What is the difference between
myArr1 => \#existingarray
and
myArr2 => [
#existingarray
]
I am assigning the #existingarray to a element in a hash map.
I mean what exactly internally happens. Is it that for the first one, it points to the same array and for the second array it creates a new array with the elements in the #existingarray
Thanks in advance
Yes, the first one takes a reference, the second one does a copy and then takes a reference.
[ ... ] is the anonymous array constructor, and turns the list inside into an arrayref.
So with #a = 1, 2, 3,
[ #a ]
is the same as
[ 1, 2, 3 ]
(the array is flattened to a list) or
do {
my #b = #a;
\#b;
}
In effect, the elements get copied.
Also,
my ($ref1, $ref2) = (\#a, [#a]);
print "$ref1 and $ref2 are " . ($ref1 eq $ref2 ? "equal" : "not equal") . "\n";
would confirm that they are not the same. And if we do
$ref1->[0] = 'a';
$ref2->[0] = 'b';
then $a[0] would equal a and not b.
You can use
perl -e 'my #a=(1); my $ra=\#a; my $rca=[#a]; $ra->[0]=2; print #a, #{$ra}, #{$rca};'
221
to see that your assumption that [#existingarray] creates a reference to a copy of #existingarray is correct (and that myArray* isn't Perl).
WRT amon's revising my perl -e "..." (fails under bash) to perl -e '...' (fails under cmd.exe): Use the quotes that work for your shell.
The square brackets make a reference to a new array with a copy of what's in #existingarray at the time of the assignment.
Related
This seems like a simple question, but Perl6/Raku isn't behaving as I'd expect. I'm trying to create a reference to an array within a hash, but am not getting the expected behavior. In Perl5, the answer would involve accessing the array by reference, but I don't see equivalent syntax for Perl6/Raku.
my $jsonstr = q:to/END/;
{
"arr" : [
"alpha","beta","delta","gamma"
]
}
END
my %json = from-json $jsonstr;
my #arr = %json{'arr'};
say "Arr length is " ~ #arr.elems; # Expect 4, get 1
say "Orig length is " ~ %json{'arr'}.elems; # Get expected value of 4
say "Arr[0] is " ~#arr[0].^name ~ " of length " ~ #arr[0].elems; # First index is array
say %json{'arr'}[0]; # Indexing into array in original location works as expected
say #arr[0][0]; # But when assigned, it needs an extra index
my #arr2 = #arr[0]; # Same issue in re-assignment here
say "Arr2[0]: " ~ #arr2[0] ~ ", length of " ~ #arr2.elems;
How do I get a new #arr variable to reference the nested array without this confusing extra [0] index layer? Is this a bug, or am I missing something in my understanding of Raku's Array/ref handling? Thanks.
When you assign the value in the key arr to the Array #arr it takes the value in %json{'arr'} which is the Array Object ["alpha","beta","delta","gamma"] and puts it into #arr so you get an Array of Array's with 1 item.
You've got a few options :
You can bind #arr to %json{"arr"} with my #arr := %json{"arr"}
Or you can pass the %json{"arr"} to a list with my (#arr) = %json{"arr"}
You have to remember in Raku Array's are Objects.
As usual, after writing+posting my question, I answered my own question.
my #arr = %json{'arr'}.Array;
I don't quite understand why this is necessary, but it gives the desired behavior.
I have two global multidimensional arrays #p and #p0e in Perl. This is part of a genetic algorith where I want to save certain keys from #p to #p0e. Modifications are then made to #p. There are several subroutines that make modifications to #p, but there's a certain subroutine where on occasion (not on every iteration) a modification to #p also leads to #p0e being modified (it receives the same keys) although #p0e should not be affected.
# this is the sub where part of #p is copied to #p0e
sub saveElite {
#p0e = (); my $i = 0;
foreach my $r (sort({$a<=>$b} keys $f{"rank"})) {
if ($i<$elN) {
$p0e[$i] = $p[$f{"rank"}{$r}]; # save chromosome
}
else {last;}
$i++;
}
}
# this is the sub that then sometimes changes #p0e
sub mutation {
for (my $i=0; $i<#p; $i++) {
for (my $j=0; $j<#{$p[$i]}; $j++) {
if (rand(1)<=$mut) { # mutation
$p[$i][$j] = mutate($p[$i][$j]);
}
}
}
}
I thought maybe I'd somehow created a reference to the original array rather than a copy, but because this unexpected behaviour doesn't happen on every iteration this shouldn't be the case.
$j = $f{"rank"}{$r};
$p0e[$i] = $p[$j];
$p[$j] is an array reference, which you can think of as pointing to a particular list of data at a particular memory address. The assignment to $p0e[$i] also tells Perl to let the $i-th row of #p0e also refer to that same block of memory. So when you later make a change to $p0e[$i][$k], you'll find the value of $p[$j][$k] has changed too.
To fix this, you'll want to assign a copy of $p[$j]. Here is one way you can do that:
$p0e[$i] = [ #{$p[$j]} ];
#{$p[$j]} deferences the array reference and [...] creates a new reference for it, so after this statement $p0e[$i] will have the same contents with the same values as $p[$j] but point to a different block of memory.
I think your problem will probably be this:
$p0e[$i] = $p[$f{"rank"}{$r}]; # save chromosome
Because it looks like #p is a multi-dimensional array.
The problem is - the way perl 'does' multi dimensional arrays is via arrays of references. So if you copy an inner array, you do so by reference.
E.g.:
#!c:\Strawberry\perl\bin
use strict;
use warnings;
use Data::Dumper;
my #list = ( [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ], );
print Dumper \#list;
my #other_list;
push ( #other_list, #list[0,1] ); #make a sub list of two rows;
print Dumper \#other_list;
### all looks good.
## but if we:
print "List:\n";
print join ("\n",#list),"\n";
print "Other List:\n";
print join ("\n", #other_list),"\n";
$list[1][1] = 9;
print Dumper \#other_list;
You will see that by changing an element in #list we also modify #other_list - and if we just print them we get:
List:
ARRAY(0x2ea384)
ARRAY(0x12cef34)
ARRAY(0x12cf024)
Other List:
ARRAY(0x2ea384)
ARRAY(0x12cef34)
Note the duplicate numbers - that means you have the same reference.
The easiest way of working around this is by using [] judicously:
push ( #other_list, [#{$list[0]}], [#{$list[1]}] ); #make a sub list of two rows;
This will then insert anonymous arrays (new ones) containing the dereferenced elements of the list.
Whilst we're at it though - please turn on strict and warnings. They will save you a lot of pain in the long run.
That's because it's an array of arrays. The first level array stores only references to the inner arrays, if you modify the inner array, it's changed in both arrays - they both refer to the same array. Clone the deep copy instead of creating a shallow one.
Perhaps I have made this more complicated than I need it to be but I am currently trying to store an array that contains, among other things, an array inside a hash in Perl.
i.e. hash -> array -> array
use strict;
my %DEVICE_INFORMATION = {}; #global hash
sub someFunction() {
my $key = 'name';
my #storage = ();
#assume file was properly opened here for the foreach-loop
foreach my $line (<DATA>) {
if(conditional) {
my #ports = ();
$storage[0] = 'banana';
$storage[1] = \#ports;
$storage[2] = '0';
$DEVICE_INFORMATION{$key} = \#storage;
}
elsif(conditional) {
push #{$DEVICE_INFORMATION{$key}[1]}, 5;
}
}#end foreach
} #end someFunction
This is a simplified version of the code I am writing. I have a subroutine that I call in the main. It parses a very specifically designed file. That file guarantees that the if statement fires before subsequent elsif statement.
I think the push call in the elsif statement is not working properly - i.e. 5 is not being stored in the #ports array that should exist in the #storage array that should be returned when I hash the key into DEVICE_INFORMATION.
In the main I try and print out each element of the #storage array to check that things are running smoothly.
#main execution
&someFunction();
print $DEVICE_INFORMATION{'name'}[0];
print $DEVICE_INFORMATION{'name'}[1];
print $DEVICE_INFORMATION{'name'}[2];
The output for this ends up being... banana ARRAY(blahblah) 0
If I change the print statement for the middle call to:
print #{$DEVICE_INFORMATION{'name'}[1]};
Or to:
print #{$DEVICE_INFORMATION{'name'}[1]}[0];
The output changes to banana [blankspace] 0
Please advise on how I can properly update the #ports array while it is stored inside the #storage array that has been hash'd into DEVICE_INFORMATION and then how I can access the elements of #ports. Many thanks!
P.S. I apologize for the length of this post. It is my first question on stackoverflow.
I was going to tell you that Data::Dumper can help you sort out Perl data structures, but Data::Dumper can also tell you about your first problem:
Here's what happens when you sign open-curly + close-curly ( '{}' ) to a hash:
use Data::Dumper ();
my %DEVICE_INFORMATION = {}; #global hash
print Dumper->Dump( [ \%DEVICE_INFORMATION ], [ '*DEVICE_INFORMATION ' ] );
Here's the output:
%DEVICE_INFORMATION = (
'HASH(0x3edd2c)' => undef
);
What you did is you assigned the stringified hash reference as a key to the list element that comes after it. implied
my %DEVICE_INFORMATION = {} => ();
So Perl assigned it a value of undef.
When you assign to a hash, you assign a list. A literal empty hash is not a list, it's a hash reference. What you wanted to do for an empty hash--and what is totally unnecessary--is this:
my %DEVICE_INFORMATION = ();
And that's unnecessary because it is exactly the same thing as:
my %DEVICE_INFORMATION;
You're declaring a hash, and that statement fully identifies it as a hash. And Perl is not going to guess what you want in it, so it's an empty hash from the get-go.
Finally, my advice on using Data::Dumper. If you started your hash off right, and did the following:
my %DEVICE_INFORMATION; # = {}; #global hash
my #ports = ( 1, 2, 3 );
# notice that I just skipped the interim structure of #storage
# and assigned it as a literal
# * Perl has one of the best literal data structure languages out there.
$DEVICE_INFORMATION{name} = [ 'banana', \#ports, '0' ];
print Data::Dumper->Dump(
[ \%DEVICE_INFORMATION ]
, [ '*DEVICE_INFORMATION' ]
);
What you see is:
%DEVICE_INFORMATION = (
'name' => [
'banana',
[
1,
2,
3
],
'0'
]
);
So, you can better see how it's all getting stored, and what levels you have to deference and how to get the information you want out of it.
By the way, Data::Dumper delivers 100% runnable Perl code, and shows you how you can specify the same structure as a literal. One caveat, you would have to declare the variable first, using strict (which you should always use anyway).
You update #ports properly.
Your print statement accesses $storage[1] (reference to #ports) in wrong way.
You may use syntax you have used in push.
print $DEVICE_INFORMATION{'name'}[0], ";",
join( ':', #{$DEVICE_INFORMATION{'name'}[1]}), ";",
$DEVICE_INFORMATION{'name'}[2], "\n";
print "Number of ports: ", scalar(#{$DEVICE_INFORMATION{'name'}[1]})),"\n";
print "First port: ", $DEVICE_INFORMATION{'name'}[1][0]//'', "\n";
# X//'' -> X or '' if X is undef
I once read the following example about "array of arrays". AOA is a two dimensional array
The following code segment is claimed to print the whole thing with refs
for $aref ( #AoA ) {
print "\t [ #$aref ],\n";
}
And the following code segment is claimed to print the whole thing with indices
for $i ( 0 .. $#AoA ) {
print "\t [ #{$AoA[$i]} ],\n";
}
What's the $aref stand for here? How to understand the definition of #$aref and #{$AoA[$i]}? Thanks.
$aref stands for "array reference", i.e. a reference for an array.
my $my_aref = \#somearray;
You can make an array from an array reference with the following syntax:
#{$my_aref}
#{$my_aref} is #somearray. (It's not a copy, it really is the same array.)
In second example, $AoA[$i] is an array reference, and you dereference it with the same syntax: #{$AoA[$i]}.
See perlreftut for more explanations and examples.
An "array of arrays" isn't actually an array of arrays. It's more an array of array references. Each element in the base array is a reference to another array. Thus, when you want to cycle through the elements in the base array, you get back array references. These are what get assigned to $aref in the first loop. They are then de-referenced by pre-pending with the # symbol, so #$aref is the array referenced by the $aref array reference.
Same sort of thing works for the second loop. $AoA[$i] is the $i-th element of the #AoA array, which is an array reference. De-referencing it by pre-pending it with the # symbol (and adding {} for clarity, and possibly for precedence) means #{$AoA[$i]} is the array referenced by the $AoA[$i] array reference.
Perl doesn't have multidimensional arrays. One places arrays into other arrays to achieve the same result.
Well, almost. Arrays (and hashes) values are scalars, so one cannot place an array into another array. What one does instead of place a reference to an array instead.
In other words, "array of arrays" is short for "array of references to arrays". Each value of the #AoA is a reference to another array, given the "illusion" of a two-dimensional array.
The reference come from the use [ ] or equivalent. [ ] creates an anonymous array, then creates a reference to that array, then returns the reference. That's where the reference comes from.
Common ways of building an AoA:
my #AoA = (
[ 'a', 'b', 'c' ],
[ 'd', 'e', 'f' ],
);
my #AoA;
push #AoA, [ 'a', 'b', 'c' ];
push #AoA, [ 'd', 'e', 'f' ];
my #AoA;
$AoA[$y][$x] = $n;
Keep in mind that
$AoA[$y][$x] = $n;
is short for
$AoA[$y]->[$x] = $n;
and it's equivalent to the following thanks to autovivification:
( $AoA[$y] //= [] )->[$x] = $n;
The whole mystery with multi-dimension structures in perl is quite easy to understand once you realize that there are only three types of variables to deal with. Scalars, arrays and hashes.
A scalar is a single value, it can contain just about anything, but
only one at the time.
An array contains a number of scalar values, ordered by a fixed
numerical index.
A hash contains scalar values, indexed by keys made of strings.
And all arrays, hashes or scalars act this way. Multi-dimension arrays are no different from single dimension.
This is also expressed very succinctly in perldata:
All data in Perl is a scalar, an array of scalars, or a hash of
scalars. A scalar may contain one single value in any of three
different flavors: a number, a string, or a reference. In general,
conversion from one form to another is transparent. Although a scalar
may not directly hold multiple values, it may contain a reference to
an array or hash which in turn contains multiple values.
For example:
my #array = (1, 2, 3);
Here, $array[0] contains 1, $array[1] contains 2, etc. Just like you would expect.
my #aoa = ( [ 1, 2, 3 ], [ 'a', 'b', 'c' ] );
Here, $array[0] contains an array reference. If you print it out, it will say something like ARRAY(0x398a84). Don't worry! That's still a scalar value. How do we know this? Because arrays can only contain scalar values.
When we do something like
for $aref ( #AoA ) {
print $aref; # prints ARRAY(0x398a84) or similar
}
It's no different from doing
for $number ( #array ) {
print $number;
}
$aref and $number are scalar values. So far, so good. Take a moment and lock this knowledge down: Arrays can only contain scalar values.
Now, the next part is simply knowing how to deal with references. This is documented in perlref and perlreftut.
A reference is a scalar value. It's an address to a location in memory. This location contains some data. In order to access the actual data, we need to dereference the reference.
As a simple example:
my #data = (1, 2, 3);
my $aref = \#data; # The backslash in front of the sigil creates a reference
print $aref; # print something like ARRAY(0xa4b6a4)
print #$aref; # prints 123
Adding a sigil in front of the reference tells perl to dereference the scalar value into the type of data the sigil represents. In this case, an array. If you choose the wrong sigil for the type of reference, perl will give an error such as:
Not a HASH reference
In the example above, we have a reference to a specific, named location. Both #$aref and #data access the same values. If we change a value in one, both are affected, because the address to the memory location is identical. Let's try it:
my #data = (1, 2, 3);
my $aref = \#data;
$$aref[1] = 'a'; # dereference to a scalar value by $ sigil
# $aref->[1] = 'a' # does the same thing, a different way
print #data; # prints 1a3
print #$aref; # prints 1a3
We can also have anonymous data. If we were only interested in building an array of arrays, we'd have no interest in the #data, and could skip it by doing this:
my $aref = [ 1, 2, 3 ];
The brackets around the list of numbers create an anonymous array. $aref still contains the same type of data: A reference. But in this case, $aref is the only way we have of accessing the data contained at the memory location. Now, let's build some more scalar values like this:
my $aref1 = [ 1, 2, 3 ];
my $aref2 = [ 'a', 'b', 'c' ];
my $aref3 = [ 'x', 'y', 'z' ];
We now have three scalar variables that contain references to anonymous arrays. What if we put these in an array?
my #aoa = ($aref1, $aref2, $aref3);
If we'd want to access $aref1, we could do print #$aref1, but we could also do
print #{$aoa[0]};
In this case, we need to use the extended form of dereferencing: #{ ... }. Because perl does not like ambiguity, it requires us to distinguish between #{$aoa[0]} (take the reference in $aoa[0] and dereference as an array) and #{$aoa}[0] (take the reference in $aoa and dereference as an array, and take that arrays first value).
Above, we could have used #{$aref}, as it is identical to #$aref.
So, if we are only interested in building an array of arrays, we are not really interested in the $aref1 scalars either. So let's cut them out of the process:
my #aoa = ( [ 1, 2, 3 ], [ 'a', 'b', 'c' ], [ 'x', 'y', 'z' ]);
Tada! That's an array of arrays.
Now, we can backtrack. To access the values inside this array, we can do
for my $scalar ( #aoa ) {
print #$scalar; # prints 123abcxyz
}
This time, I used a different variable name, just to make a point. This loop takes each value from #aoa -- which still is only a scalar value -- dereferences it as an array, and prints it.
Or we can access #aoa via its indexes
for my $i ( 0 .. $#aoa ) {
print #{$aoa[$i]};
}
And that's all there is to it!
Given three scalars, what is the perl syntax to fill a hash in which one of the scalars is the key, another determines which of two arrays is filled, and the third is appended to one of the arrays? For example:
my $weekday = "Monday";
my $kind = "Good";
my $event = "Birthday";
and given only the scalars and not their particular values, obtained inside a loop, I want a hash like:
my %Weekdays = {
'Monday' => [
["Birthday", "Holiday"], # The Good array
["Exam", "Workday"] # The Bad array
]
'Saturday' => [
["RoadTrip", "Concert", "Movie"],
["Yardwork", "VisitMIL"]
]
}
I know how to append a value to an array in a hash, such as if the key is a single array:
push( #{ $Weekdays{$weekday} }, $event);
Used in a loop, that could give me:
%Weekdays = {
'Monday' => [
'Birthday',
'Holiday',
'Exam',
'Workday'
]
}
I suppose the hash key is the particular weekday, and the value should be a two dimensional array. I don't know the perl syntax to, say, push Birthday into the hash as element [0][0] of the weekday array, and the next time through the loop, push another event in as [0][1] or [1][0]. Similarly, I don't know the syntax to access same.
Using your variables, I'd write it like this:
push #{ $Weekdays{ $weekday }[ $kind eq 'Good' ? 0 : 1 ] }, $event;
However, I'd probably just make the Good/Bad specifiers keys as well. And given my druthers:
use autobox::Core;
( $Weekdays{ $weekday }{ $kind } ||= [] )->push( $event );
Note that the way I've written it here, neither expression cares whether or not an array exists before we start.
Is there some reason that
push #{ $Weekdays{Monday}[0] }, "whatever";
isn’t working for you?