(...); doesn't end up having any visible permanent effects.
You'll need to call $t->Reorder() within your SortKeyByFunc method to have any lasting impact on the array. I haven't tried it, but something like:
$t->Reorder(#sorted);
at the end of your method may be sufficient.

three questions on a Perl function

I am trying to use an existing Perl program, which includes the following function of GetItems. The way to call this function is listed in the following.
I have several questions for this program:
what does foreach my $ref (#_) aim to do? I think #_ should be related to the parameters passed, but not quite sure.
In my #items = sort { $a <=> $b } keys %items; the "items" on the left side should be different from the "items" on the right side? Why do they use the same name?
What does $items{$items[$i]} = $i + 1; aim to do? Looks like it just sets up the value for the hash $items sequentially.
$items = GetItems($classes, $pVectors, $nVectors, $uVectors);
######################################
sub GetItems
######################################
{
my $classes = shift;
my %items = ();
foreach my $ref (#_)
{
foreach my $id (keys %$ref)
{
foreach my $cui (keys %{$ref->{$id}}) { $items{$cui} = 1 }
}
}
my #items = sort { $a <=> $b } keys %items;
open(VAL, "> $classes.items");
for my $i (0 .. $#items)
{
print VAL "$items[$i]\n";
$items{$items[$i]} = $i + 1;
}
close VAL;
return \%items;
}
When you enter a function, #_ starts out as an array of (aliases to) all the parameters passed into the function; but the my $classes = shift removes the first element of #_ and stores it in the variable $classes, so the foreach my $ref (#_) iterates over all the remaining parameters, storing (aliases to) them one at a time in $ref.
Scalars, hashes, and arrays are all distinguished by the syntax, so they're allowed to have the same name. You can have a $foo, a #foo, and a %foo all at the same time, and they don't have to have any relationship to each other. (This, together with the fact that $foo[0] refers to #foo and $foo{'a'} refers to %foo, causes a lot of confusion for newcomers to the language; you're not alone.)
Exactly. It sets each element of %items to a distinct integer ranging from one to the number of elements, proceeding in numeric (!) order by key.
foreach my $ref (#_) loops through each hash reference passed as a parameter to GetItems. If the call looks like this:
$items = GetItems($classes, $pVectors, $nVectors, $uVectors);
then the loop processes the hash refs in $pVector, $nVectors, and $uVectors.
#items and %items are COMPLETELY DIFFERENT VARIABLES!! #items is an array variable and %items is a hash variable.
$items{$items[$i]} = $i + 1 does exactly as you say. It sets the value of the %items hash whose key is $items[$i] to $i+1.
Here is an (nearly) line by line description of what is happening in the subroutine
Define a sub named GetItems.
sub GetItems {
Store the first value in the default array #_, and remove it from the array.
my $classes = shift;
Create a new hash named %items.
my %items;
Loop over the remaining values given to the subroutine, setting $ref to the value on each iteration.
for my $ref (#_){
This code assumes that the previous line set $ref to a hash ref. It loops over the unsorted keys of the hash referenced by $ref, storing the key in $id.
for my $id (keys %$ref){
Using the key ($id) given by the previous line, loop over the keys of the hash ref at that position in $ref. While also setting the value of $cui.
for my $cui (keys %{$ref->{$id}}) {
Set the value of %item at position $cui, to 1.
$items{$cui} = 1;
End of the loops on the previous lines.
}
}
}
Store a sorted list of the keys of %items in #items according to numeric value.
my #items = sort { $a <=> $b } keys %items;
Open the file named by $classes with .items appended to it. This uses the old-style two arg form of open. It also ignores the return value of open, so it continues on to the next line even on error. It stores the file handle in the global *VAL{IO}.
open(VAL, "> $classes.items");
Loop over a list of indexes of #items.
for my $i (0 .. $#items){
Print the value at that index on it's own line to *VAL{IO}.
print VAL "$items[$i]\n";
Using that same value as an index into %items (which it is a key of) to the index plus one.
$items{$items[$i]} = $i + 1;
End of loop.
}
Close the file handle *VAL{IO}.
close VAL;
Return a reference to the hash %items.
return \%items;
End of subroutine.
}
I have several questions for this program:
What does foreach my $ref (#_) aim to do? I think #_ should be related to the parameters passed, but not quite sure.
Yes, you are correct. When you pass parameters into a subroutine, they automatically are placed in the #_ array. (Called a list in Perl). The foreach my $ref (#_) begins a loop. This loop will be repeated for each item in the #_ array, and each time, the value of $ref will be assigned the next item in the array. See Perldoc's Perlsyn (Perl Syntax) section about for loops and foreach loops. Also look at Perldoc's Perlvar (Perl Variables) section of General variables for information about special variables like #_.
Now, the line my $classes = shift; is removing the first item in the #_ list and putting it into the variable $classes. Thus, the foreach loop will be repeated three times. Each time, $ref will be first set to the value of $pVectors, $nVectors, and finally $uVectors.
By the way, these aren't really scalar values. In Perl, you can have what is called a reference. This is the memory location of the data structure you're referencing. For example, I have five students, and each student has a series of tests they've taken. I want to store all the values of each test in a hash keyed by the student's ID.
Normally, each entry in the hash can only contain a single item. However, what if this item refers to a list that contains the student's grades?
Here's the list of student #100's grade:
#grades = (100, 93, 89, 95, 74);
And here's how I set Student 100's entry in my hash:
$student{100} = \#grades;
Now, I can talk about the first grade of the year for Student #100 as $student{100}[0]. See the Perldoc's Mark's very short tutorial about references.
In my #items = sort { $a <=> $b } keys %items; the "items" on the left side should be different from the "items" on the right side? Why do they use the same name?
In Perl, you have three major types of variables: Lists (what some people call Arrays), Hashes (what some people call Keyed Arrays), and Scalars. In Perl, it is perfectly legal to have different variable types have the same name. Thus, you can have $var, %var, and #var in your program, and they'll be treated as completely separate variables1.
This is usually a bad thing to do and is highly discouraged. It gets worse when you think of the individual values: $var refers to the scalar while $var[3] refers to the list, and $var{3} refers to the hash. Yes, it can be very, very confusing.
In this particular case, he has a hash (a keyed array) called %item, and he's converting the keys in this hash into a list sorted by the keys. This syntax could be simplified from:
my #items = sort { $a <=> $b } keys %items;
to just:
my #items = sort keys %items;
See the Perldocs on the sort function and the keys function.
What does $items{$items[$i]} = $i + 1; aim to do? Looks like it just sets up the value for the hash $items sequentially.
Let's look at the entire loop:
foreach my $i (0 .. $#items)
{
print VAL "$items[$i]\n";
$items{$items[$i]} = $i + 1;
}
The subroutine is going to loop through this loop once for each item in the #items list. This is the sorted list of keys to the old %items hash. The $#items means the largest index in the item list. For example, if #items = ("foo", "bar", and "foobar"), then $#item would be 2 because the last item in this list is $item[2] which equals foobar.
This way, he's hitting the index of each entry in #items. (REMEMBER: This is different from %item!).
The next line is a bit tricky:
$items{$items[$i]} = $i + 1;
Remember that $item{} refers to the old %items hash! He's creating a new %items hash. This is being keyed by each item in the #items list. And, the value is being set to the index of that item plus 1. Let's assume that:
#items = ("foo", "bar", "foobar")
In the end, he's doing this:
$item{foo} = 1;
$item{bar} = 2;
$item{foobar} = 3;
1 Well, this isn't 100% true. Perl stores each variable in a kind of hash structure. In memory, $var, #var, and %var will be stored in the same hash entry in memory, but in positions related to each variable type. 99.9999% of the time, this matters not one bit. As far as you are concerned, these are three completely different variables.
However, there are a few rare occasions where a programmer will take advantage of this when they futz directly with memory in Perl.
I want to show you how I would write that subroutine.
Bur first, I want to show you some of the steps of how, and why, I changed the code.
Reduce the number of for loops:
First off this loop doesn't need to set the value of $items{$cui} to anything in particular. It also doesn't have to be a loop at all.
foreach my $cui (keys %{$ref->{$id}}) { $items{$cui} = 1 }
This does practically the same thing. The only real difference is it sets them all to undef instead.
#items{ keys %{$ref->{$id}} } = ();
If you really needed to set the values to 1. Note that (1)x#keys returns a list of 1's with the same number of elements in #keys.
my #keys = keys %{$ref->{$id}};
#items{ #keys } = (1) x #keys;
If you are going to have to loop over a very large number of elements then a for loop may be a good idea, but only if you have to set the value to something other than undef. Since we are only using the loop variable once, to do something simple; I would use this code:
$items{$_} = 1 for keys %{$ref->{$id}};
Swap keys with values:
On the line before that we see:
foreach my $id (keys %$ref){
In case you didn't notice $id was used only once, and that was for getting the associated value.
That means we can use values and get rid of the %{$ref->{$id}} syntax.
for my $hash (values %$ref){
#items{ keys %$hash } = ();
}
( $hash isn't a good name, but I don't know what it represents. )
3 arg open:
It isn't recommended to use the two argument form of open, or to blindly use the bareword style of filehandles.
open(VAL, "> $classes.items");
As an aside, did you know there is also a one argument form of open. I don't really recommend it though, it's mostly there for backward compatibility.
our $VAL = "> $classes.items";
open(VAL);
The recommend way to do it, is with 3 arguments.
open my $val, '>', "$classes.items";
There may be some rare edge cases where you need/want to use the two argument version though.
Put it all together:
sub GetItems {
# this will cause open and close to die on error (in this subroutine only)
use autodie;
my $classes = shift;
my %items;
for my $vector_hash (#_){
# use values so that we don't have to use $ref->{$id}
for my $hash (values %$ref){
# create the keys in %items
#items{keys %$hash} = ();
}
}
# This assumes that the keys of %items are numbers
my #items = sort { $a <=> $b } keys %items;
# using 3 arg open
open my $output, '>', "$classes.items";
my $index; # = 0;
for $item (#items){
print {$output} $item, "\n";
$items{$item} = ++$index; # 1...
}
close $output;
return \%items;
}
Another option for that last for loop.
for my $index ( 1..#items ){
my $item = $items[$index-1];
print {$output} $item, "\n";
$items{$item} = $index;
}
If your version of Perl is 5.12 or newer, you could write that last for loop like this:
while( my($index,$item) = each #items ){
print {$output} $item, "\n";
$items{$item} = $index + 1;
}

Why is Perl foreach variable assignment modifying the values in the array?

OK, I have the following code:
use strict;
my #ar = (1, 2, 3);
foreach my $a (#ar)
{
$a = $a + 1;
}
print join ", ", #ar;
and the output?
2, 3, 4
What the heck? Why does it do that? Will this always happen? is $a not really a local variable? What where they thinking?
Perl has lots of these almost-odd syntax things which greatly simplify common tasks (like iterating over a list and changing the contents in some way), but can trip you up if you're not aware of them.
$a is aliased to the value in the array - this allows you to modify the array inside the loop. If you don't want to do that, don't modify $a.
See perldoc perlsyn:
If any element of LIST is an lvalue, you can modify it by modifying VAR inside the loop. Conversely, if any element of LIST is NOT an lvalue, any attempt to modify that element will fail. In other words, the foreach loop index variable is an implicit alias for each item in the list that you're looping over.
There is nothing weird or odd about a documented language feature although I do find it odd how many people refuse check the docs upon encountering behavior they do not understand.
$a in this case is an alias to the array element. Just don't have $a = in your code and you won't modify the array. :-)
If I remember correctly, map, grep, etc. all have the same aliasing behaviour.
As others have said, this is documented.
My understanding is that the aliasing behavior of #_, for, map and grep provides a speed and memory optimization as well as providing interesting possibilities for the creative. What happens is essentially, a pass-by-reference invocation of the construct's block. This saves time and memory by avoiding unnecessary data copying.
use strict;
use warnings;
use List::MoreUtils qw(apply);
my #array = qw( cat dog horse kanagaroo );
foo(#array);
print join "\n", '', 'foo()', #array;
my #mapped = map { s/oo/ee/g } #array;
print join "\n", '', 'map-array', #array;
print join "\n", '', 'map-mapped', #mapped;
my #applied = apply { s/fee//g } #array;
print join "\n", '', 'apply-array', #array;
print join "\n", '', 'apply-applied', #applied;
sub foo {
$_ .= 'foo' for #_;
}
Note the use of List::MoreUtils apply function. It works like map but makes a copy of the topic variable, rather than using a reference. If you hate writing code like:
my #foo = map { my $f = $_; $f =~ s/foo/bar/ } #bar;
you'll love apply, which makes it into:
my #foo = apply { s/foo/bar/ } #bar;
Something to watch out for: if you pass read only values into one of these constructs that modifies its input values, you will get a "Modification of a read-only value attempted" error.
perl -e '$_++ for "o"'
the important distinction here is that when you declare a my variable in the initialization section of a for loop, it seems to share some properties of both locals and lexicals (someone with more knowledge of the internals care to clarify?)
my #src = 1 .. 10;
for my $x (#src) {
# $x is an alias to elements of #src
}
for (#src) {
my $x = $_;
# $_ is an alias but $x is not an alias
}
the interesting side effect of this is that in the first case, a sub{} defined within the for loop is a closure around whatever element of the list $x was aliased to. knowing this, it is possible (although a bit odd) to close around an aliased value which could even be a global, which I don't think is possible with any other construct.
our #global = 1 .. 10;
my #subs;
for my $x (#global) {
push #subs, sub {++$x}
}
$subs[5](); # modifies the #global array
Your $a is simply being used as an alias for each element of the list as you loop over it. It's being used in place of $_. You can tell that $a is not a local variable because it is declared outside of the block.
It's more obvious why assigning to $a changes the contents of the list if you think about it as being a stand in for $_ (which is what it is). In fact, $_ doesn't exist if you define your own iterator like that.
foreach my $a (1..10)
print $_; # error
}
If you're wondering what the point is, consider the case:
my #row = (1..10);
my #col = (1..10);
foreach (#row){
print $_;
foreach(#col){
print $_;
}
}
In this case it is more readable to provide a friendlier name for $_
foreach my $x (#row){
print $x;
foreach my $y (#col){
print $y;
}
}
Try
foreach my $a (#_ = #ar)
now modifying $a does not modify #ar.
Works for me on v5.20.2

Can't dereference scalar in Perl

Can anyone tell me why this prints REF(*) instead of 0?
$a = 0;
$a = \$a;
print $$a . "\n";
Erm, because $a doesn't contain 0 any more, but instead a reference to itself. You overwrote it in the second line!
If you want to set $a to a reference to $a's original value then you can write
$a = \"$a"
which works by using an expression that happens to evaluate to the value of $a, and taking a reference to that. This will fail if $a starts out as something other than a simple string or number, such as a reference or any value which changes when it is stringified. Then you could write an identity function such as
sub i { shift }
and use
$a = \i($a)
Your last assignment makes $a to reference to itself, so dereferencing it makes sense as chasing own tail in circular manner,
use warnings;
use Data::Dumper;
$a = 0;
$a = \$a;
print Dumper $a;
output
$VAR1 = \$VAR1;
You might want to make reference to a list of values which does what you want,
($a) = \map { $_ } $a;
print Dumper $a;
output
$VAR1 = \0;