Can someone finish this for me and explain what you did?
my %hash;
#$hash{list_ref}=[1,2,3];
#my #array=#{$hash{list_ref}};
$hash{list_ref}=\[1,2,3];
my #array=???
print "Two is $array[1]";
#array = #{${$hash{list_ref}}};
(1,2,3) is a list.
[1,2,3] is a reference to a list an array (technically, there's no such thing in Perl as a reference to a list).
\[1,2,3] is a reference to a reference to an array.
$hash{list_ref} is a reference to a reference to an array.
${$hash{list_ref}} is a reference to an array.
#{${$hash{list_ref}}} is an array.
Since a reference is considered a scalar, a reference to a reference is a scalar reference, and the scalar dereferencing operator ${...} is used in the middle step.
Others have pretty much already answered the question, but more generally, if you are ever confused about a data structure, use Data::Dumper. This will print out the structure of the mysterious blob of data, and help you parse it.
use strict; #Always, always, always
use warnings; #Always, always, always
use feature qw(say); #Nicer than 'print'
use Data::Dumper; #Calling in the big guns!
my $data_something = \[1,2,3];
say Dumper $data_something;
say Dumper ${ $data_something };
Let's see what it prints out...
$ test.pl
$VAR1 = \[
1,
2,
3
];
$VAR1 = [
1,
2,
3
];
From the first dump, it appears that $data_something is a plain scalar reference to an array reference. That lead me to add the second Dumper after I ran the program the first time. That showed me that ${ $data_something } is now a reference to an array.
I can now access that array like this:
use strict; #Always, always, always
use warnings; #Always, always, always
use feature qw(say); #Nicer than 'print'
use Data::Dumper; #Calling in the big guns!
my $data_something = \[1,2,3];
# Double dereference
my #array = #{ ${ $data_something } }; #Could be written as #$$data_something
for my $element (#array) {
say "Element is $element";
}
And now...
$ test.pl
Element is 1
Element is 2
Element is 3
It looks like you meant:
my $hash{list_ref} = [1,2,3];
and not:
$hash{list_ref} = \[1,2,3];
That latter one got you an scalar reference of a array reference which really doesn't do you all that much good except add confusion to the situation.
Then, all you had to do to refer to a particular element is $hash{list_ref}->[0]. This is just a shortcut for ${ $hash{list_ref} }[0]. It's easier to read and understand.
use strict;
use warnings;
use feature qw(say);
my %hash;
$hash{list_ref} = [1, 2, 3];
foreach my $element (0..2) {
say "Element is " . $hash{list_ref}->[$element];
}
And...
$ test.pl
Element is 1
Element is 2
Element is 3
So, next time you are confused about what a particular data structure looks like (and it happens to the best of us. Well... not the best of us, It happens to me), use Data::Dumper.
my %hash;
#$hash{list_ref}=[1,2,3];
#Putting the list in square brackets makes it a reference so you don't need '\'
$hash{list_ref}=[1,2,3];
#If you want to use a '\' to make a reference it is done like this:
# $something = \(1,2,3); # A reference to a list
#
# or (as I did above)...
#
# $something = [1,2,3]; # Returns a list reference
#
# They are the same (for all intent and purpose)
print "Two is $hash{list_ref}->[1]\n";
# To make it completely referenced do this:
#
# $hash = {};
# $hash->{list_ref} = [1,2,3];
#
# print $hash->{list_ref}[1] . "\n";
To get at the array (as an array or list) do this:
my #array = #{ $hash{list_ref} }
[ EXPR ]
creates an anonymous array, assigns the value returned by EXPR to it, and returns a reference to it. That means it's virtually the same as
do { my #anon = ( EXPR ); \#anon }
That means that
\[ EXPR ]
is virtually the same as
do { my #anon = ( EXPR ); \\#anon }
It's not something one normally sees.
Put differently,
1,2,3 returns a list of three elements (in list context).
(1,2,3) same as previous. Parens simply affect precedence.
[1,2,3] returns a reference to an array containing three elements.
\[1,2,3] returns a reference to a reference to an array containing three elements.
In practice:
my #data = (1,2,3);
print #data;
my $data = [1,2,3]; $hash{list_ref} = [1,2,3];
print #{ $data }; print #{ $hash{list_ref} };
my $data = \[1,2,3]; $hash{list_ref} = \[1,2,3];
print #{ ${ $data } }; print #{ ${ $hash{list_ref} } };
Related
I'm new to Perl and trying to get used to data structures and references in Perl.
I learned
key %hash returns an array of the keys in %hash
\{ #array } returns the reference to the #array
So I combined these two and wrote something like this,
use strict;
use warnings;
use Data::Dumper;
my $hash = {
key1 => 'value1',
key2 => 'value2'
};
my $keys = \{ keys %$hash }; # Supposed to be an array reference?
print Dumper $keys; # Output 1
print Dumper $keys->[0]; # Output 2
which yielded the error Not an ARRAY reference at the line of Output 2. Also, Output 1 shows something that looks like a hash reference though it's supposed to be an array reference.
What's wrong with my code?
Similarly, the following code didn't work with the same error.
use strict;
use warnings;
my $array = [1, 2, 3, 4, 5];
my $first_two = \{ #{ $array }[0..1] }; # Isn't it an array ref?
my $first = $first_two->[0];
I guess I misunderstand something about array references.
The problem you have is this is not correct: '\{ #array } returns the reference to the #array' . Instead, the \ is simply prepended to an existing variable, like this: \#array. Braces {} are used to create anonymous hash references, and brackets [] are used to create anonymous array references.
In your example, what you want to do is either (1) store the keys as an array and then use \ to get a reference:
my #keys = keys %$hash;
my $keys = \#keys;
Or (2) use an anonymous array reference:
my $keys = [ keys %$hash ];
Here's a good "reference" ;) https://perldoc.perl.org/perlref.html#Making-References
I'm trying to understand the 'Common Mistake' section in the perldsc documentation. What is the author trying to convey when he mentions:
The two most common mistakes made in constructing something like an array of arrays is either accidentally counting the number of elements or else taking a reference to the same memory location repeatedly. Here's the case where you just get the count instead of a nested array:
for my $i (1..10) {
my #array = somefunc($i);
$AoA[$i] = #array; # WRONG!
}
From what I understand is that when it iterate it will take the first value of (1..10) which is 1 and will pass it to a function like this:
my #array = somefunc(1);
Since that function is not defined, I'll create the logic.
sub somefunc {
my $a = shift;
print $a * $a;
}
which will essentially do this:
1 * 1
and the result is '1'.
To my understanding my #array will look like:
#array = ('1');
And the next line will do:
$AoA[$i] = #array;
I'm assuming that $AoA[$1] is an anonymous array ( he/she didn't declare with 'my', btw) and the #array will be the first element of the this anonymous array which the author said it' WRONG. And the for each loop with iterate to '2'.
somefunc(2);
Which will be '4' and passed to:
$AoA[$i] = #array
What is the point of the author with this code which is wrong. I'm trying to understand what is wrong but more importantly, I'm trying to understand the code. Any help will be appreciated.
UPDATE
I think I understand why this is a common mistake because when I use print and Dumper, I can visually see what the author is trying to convey, here is the revised code.
#!/usr/bin/perl -w
use strict;
use Data::Dumper;
for my $i (1..10) {
my #AoA;
my #array = somefunc($i);
print "The array is Dumper(#array)\n";
$AoA[$i] = #array; # WRONG!
print Dumper($AoA[$i]);
}
sub somefunc {
my $a = shift;
return $a * $a;
}
In the Common Mistakes paragraph of perldoc perldsc, he/she states
Here's the case where you just get the count instead of a nested array:
Below is the output of the Dumper.
The array is Dumper(1)
$VAR1 = 1;
The array is Dumper(4)
$VAR1 = 1;
The array is Dumper(9)
$VAR1 = 1;
The array is Dumper(16)
$VAR1 = 1;
The array is Dumper(25)
$VAR1 = 1;
The array is Dumper(36)
$VAR1 = 1;
The array is Dumper(49)
$VAR1 = 1;
The array is Dumper(64)
$VAR1 = 1;
The array is Dumper(81)
$VAR1 = 1;
The array is Dumper(100)
$VAR1 = 1;
So I'm assuming that the repeated
$VAR1 = 1;
is the count and not the nested array.
The author did indicate that if the count is what I truly want then to rewrite the code like this:
#!/usr/bin/perl -w
use strict;
use Data::Dumper;
for my $i (1..10) {
my #count;
my #array = somefunc($i);
print "The array is Dumper(#array)\n";
$count[$i] = scalar #array;
print Dumper($count[$i]);
}
sub somefunc {
my $a = shift;
return $a * $a;
}
But the documentation didn't tell me how to get the nested array?
UPDATE
Correct me if I am wrong but I rewrote the code to get the nested array:
#!/usr/bin/perl -w
use strict;
use Data::Dumper;
my #count;
my #new_array;
for my $i (1..10) {
#my #count;
my #array = somefunc($i);
push #new_array, [#array];
}
sub somefunc {
my $a = shift;
return $a * $a;
}
print Dumper(\#new_array);
Which printed
$VAR1 = [
[
1
],
[
4
],
[
9
],
[
16
],
[
25
],
[
36
],
[
49
],
[
64
],
[
81
],
[
100
]
];
In the following statement:
$AoA[$i] = #array;
the #array is referenced in a scalar context, yielding a number of its elements. The context is imposed by LHS, that is $AoA[$i] which is a single element of the #AoA array.
In Perl, there are no array of arrays in a strict sense. These are emulated essentially by either "flatten" arrays or array with references. For the latter, you would need to use take reference operator as in:
$AoA[$i] = \#array;
For the starter, you may find, that Data::Dumper is very handy in examining complex data stuctures such as arrayrefs and hashrefs.
Perl is polymorphic, which means that it deals with different data types transparently, and makes what is usually a pretty good guess on how to deal with something. This makes the programmer's work much easier because it is not strongly typed like other languages.
So for example if $i is the number 4, you can do this:
print $i + 1;
and you will see a 5 - pretty logical, right?
and if you do this:
print "I am " , $i , " years old";
You will see "I am 4 years old" - in this case perl says "you are operating in list context, so I will treat $i as a string. No need to convert the number into a string as many other languages insist.
So when you assign
$AoA[$i] = #array;
The way it treats this depends on the context. In scalar context, it will set $AoA[$i] to be the length of the array.
For more information about scalar vs list context, read this answer:
http://perl.plover.com/context.html
Your example isn't very useful in understanding what is going on here as your subroutine always returns "1" - the result of calling print(). If you replace the print() with return() then you will at get different values (1, 4, 9, etc).
But the next line of code:
$AoA[$i] = #array;
Will always assign 1 to the element of #Aoa. That's because You are assigning an array (#array) to a scalar variable ($AoA[$i]) and when you evaluate an array in a scalar context, you get the number of elements in the array.
Now, as your #array only ever has a single element, you could do this:
$AoA[$i] = $array[0];
But that's not really building an array of arrays. What you really want to do is to get a reference to an array.
$AoA[$i] = \#array;
This would be more useful if your subroutine returned more than one value.
sub somefunc {
# Used $x instead of $a as $a has a special meaning in Perl
my $x = shift;
return ($x * $x, $x * $x * $x);
}
for my $i (1..10) {
my #array = somefunc($i);
$AoA[$i] = \#array;
}
As useful tool for exploring this is Data::Dumper. Try adding:
use Data::Dumper;
To the top of your code and:
print Dumper #AoA;
After the foreach loop to see the different data structures that you get back.
Consider:
sub abc()
{
}
abc(#array, $a);
How do I access #array and $a in subroutine abc()?
I know about $_[0] and $_[1], but I wasn't sure if I can use it for arrays.
You access a sub's arguments with the #_ array. The first argument is $_[0], the second - $_[1], etc. In this particular case, your array will be unrolled to list of its elements, so $_[0] is $array[0], $_[1] is $array[1] and then after all those elements, last element of #_ will be the value of $a.
If you want to avoid unrolling that always happens when you use an array in a list context, use a reference to the array instead. References to arrays and hashes are created using \. So call your function like:
abc(\#array, $a);
After that, $_[0] will have reference to #array and $_[1] will be $a. To access array elements through reference, use -> operator. $_[0]->[2] is same as $array[2]. Actually you can even drop -> as long as it is between brackets, so $_[0][2] will work too. See more details on references in perlref.
You have two options:
Pass the scalar variable first (the dirty way)
abc($a, #array);
Then receive the parameters in subroutine as
my ($a, #array) = #_;
Pass your array as reference by adding a backslash before the array variable (recommended)
abc(\#array, $a);
Then receive the parameters in subroutine as
my ($array_ref, $a) = #_;
And dereference the $array_ref
my #array = #$array_ref;
More information about perlref.
The other answers explained the two basic approaches. However, it is important to note that there is a big difference between the two: When you pass an array by reference, any changes you make to it also change the original array. Here is an example:
use warnings;
use strict;
my #array = (1, 2, 3, 4, 5);
sub by_ref
{
my $array_ref = $_[0];
#$array_ref = (0, 0, 0);
print "Array inside by_ref: #$array_ref\n";
}
sub by_val
{
my #array_copy = #_;
#array_copy = (0,0,0);
print "Array inside by_val: #array_copy\n";
}
by_val(#array);
print "Original array after calling by_val: #array\n";
by_ref(\#array);
print "Original array after calling by_ref: #array\n";
If you do pass by reference, you need to keep this behavior in mind, making a copy of the referenced array if you don't want changes made in your sub to affect the original.
It would be nice if you pass the array reference instead of an array as mentioned by Oleg V. Volkov like
sub abc()
{
my ( $array, $a ) = #_; #receiving the paramters
my #arr = #{$array}; # dereferencing the array
}
abc(\#array,$a);
my %hash = #array;
transforms an array into a hash, but how can do the same with $hash{something}?
$hash{something} = { #array };
The {} around it creates a hashref.
When you write:
my %hash = #array;
What Perl sees is:
my %hash = ($array[0], $array[1], ... $array[$#array]);
So the #array is expanded into a list, and that list is assigned to the plural %hash. The list must have an even number of elements, or you will get a warning (assuming you are using use warnings; in your script, which you always should. use strict; as well).
Broken down even more, it is:
my %hash;
$hash{$array[0]} = $array[1];
$hash{$array[2]} = $array[3];
...
$hash{$array[$#array - 1]} = $array[$#array];
So with the translation from #array to %hash explained, to insert this hash into a hash of hashes at a particular key:
$HoH{key} = \%hash;
Where the \ character takes a reference to %hash. If %hash is not needed elsewhere you could constrain it with a block:
{
my %hash = #array;
$HoH{key} = \%hash;
}
Using a do {...} makes it a little shorter:
$HoH{key} = do {my %hash = #array; \%hash};
But not much shorter. If the above seems tedious to you, it should. In Perl, the above construct can be reduced to:
$HoH{key} = { #array };
Where the {...} denotes the anonymous hash reference constructor, equivalent to
do {my %hash = (...); \%hash}
In addition to {...}, Perl provides the [...] construct to create anonymous array references. The ... of both constructs sees list context.
In this question the poster asked how to do the following in one line:
sub my_sub {
my $ref_array = shift;
my #array = #$ref_array;
}
which with my knowledge of the basic Perl magic I would avoid by simply using something like:
sub my_sub {
my $ref_array = shift;
for (#$ref_array) {
#do somthing with $_ here
};
#use $ref_array->[$element] here
}
However in this answer one of SO's local monks tchrist suggested:
sub my_sub {
local *array = shift();
#use #array here
}
When I asked
In trying to learn the mid-level Perl
magic, can I ask, what is it that you
are setting to what here? Are you
setting a reference to #array to the
arrayref that has been passed in? How
do you know that you create #array and
not %array or $array? Where can I
learn more about this * operator
(perlop?). Thanks!
I was suggested to ask it as a new post, though he did give nice references. Anyway, here goes? Can someone please explain what gets assigned to what and how come #array gets created rather than perhaps %array or $array? Thanks.
Assignment to a glob
*glob = VALUE
contains some magic that depends on the type of VALUE (i.e., return value of, say, Scalar::Util::reftype(VALUE)). If VALUE is a reference to a scalar, array, hash, or subroutine, then only that entry in the symbol table will be overwritten.
This idiom
local *array = shift();
#use #array here
works as documented when the first argument to the subroutine is an array reference. If the first argument was instead, say, a scalar reference, then only $array and not #array would be affected by the assignment.
A little demo script to see what is going on:
no strict;
sub F {
local *array = shift;
print "\#array = #array\n";
print "\$array = $array\n";
print "\%array = ",%array,"\n";
print "------------------\n";
}
$array = "original scalar";
%array = ("original" => "hash");
#array = ("orignal","array");
$foo = "foo";
#foo = ("foo","bar");
%foo = ("FOO" => "foo");
F ["new","array"]; # array reference
F \"new scalar"; # scalar reference
F {"new" => "hash"}; # hash reference
F *foo; # typeglob
F 'foo'; # not a reference, but name of assigned variable
F 'something else'; # not a reference
F (); # undef
Output:
#array = new array
$array = original scalar
%array = originalhash
------------------
#array = orignal array
$array = new scalar
%array = originalhash
------------------
#array = orignal array
$array = original scalar
%array = newhash
------------------
#array = foo bar
$array = foo
%array = FOOfoo
------------------
#array = foo bar
$array = foo
%array = FOOfoo
------------------
#array =
$array =
%array =
------------------
#array = orignal array
$array = original scalar
%array = originalhash
------------------
Additional doc at perlmod and perldata. Back in the days before references were a part of Perl, this idiom was helpful for passing arrays and hashes into subroutines.
With my admittedly less-than-wizard knowledge of Perl, I'll venture an answer. The * operator assigns the symbol table entry. As I understand it, #array, %array, and $array all refer to the same symbol table entry for the string 'array', but to different fields in that entry: the ARRAY, HASH, and SCALAR fields. So assigning local *array = shift; actually assigns the entire local symbol table entry for 'array' (including the ARRAY, HASH, and SCALAR fields) to what was passed used in the caller.