In Programming Perl (the book) I read that I can create a dictionary where the entries hold an array as follows:
$wife{"Jacob"} = ["Leah", "Rachel", "Bilhah", "Zilpah"];
Say that I want to grab the contents of $wife{"Jacob"} in a list. How can I do that?
If I try:
$key = "Jacob";
say $wife{$key};
I get:
ARRAY (0x56d5df8)
which makes me believe that I am getting a reference, and not the actual list.
See
perllol,
perldsc and
perlreftut
for information on using complex data structures and references.
Essentially, a hash can only have scalars as values, but references are scalars, Therefore, you are saving an arrayref inside the hash, and have to dereference it to an array.
To dereference a reference, use the #{...} syntax.
say #{$wife{Jacob}};
or
say "#{$wife{Jacob}}"; # print elements with spaces in between
I guess by this time you must be knowing that
$ refers to a scalar
and # refers to an array.
since you yourself said that the value for that key is an array,then you should
say #wife{$key};
instead of
say $wife{$key};
Related
Below is code in which I need help.
#!/usr/bin/perl -w
use strict;
use Data::Dumper;
my #arrayElements = ('Array Functions');
print join(", ", #arrayElements);
### Output => Array Functions
my %hashElements = ();
I want to assign the content of #arrayElements to $hashElements{Item}
Missing some core concepts or trying wrong and been a while struggling with this.
You seem to be missing some core concepts of Perl (or programming in general). If you are learning Perl through a book or online tutorial, I suggest you re-read the chapters on arrays and hashes.
Let's look at the things involved here. You have:
#arrayElements, which is an array. It contains a list with one elements, the string 'Array Functions'.
%hashElements, which is a hash. It's empty.
$hashElements{Item}, which is a scalar value. You want to set this.
You say you want $hashElements{Item} to have the value 'Array Functions', which you have as the first element in your array #arrayElements.
$hashElements{Item} = $arrayElements[0];
And that's it. Both $hashElements{Item} and $arrayElements[0] are scalar values. That's why their sigils (the sign at the front) changes from an # (for array) or % (for hash) to a $. You can distinguish whether the value came from a hash or an array by the brackets used to access the elements. [] is for arrays, and {} is for hashes.
You cannot do the following though.
$hashElements{Item} = #arrayElements;
Because $hashElements{Item} is a scalar, the thing on the right hand side of the assignment will be treated in scalar context. An array in scalar context gets converted to the number of elements in the array, so this would assign 1. That's not what you want.
You should really read up more about this, and also pick better names for your variables. Your example is very confusing. In general, we don't do $CamelCase for variable names in Perl, but instead use $snake_case, which is easier to read and type.
Take a look at the following resources to learn more about the concepts I've mentioned above.
Perl Maven, perldata, perldsc
I have to some debugging on an existing script without having a lot of knowledge about perl.
This script uses data types like these to store all the fields from a file:
${$LineRefs->{FIELD_NAME}}
I've been trying to figure out how to find all possible fields separately by iterating over this scalar/hash/array or whatever it may be but I have no clue how.
Could anyone please point me in the right direction?
It's certainly very odd
$LineRefs is a reference to a hash which has an element with key FIELD_NAME whose value is a reference to a scalar
Like this
use v5.14;
my $LineRefs = {
FIELD_NAME => \99,
};
print ${ $LineRefs->{FIELD_NAME} }, "\n";
output
99
References to hashes and arrays are common because they allow a large data structure to be represented by a single scalar. But references to scalars are far less useful because they just replace a scalar by another scalar
I'm sorry, thanks #glennjackman I read the question too hastily and assumed it was about why a hash element was being dereferenced as a scalar
I've been trying to figure out how to find all possible fields separately by iterating over this scalar/hash/array or whatever it may be but I have no clue how
You're dealing with a hash, which is like an array but indexed by strings (keys) instead of integers (indexes)
You can use keys, values, or each to iterate over a hash
You can print all the keys and their values like this. Since your variable $LineRefs is a hash reference you need to dereference it as %$LineRefs
for my $key ( keys %$LineRefs ) {
my $value = $LineRefs->{$key};
print "$key => $value\n";
}
If your hash values really are references to scalars then you will see things like SCALAR(0x640448) printed for the values
I want to make an multidimensional array, but as i will declare it, i don't know how many element it will have, so I tried this:
my #multarray = [ ][ ];
Is it good?
Perl isn't C, and you don't need to initialise any sort of array. Just my #multarray is fine.
Observe
use strict;
use warnings;
my #data;
$data[2][2] = 99;
print $data[2][2], "\n";
The section in perldoc perlol on Declaration and Access of Arrays of Arrays will be of help here.
output
99
Perl doesn't support multidimensional arrays directly; an array is always just a sequence of scalars. But you can create references to arrays, and store the references in arrays, and that allows you to simulate multidimensional arrays.
The square bracket notation is useful here to create an array literal and return a reference to it. For example, the following creates an array of ('a','b',1234) and returns a reference to it:
my $ref_array = ['a','b',1234];
Here's an example of how to create (what we might call) a multidimensional array literal, of dimension 3x2:
my $multarray = [['a','b'],['c','d'],['e','f']];
So you could access the 'c' (for example) with:
print($multarray->[1]->[0]);
Now, you say you don't know what the final dimensions will be. That's fine; in Perl, you can conceptually view arrays as being infinite in size; all elements that haven't been assigned yet will be returned as undef, whether or not their index is less than or equal to the greatest index that has thus far been assigned. So here's what you can do:
my $multarray = [];
Now you can read and write directly to any element at any index and at any depth level:
$multarray->[23]->[19234]->[3] = 'a';
print($multarray->[23]->[19234]->[3]); ## prints 'a'
The syntax I've been using is the "explicit" syntax, where you explicitly create and dereference all the array references that are being manipulated. But for convenience, Perl allows you to omit the dereference tokens when you bracket-index an array reference, except when dereferencing the top-level array reference, which must always be explicitly dereferenced with the -> token:
$multarray->[23][19234][3] = 'a';
print($multarray->[23][19234][3]); ## prints 'a'
Finally, if you prefer to work with array variables (unlike me; I actually prefer to work with scalar references all the time), you can work with a top-level array variable instead of an array reference, and in that case you can escape having to use the dereference token entirely:
my #multarray;
$multarray[23][19234][3] = 'a';
print($multarray[23][19234][3]); ## prints 'a'
But even if you use that somewhat deceptively concise syntax, it's good to have an understanding of what's going on behind-the-scenes; it's still a nested structure of array references.
I'm having troubles with a Perl script. I try to store a hash in a hash. The script is trivial:
use Data::Dumper;
my %h1=();
$h1{name}="parent";
my %h2=();
$h2{name}="child";
$h1{nested}=%h2; # store hash h2 in hash h1
print "h2:\n";
print Dumper(%h2); # works
print "h1{nested}:\n";
print Dumper($h1{nested}); # fails
The results:
h2:
$VAR1 = 'name';
$VAR2 = 'child';
h1{nested}:
$VAR1 = '1/8';
Why is the $h1{nested} not dumped as a hash, but as some kind of weird scalar (1/8)?
PS: even if this question sounds trivial - I searched SO but did not find that it was asked before.
PPS: my Perl is v5.10.1 (*) built for x86_64-linux-gnu-thread-multi
(with 53 registered patches, see perl -V for more detail)
You can only store a hashref in a hash:
$h1{nested}=\%h2;
and then you would access %h2's name by doing
$h1{nested}->{name}
In your code, %h2 is forced to scalar context, which shows you that "1/8" value, and stores that.
In perl the values stored in a list (hash or array) are always scalars. Given this, the only way to store a hash inside another hash is to store a reference to it.
$h1{'nested'} = \%h2;
or also
$h1{'nested'} = { 'name'=>'child' };
(the braces in the right hand side is a reference to an anonymous hash).
BTW, to not quote the literals in the keys is usually considered bad practice, see here
Why is the $h1{nested} not dumped as a hash, but as some kind of weird scalar (1/8)?
Because you're storing it in a scalar context!
When you do this:
$h1{nested} = %h2;
You're storing a scalar. Since %h2 is a hash, you're given the ol' fraction string. According to the Perldoc website
If you evaluate a hash in scalar context, it returns false if the hash is empty. If there are any key/value pairs, it returns true; more precisely, the value returned is a string consisting of the number of used buckets and the number of allocated buckets, separated by a slash.
That explains the 1/8 you're getting.
What you need to do is store the hash as a reference in the other hash. As others pointed out, it should be:
$h1{nested} = \%h2;
The backslash before the hash's name gives you the memory location where the hash is stored. You can use the curly braces, but I prefer the backslash notation.
Take a look at perldoc prelreftut on your computer (or on the webpage I've linked to). This will tell you how to make such things as a list of lists, hashes or hashes, lists of hashes, and hashes of lists. Just a word o` warning: If you get too complex, it'll be hard to maintain, so once you've had your fun, take a look at perldoc's Perl Object Orientation Programming Tutorial.
The perldoc command contains lots of Perl documentation including for all Perl function, Perl modules installed on your system, and even basic information about the Perl language.
I'm trying to figure out Perl subroutines and how they work.
From perlsub I understand that subroutines are call-by-reference and that an assignment (like my(#copy) = #_;) is needed to turn them into call-by-value.
In the following, I see that change is called-by-reference because "a" and "b" are changed into "x" and "y". But I'm confused about why the array isn't extended with an extra element "z"?
use strict;
use Data::Dumper;
my #a = ( "a" ,"b" );
change(#a);
print Dumper(\#a);
sub change
{
#_[0] = "x";
#_[1] = "y";
#_[2] = "z";
}
Output:
$VAR1 = [
'x',
'y'
];
In the following, I pass a hash instead of an array. Why isn't the key changed from "a" to "x"?
use strict;
use Data::Dumper;
my %a = ( "a" => "b" );
change(%a);
print Dumper(\%a);
sub change
{
#_[0] = "x";
#_[1] = "y";
}
Output:
$VAR1 = {
'a' => 'y'
};
I know the real solution is to pass the array or hash by reference using \#, but I'd like to understand the behaviour of these programs exactly.
Perl always passes by reference. It's just that sometimes the caller passes temporary scalars.
The first thing you have to realise is that the arguments of subs can be one and only one thing: a list of scalars.* One cannot pass arrays or hashes to them. Arrays and hashes are evaluated, returning a list of their content. That means that
f(#a)
is the same** as
f($a[0], $a[1], $a[2])
Perl passes by reference. Specifically, Perl aliases each of the arguments to the elements of #_. Modifying the elements #_ will change the scalars returned by $a[0], etc. and thus will modify the elements of #a.
The second thing of importance is that the key of an array or hash element determines where the element is stored in the structure. Otherwise, $a[4] and $h{k} would require looking at each element of the array or hash to find the desired value. This means that the keys aren't modifiable. Moving a value requires creating a new element with the new key and deleting the element at the old key.
As such, whenever you get the keys of an array or hash, you get a copy of the keys. Fresh scalars, so to speak.
Back to the question,
f(%h)
is the same** as
f(
my $k1 = "a", $h{a},
my $k2 = "b", $h{b},
my $k2 = "c", $h{c},
)
#_ is still aliased to the values returned by %h, but some of those are just temporary scalars used to hold a key. Changing those will have no lasting effect.
* — Some built-ins (e.g. grep) are more like flow control statements (e.g. while). They have their own parsing rules, and thus aren't limited to the conventional model of a sub.
** — Prototypes can affect how the argument list is evaluated, but it will still result in a list of scalars.
Perl's subroutines accept parameters as flat lists of scalars. An array passed as a parameter is for all practical purposes a flat list too. Even a hash is treated as a flat list of one key followed by one value, followed by one key, etc.
A flat list is not passed as a reference unless you do so explicitly. The fact that modifying $_[0] modifies $a[0] is because the elements of #_ become aliases for the elements passed as parameters. Modifying $_[0] is the same as modifying $a[0] in your example. But while this is approximately similar to the common notion of "pass by reference" as it applies to any programming language, this isn't specifically passing a Perl reference; Perl's references are different (and indeed "reference" is an overloaded term). An alias (in Perl) is a synonym for something, where as a reference is similar to a pointer to something.
As perlsyn states, if you assign to #_ as a whole, you break its alias status. Also note, if you try to modify $_[0], and $_[0] happens to be a literal instead of a variable, you'll get an error. On the other hand, modifying $_[0] does modify the caller's value if it is modifiable. So in example one, changing $_[0] and $_[1] propagates back to #a because each element of #_ is an alias for each element in #a.
Your second example is a little tricky. Hash keys are immutable. Perl doesn't provide a way to modify a hash key, aside from deleting it. That means that $_[0] is not modifiable. When you attempt to modify $_[0] Perl cannot comply with that request. It probably ought to throw a warning, but doesn't. You see, the flat list passed to it consists of unmodifiable-key followed by modifiable-value, etc. This is mostly a non-issue. I cannot think of any reason to modify individual elements of a hash in the way you're demonstrating; since hashes have no particular order you wouldn't have simple control over which elements in #_ propagate back to which values in %a.
As you pointed out, the proper protocol is to pass \#a or \%a, so that they can be referred to as $_[0]->{element} or $_[0]->[0]. Even though the notation is a little more complicated, it becomes second nature after awhile, and is much clearer (in my opinion) as to what is going on.
Be sure to have a look at the perlsub documentation. In particular:
Any arguments passed in show up in the array #_. Therefore, if you called a function with two arguments, those would be stored in $_[0] and $_[1]. The array #_ is a local array, but its elements are aliases for the actual scalar parameters. In particular, if an element $_[0] is updated, the corresponding argument is updated (or an error occurs if it is not updatable). If an argument is an array or hash element which did not exist when the function was called, that element is created only when (and if) it is modified or a reference to it is taken. (Some earlier versions of Perl created the element whether or not the element was assigned to.) Assigning to the whole array #_ removes that aliasing, and does not update any arguments.
(Note that use warnings is even more important than use strict.)
#_ itself isn't a reference to anything, it is an array (really, just a view of the stack, though if you do something like take a reference to it, it morphs into a real array) whose elements each are an alias to a passed parameter. And those passed parameters are the individual scalars passed; there is no concept of passing an array or hash (though you can pass a reference to one).
So shifts, splices, additional elements added, etc. to #_ don't affect anything passed, though they may change the index of or remove from the array one of the original aliases.
So where you call change(#a), this puts two aliases on the stack, one to $a[0] and one to $a[1]. change(%a) is more complicated; %a flattens out into an alternating list of keys and values, where the values are the actual hash values and modifying them modifies what's stored in the hash, but where the keys are merely copies, no longer associated with the hash.
Perl does not pass the array or hash itself by reference, it unfurls the entries (the array elements, or the hash keys and values) into a list and passes this list to the function. #_ then allows you to access the scalars as references.
This is roughly the same as writing:
#a = (1, 2, 3);
$b = \$a[2];
${$b} = 4;
#a now [1, 2, 4];
You'll note that in the first case you were not able to add an extra item to #a, all that happened was that you modified the members of #a that already existed. In the second case, the hash keys don't really exist in the hash as scalars, so these need to be created as copies in temporary scalars when the expanded list of the hash is created to be passed into the function. Modifying this temporary scalar will not modify the hash key, as it is not the hash key.
If you want to modify an array or hash in a function, you will need to pass a reference to the container:
change(\%foo);
sub change {
$_[0]->{a} = 1;
}
Firstly, you are confusing the # sigil as indicating an array. This is actually a list. When you call Change(#a) you are passing the list to the function, not an array object.
The case with the hash is slightly different. Perl evaluates your call into a list and passes the values as a list instead.