An array in perl is dereferenced like so,
my #array = #{$array_reference};
When trying to assign an array to a dereference without the '#', like,
my #array = {$array_reference};
Perl throws the error, 'Odd number of elements in anonymous hash at ./sand.pl line 22.' We can't assign it to an array variable becauase Perl is confused about the type.
So how can we perform...
my $lastindex = $#{$array_reference};
if Perl struggles to understand that '{$array_reference}' is an array type? It would make more sense to me if this looked like,
my $lastindex = $##{$array_reference};
(despite looking much uglier).
tl;dr: It's $#{$array_reference} to match the syntax of $#array.
{} is overloaded with many meanings and that's just how Perl is.
Sometimes {} creates an anonymous hash. That's what {$array_reference} is doing, trying to make a hash where the key is the stringification of $array_reference, something like "ARRAY(0x7fb21e803280)" and there is no value. Because you're trying to create a hash with a key and no value you get an "odd number of elements" warning.
Sometimes {...} is a block like sub { ... } or if(...) { ... }, or do {...} and so on.
Sometimes it's a bare block like { local $/; ... }.
Sometimes it's indicating the key of a hash like $hash{key} or $hash->{key}.
Preceeded with certain sigils {} makes dereferencing explicit. While you can write $#$array_reference or #$array_reference sometimes you want to dereference something that isn't a simple scalar. For example, if you had a function that returned an array reference you could get its size in one line with $#{ get_array_reference() }. It's $#{$array_reference} to match the syntax of $#array.
$#{...} dereferences an array and gets the index. #{...} dereferences an array. %{...} dereferences a hash. ${...} dereferences a scalar. *{...} dereferences a glob.
You might find the section on Variable Names and Sigils in Modern Perl helpful to see the pattern better.
It would make more sense to me if this looked like...
There's a lot of things like that. Perl has been around since 1987. A lot of these design decisions were made decades ago. The code for deciding what {} means is particularly complex. That there is a distinction between an array and an array reference at all is a bit odd.
$array[$index]
#array[#indexes]
#array
$#array
is equivalent to
${ \#array }[$index]
#{ \#array }[#indexes]
#{ \#array }
$#{ \#array }
See the pattern? Wherever the NAME of an array isused, you can use a BLOCK that returns a reference to an array instead. That means you can use
${ $ref }[$index]
#{ $ref }[#indexes]
#{ $ref }
$#{ $ref }
This is illustrated in Perl Dereferencing Syntax.
Note that you can omit the curlies if the BLOCK contains nothing but a simple scalar.
$$ref[$index]
#$ref[#indexes]
#$ref
$#$ref
There's also an "arrow" syntax which is considered clearer.
$ref->[$index]
$ref->#[#indexes]
$ref->#*
$ref->$#*
Perl is confused about the type
Perl struggles to understand that '{$array_reference}' is an array type
Well, it's not an array type. Perl doesn't "struggle"; you just have wrong expectations.
The general rule (as explained in perldoc perlreftut) is: You can always use a reference in curly braces in place of a variable name.
Thus:
#array # a whole array
#{ $array_ref } # same thing with a reference
$array[$i] # an array element
${ $array_ref }[$i] # same thing with a reference
$#array # last index of an array
$#{ $array_ref } # same thing with a reference
On the other hand, what's going on with
my #array = {$array_reference};
is that you're using the syntax for a hash reference constructor, { LIST }. The warning occurs because the list in question is supposed to have an even number of elements (for keys and values):
my $hash_ref = {
key1 => 'value1',
key2 => 'value2',
};
What you wrote is treated as
my #array = ({
$array_reference => undef,
});
i.e. an array containing a single element, which is a reference to a hash containing a single key, which is a stringified reference (and whose value is undef).
The syntactic difference between a dereference and a hashref constructor is that a dereference starts with a sigil (such as $, #, or %) whereas a hashref constructor starts with just a bare {.
Technically speaking the { } in the dereference syntax form an actual block of code:
print ${
print "one\n"; # yeah, I just put a statement in the middle of an expression
print "two\n";
["three"] # the last expression in this block is implicitly returned
# (and dereferenced by the surrounding $ [0] construct outside)
}[0], "\n";
For (hopefully) obvious reasons, no one actually does this in real code.
The syntax is
my $lastindex = $#$array_reference;
which assigns to $lastindex the index of the last element of the anonymous array which reference is in the variable $array_reference.
The code
my #ary = { $ra }; # works but you get a warning
doesn't throw "an error" but rather a warning. In other words, you do get #ary with one element, a reference to an anonymous hash. However, a hash need have an even number of elements so you also get a warning that that isn't so.
Your last attempt dereferences the array with #{$array_reference} -- which returns a list, not an array variable. A "list" is a fleeting collection of scalars in memory (think of copying scalars on stack to go elsewhere); there is no notion of "index" for such a thing. For this reason a $##{$ra} isn't even parsed as intended and is a syntax error.
The syntax $#ary works only with a variable #ary, and then there is the $#$arrayref syntax. You can in general write $#{$arrayref} since the curlies allow for an arbitrary expression that evaluates to an array reference but there is no reason for that since you do have a variable with an array reference.
I'd agree readily that much of this syntax takes some getting-used-to, to put it that way.
Related
I use #_ in a subroutine to get a parameter which is assigned as a reference of an array, but the result dose not showing as an array reference.
My code is down below.
my #aar = (9,8,7,6,5);
my $ref = \#aar;
AAR($ref);
sub AAR {
my $ref = #_;
print "ref = $ref";
}
This will print 1 , not an array reference , but if I replace #_ with shift , the print result will be a reference.
can anyone explain why I can't get a reference using #_ to me ?
This is about context in Perl. It is a crucial aspect of the language.
An expression like
my $var = #ary;
attempts to assign an array to a scalar.
That doesn't make sense as it stands and what happens is that the right-hand side is evaluated to the number of elements of the array and that is assigned to $var.
In order to change that behavior you need to provide the "list context" to the assignment operator.† In this case you'd do
my ($var) = #ary;
and now we have an assignment of a list (of array elements) to a list (of variables, here only $var), where they are assigned one for one. So here the first element of #ary is assigned to $var. Please note that this statement plays loose with the elusive notion of the "list."
So in your case you want
my ($ref) = #_;
and the first element from #_ is assigned to $ref, as needed.
Alternatively, you can remove and return the first element of #_ using shift, in which case the scalar-context assignment is fine
my $ref = shift #_;
In this case you can also do
my $ref = shift;
since shift by default works on #_.
This is useful when you want to remove the first element of input as it's being assigned so that the remaining #_ is well suited for further processing. It is often done in object-oriented code.
It is well worth pointing out that many operators and builtin facilities in Perl act differently depending on what context they are invoked in.
For some specifics, just a few examples: the regex match operator returns true/false (1/empty string) in scalar context but the actual matches in list context,‡ readdir returns a single entry in scalar context but all of them in list context, while localtime shows a bit more distinct difference. This context-sensitive behavior is in every corner of Perl.
User level subroutines can be made to behave that way via wantarray.
†
See Scalar vs List Assignment Operator
for a detailed discussion
‡
See it in perlretut and in perlop for instance
When you assign an array to a scalar, you're getting the size of the array. You pass one argument (a reference to an array) to AAR, that's why you get 1.
To get the actual parameters, place the local variable in braces:
sub AAR {
my ($ref) = #_;
print "ref = $ref\n";
}
This prints something like ref = ARRAY(0x5566c89a4710).
You can then use the reference to access the array elements like this:
print join(", ", #{$ref});
I'm converting perl script to python.
I haven't used perl language before so many things in perl confused me.
For example, below, opt was declared as a scalar first but declared again as a hash. %opt = ();
Is it possible to declare a scalar and a hash with the same name in perl?
As I know, foreach $opt (#opts) means that scalar opt gets values of array opts one by one. opt is an array at this time???
In addition, what does $opt{$opt} mean?
opt outside $opt{$opt} is a hash and opt inside $opt{$opt} is a scalar?
I'm so confused, please help me ...
sub ParseOptions
{
local (#optval) = #_;
local ($opt, #opts, %valFollows, #newargs);
while (#optval) {
$opt = shift(#optval);
push(#opts,$opt);
$valFollows{$opt} = shift(#optval);
}
#optArgs = ();
%opt = ();
arg: while (defined($arg = shift(#ARGV))) {
foreach $opt (#opts) {
if ($arg eq $opt) {
push(#optArgs, $arg);
if ($valFollows{$opt}) {
if (#ARGV == 0) {
&Usage();
}
$opt{$opt} = shift(#ARGV);
push(#optArgs, $opt{$opt});
} else {
$opt{$opt} = 1;
}
next arg;
}
}
push(#newargs,$arg);
}
#ARGV = #newargs;
}
In Perl types SCALAR, ARRAY, and HASH are distinguished by the leading sigil in the variable name, $, #, and % respectively, but otherwise they may bear the same name. So $var and #var and %var are, other than belonging to the same *var typeglob, completely distinct variables.
A key for a hash must be a scalar, let's call it $key. A value in a hash must also be a scalar, and the one corresponding to $key in the hash %h is $h{$key}, the leading $ indicating that this is now a scalar. Much like an element of an array is a scalar, thus $ary[$index].
In foreach $var (#ary) the scalar $var does not really "get values" of array elements but rather aliases them. So if you change it in the loop you change the element of the array.
(Note, you have the array #opts, not #opt.)
A few comments on the code, in the light of common practices in modern Perl
One must always have use warnings; on top. The other one that is a must is use strict, as it enforces declaration of variables with my (or our), promoting all kinds of good behavior. This restricts the scope of lexical (my) variables to the nearest block enclosing the declaration
A declaration may be done inside loops as well, allowing while (my $var = EXPR) {} and foreach my $var (LIST) {}, and then $var is scoped to the loop's block (and is also usable in the rest of the condition)
The local is a different beast altogether, and in this code those should be my instead
Loop labels like arg: are commonly typed in block capitals
The & in front of a function name has rather particular meanings† and is rarely needed. It is almost certainly not needed in this code
Assignment of empty list like my #optArgs = () when the variable is declared (with my) is unneeded and has no effect (with a global #name it may be needed to clear it from higher scope)
In this code there is no need to introduce variables first since they are global and thus created the first time they are used, much like in Python – Unless they are used outside of this sub, in which case they may need to be cleared. That's the thing with globals, they radiate throughout all code
Except for the last two these are unrelated to your Python translation task, for this script.
† It suppresses prototypes; informs interpreter (at runtime) that it's a subroutine and not a "bareword", if it hasn't been already defined (can do that with () after the name); ensures that the user-defined sub is called, if there is a builtin of the same name; passes the caller's #_ (when without parens, &name); in goto &name and defined ...
See, for example, this post and this post
First, to make it clear, in perl, as in shell, you could surround variable names in curly brackets {} :
${bimbo} and $bimbo are the same scalar variable.
#bimbo is an array pointer ;
%bimbo is a hash pointer ;
$bimbo is a scalar (unique value).
To address an array or hash value, you will have to use the '$' :
$bimbo{'index'} is the 'index' value of hash %bimbo.
If $i contains an int, such as 1 for instance,
$bimbo[$i] is the second value of the array #bimbo.
So, as you can see, # or % ALWAYS refers to the whole array, as $bimbo{} or $bimbo[] refers to any value of the hash or array.
${bimbo[4]} refers to the 5th value of the array #bimbo ; %{bimbo{'index'}} refers to the 'index' value of array %bimbo.
Yes, all those structures could have the same name. This is one of the obvious syntax of perl.
And, euhm… always think in perl as a C++ edulcored syntax, it is simplified, but it is C.
Lets say I have a large hash and I want to iterate over the contents of it contents. The standard idiom would be something like this:
while(($key, $value) = each(%{$hash_ref})){
///do something
}
However, if I understand my perl correctly this is actually doing two things. First the
%{$hash_ref}
is translating the ref into list context. Thus returning something like
(key1, value1, key2, value2, key3, value3 etc)
which will be stored in my stacks memory. Then the each method will run, eating the first two values in memory (key1 & value1) and returning them to my while loop to process.
If my understanding of this is right that means that I have effectively copied my entire hash into my stacks memory only to iterate over the new copy, which could be expensive for a large hash, due to the expense of iterating over the array twice, but also due to potential cache hits if both hashes can't be held in memory at once. It seems pretty inefficient. I'm wondering if this is what really happens, or if I'm either misunderstanding the actual behavior or the compiler optimizes away the inefficiency for me?
Follow up questions, assuming I am correct about the standard behavior.
Is there a syntax to avoid copying of the hash by iterating over it values in the original hash? If not for a hash is there one for the simpler array?
Does this mean that in the above example I could get inconsistent values between the copy of my hash and my actual hash if I modify the hash_ref content within my loop; resulting in $value having a different value then $hash_ref->($key)?
No, the syntax you quote does not create a copy.
This expression:
%{$hash_ref}
is exactly equivalent to:
%$hash_ref
and assuming the $hash_ref scalar variable does indeed contain a reference to a hash, then adding the % on the front is simply 'dereferencing' the reference - i.e. it resolves to a value that represents the underlying hash (the thing that $hash_ref was pointing to).
If you look at the documentation for the each function, you'll see that it expects a hash as an argument. Putting the % on the front is how you provide a hash when what you have is a hashref.
If you wrote your own subroutine and passed a hash to it like this:
my_sub(%$hash_ref);
then on some level you could say that the hash had been 'copied', since inside the subroutine the special #_ array would contain a list of all the key/value pairs from the hash. However even in that case, the elements of #_ are actually aliases for the keys and values. You'd only actually get a copy if you did something like: my #args = #_.
Perl's builtin each function is declared with the prototype '+' which effectively coerces a hash (or array) argument into a reference to the underlying data structure.
As an aside, starting with version 5.14, the each function can also take a reference to a hash. So instead of:
($key, $value) = each(%{$hash_ref})
You can simply say:
($key, $value) = each($hash_ref)
No copy is created by each (though you do copy the returned values into $key and $value through assignment). The hash itself is passed to each.
each is a little special. It supports the following syntaxes:
each HASH
each ARRAY
As you can see, it doesn't accept an arbitrary expression. (That would be each EXPR or each LIST). The reason for that is to allow each(%foo) to pass the hash %foo itself to each rather than evaluating it in list context. each can do that because it's an operator, and operators can have their own parsing rules. However, you can do something similar with the \% prototype.
use Data::Dumper;
sub f { print(Dumper(#_)); }
sub g(\%) { print(Dumper(#_)); } # Similar to each
my %h = (a=>1, b=>2);
f(%h); # Evaluates %h in list context.
print("\n");
g(%h); # Passes a reference to %h.
Output:
$VAR1 = 'a'; # 4 args, the keys and values of the hash
$VAR2 = 1;
$VAR3 = 'b';
$VAR4 = 2;
$VAR1 = { # 1 arg, a reference to the hash
'a' => 1,
'b' => 2
};
%{$h_ref} is the same as %h, so all of the above applies to %{$h_ref} too.
Note that the hash isn't copied even if it is flattened. The keys are "copied", but the values are returned directly.
use Data::Dumper;
my %h = (abc=>"def", ghi=>"jkl");
print(Dumper(\%h));
$_ = uc($_) for %h;
print(Dumper(\%h));
Output:
$VAR1 = {
'abc' => 'def',
'ghi' => 'jkl'
};
$VAR1 = {
'abc' => 'DEF',
'ghi' => 'JKL'
};
You can read more about this here.
I was reviewing this question, esp the response from Mr Eric Strom, and had a question regarding a portion of the more "magical" element within. Please review the linked question for the context as I'm only trying to understand the inner portion of this block:
for (qw($SCALAR #ARRAY %HASH)) {
my ($sigil, $type) = /(.)(.+)/;
if (my $ref = *$glob{$type}) {
$vars{$sigil.$name} = /\$/ ? $$ref : $ref
}
}
So, it loops over three words, breaking each into two vars, $sigil and $type. The if {} block is what I am not understanding. I suspect the portion inside the ( .. ) is getting a symbolic reference to the content within $glob{$type}... there must be some "magic" (some esoteric element of the underlying mechanism that I don't yet understand) relied upon there to determine the type of the "pointed-to" data?
The next line is also partly baffling. Appears to me that we are assigning to the vars hash, but what is the rhs doing? We did not assign to $_ in the last operation ($ref was assigned), so what is being compared to in the /\$/ block? My guess is that, if we are dealing with a scalar (though I fail to discern how we are), we deref the $ref var and store it directly in the hash, otherwise, we store the reference.
So, just looking for a little tale of what is going on in these three lines. Many thanks!
You have hit upon one of the most arcane parts of the Perl language, and I can best explain by referring you to Symbol Tables and Typeglobs from brian d foy's excellent Mastering Perl. Note also that there are further references to the relevant sections of Perl's own documentation at the bottom of the page, the most relevant of which is Typeglobs and Filehandles in perldata.
Essentially, the way perl symbol tables work is that every package has a "stash" -- a "symbol table hash" -- whose name is the same as the package but with a pair of trailing semicolons. So the stash for the default package main is called %main::. If you run this simple program
perl -E"say for keys %main::"
you will see all the familiar built-in identifiers.
The values for the stash elements are references to typeglobs, which again are hashes but have keys that correspond to the different data types, SCALAR, ARRAY, HASH, CODE etc. and values that are references to the data item with that type and identifier.
Suppose you define a scalar variable $xx, or more fully, $main:xx
our $xx = 99;
Now the stash for the main package is %main::, and the typeglob for all data items with the identifier xx is referenced by $main::{xx} so, because the sigil for typeglobs is a star * in the same way that scalar identifiers have a dollar $, we can dereference this as *{$main::{xx}}. To get the reference to the scalar variable that has the identifier xx, this typeglob can be indexed with the SCALAR string, giving *{$main::{xx}}{SCALAR}. Once more, this is a reference to the variable we're after, so to collect its value it needs dereferencing once again, and if you write
say ${*{$main::{xx}}{SCALAR}};
then you will see 99.
That may look a little complex when written in a single statement, but it is fairly stratighforward when split up. The code in your question has the variable $glob set to a reference to a typeglob, which corresponds to this with respect to $main::xx
my $type = 'SCALAR';
my $glob = $main::{xx};
my $ref = *$glob{$type};
now if we say $ref we get SCALAR(0x1d12d94) or similar, which is a reference to $main::xx as before, and printing $$ref will show 99 as expected.
The subsequent assignment to #vars is straightforward Perl, and I don't think you should have any problem understanding that once you get the principle that a packages symbol table is a stash of typglobs, or really just a hash of hashes.
The elements of the iteration are strings. Since we don't have a lexical variable at the top of the loop, the element variable is $_. And it retains that value throughout the loop. Only one of those strings has a literal dollar sign, so we're telling the difference between '$SCALAR' and the other cases.
So what it is doing is getting 3 slots out of a package-level typeglob (sometimes shortened, with a little ambiguity to "glob"). *g{SCALAR}, *g{ARRAY} and *g{HASH}. The glob stores a hash and an array as a reference, so we simply store the reference into the hash. But, the glob stores a scalar as a reference to a scalar, and so needs to be dereferenced, to be stored as just a scalar.
So if you had a glob *a and in your package you had:
our $a = 'boo';
our #a = ( 1, 2, 3 );
our %a = ( One => 1, Two => 2 );
The resulting hash would be:
{ '$a' => 'boo'
, '%a' => { One => 1, Two => 2 }
, '#a' => [ 1, 2, 3 ]
};
Meanwhile the glob can be thought to look like this:
a =>
{ SCALAR => \'boo'
, ARRAY => [ 1, 2, 3 ]
, HASH => { One => 1, Two => 2 }
, CODE => undef
, IO => undef
, GLOB => undef
};
So to specifically answer your question.
if (my $ref = *$glob{$type}) {
$vars{$sigil.$name} = /\$/ ? $$ref : $ref
}
If a slot is not used it is undef. Thus $ref is assigned either a reference or undef, which evaluates to true as a reference and false as undef. So if we have a reference, then store the value of that glob slot into the hash, taking the reference stored in the hash, if it is a "container type" but taking the value if it is a scalar. And it is stored with the key $sigil . $name in the %vars hash.
I'm just trying to learn a bit of Perl and have come across this:
foreach $element (#{$records})
{
do something;
}
To my newbie eyes, this reads:
"for each element in an array named #{$records}, do something"
but, since that seems an unlikely name for an array (with "#{$" altogether), I imagine it isn't that simple?
I've also come across "%$" used together.
I know % signifies a hash and $ signifies a scalar but don't know what they mean together.
Can anyone shed any light on these?
In Perl you can have a reference (a pointer) to a data structure:
# an array
my #array;
# a reference to an array
my $ref = \#array;
When you have a reference to be able to use the array you need to dereference it
#{ $ref }
If you need to access an element as in
$array[0]
you can do the same with a reference
${$ref}[0]
The curly brackets {} are optional and you can also use
$$ref[0]
#$ref
but I personally find them less readable.
The same applies to every other type (as %$ for a hash reference).
See man perlref for the details and man perlreftut for a tutorial.
Edit
The arrow operator -> can also be used to dereference an array or an hash
$array_ref->[0]
or
$hash_ref->{key}
See man perlop for details
If you have a reference to an array or a hash, you would use a scalar to hold the reference:
my $href = \%hash;
my $aref = \#array;
When you want to de-reference these references, you would use the symbol appropriate for the reference type:
for my $element (#$aref) {
}
for my $key (keys %$href) {
}