$m{$LoL,$b}
notation borrowed from awk in which the keys are actually more like a single concatenated string "$LoL$b"
, but traversal and sorting were difficult. More desperate programmers even
hacked Perl's internal symbol table directly, a strategy that proved hard
to develop and maintain--to put it mildly.
The 5.0 release of Perl let us have complex data structures. You may now write something like this and all of a sudden, you'd have a array with three dimensions!
for $x (1 .. 10) { for $y (1 .. 10) { for $z (1 .. 10) { $LoL[$x][$y][$z] = $x ** $y + $z; } } }
Alas, however simple this may appear, underneath it's a much more elaborate construct than meets the eye!
How do you print it out? Why can't you say just print @LoL
? How do you sort it? How can you pass it to a function or get one of these
back from a function? Is is an object? Can you save it to disk to read back
later? How do you access whole rows or columns of that matrix? Do all the
values have to be numeric?
As you see, it's quite easy to become confused. While some small portion of the blame for this can be attributed to the reference-based implementation, it's really more due to a lack of existing documentation with examples designed for the beginner.
This document is meant to be a detailed but understandable treatment of the many different sorts of data structures you might want to develop. It should also serve as a cookbook of examples. That way, when you need to create one of these complex data structures, you can just pinch, pilfer, or purloin a drop-in example from here.
Let's look at each of these possible constructs in detail. There are separate sections on each of the following:
But for now, let's look at some of the general issues common to all of these types of data structures.
@ARRAY
s and %HASH
es are all internally one-dimensional. They can hold only scalar values
(meaning a string, number, or a reference). They cannot directly contain
other arrays or hashes, but instead contain references to other arrays or hashes.
You can't use a reference to a array or hash in quite the same way that you would a real array or hash. For C or C++ programmers unused to distinguishing between arrays and pointers to the same, this can be confusing. If so, just think of it as the difference between a structure and a pointer to a structure.
You can (and should) read more about references in the perlref
man page. Briefly, references are rather like pointers that know what they
point to. (Objects are also a kind of reference, but we won't be needing
them right away--if ever.) This means that when you have something which
looks to you like an access to a two-or-more-dimensional array and/or hash,
what's really going on is that the base type is merely a one-dimensional
entity that contains references to the next level. It's just that you can use it as though it were a two-dimensional one. This is actually the way almost all
C multidimensional arrays work as well.
$list[7][12] # array of arrays $list[7]{string} # array of hashes $hash{string}[7] # hash of arrays $hash{string}{'another string'} # hash of hashes
Now, because the top level contains only references, if you try to print
out your array in with a simple print
function, you'll get
something that doesn't look very nice, like this:
@LoL = ( [2, 3], [4, 5, 7], [0] ); print $LoL[1][2]; 7 print @LoL; ARRAY(0x83c38)ARRAY(0x8b194)ARRAY(0x8b1d0)
That's because Perl doesn't (ever) implicitly dereference your variables.
If you want to get at the thing a reference is referring to, then you have
to do this yourself using either prefix typing indicators, like
${$blah}
, @{$blah}
, @{$blah[$i]}
, or else postfix pointer arrows, like $a->[3]
, $h->{fred}
, or even $ob->method->[3]
.
for $i (1..10) { @list = somefunc($i); $LoL[$i] = @list; # WRONG! }
That's just the simple case of assigning a list to a scalar and getting its element count. If that's what you really and truly want, then you might do well to consider being a tad more explicit about it, like this:
for $i (1..10) { @list = somefunc($i); $counts[$i] = scalar @list; }
Here's the case of taking a reference to the same memory location again and again:
for $i (1..10) { @list = somefunc($i); $LoL[$i] = \@list; # WRONG! }
So, what's the big problem with that? It looks right, doesn't it? After all, I just told you that you need an array of references, so by golly, you've made me one!
Unfortunately, while this is true, it's still broken. All the references in
@LoL
refer to the very same place, and they will therefore all hold whatever was last in @list! It's similar to the problem demonstrated in the following
C program:
#include <pwd.h> main() { struct passwd *getpwnam(), *rp, *dp; rp = getpwnam("root"); dp = getpwnam("daemon");
printf("daemon name is %s\nroot name is %s\n", dp->pw_name, rp->pw_name); }
Which will print
daemon name is daemon root name is daemon
The problem is that both rp
and dp
are pointers to the same location in memory! In
C, you'd have to remember to malloc
yourself some new memory. In Perl, you'll want to use the array constructor []
or the hash constructor {}
instead. Here's the right way to do the preceding broken code fragments:
for $i (1..10) { @list = somefunc($i); $LoL[$i] = [ @list ]; }
The square brackets make a reference to a new array with a copy
of what's in @list
at the time of the assignment. This is what
you want.
Note that this will produce something similar, but it's much harder to read:
for $i (1..10) { @list = 0 .. $i; @{$LoL[$i]} = @list; }
Is it the same? Well, maybe so--and maybe not. The subtle difference is
that when you assign something in square brackets, you know for sure it's
always a brand new reference with a new copy of the data. Something else could be going on in this new case with the @{$LoL[$i]}}
dereference on the left-hand-side of the assignment. It all depends on
whether $LoL[$i]
had been undefined to start with, or whether it already contained a
reference. If you had already populated @LoL
with references,
as in
$LoL[3] = \@another_list;
Then the assignment with the indirection on the left-hand-side would use the existing reference that was already there:
@{$LoL[3]} = @list;
Of course, this would have the ``interesting'' effect of clobbering @another_list. (Have you ever noticed how when a programmer says something is ``interesting'', that rather than meaning ``intriguing'', they're disturbingly more apt to mean that it's ``annoying'', ``difficult'', or both? :-)
So just remember always to use the array or hash constructors with []
or {}
, and you'll be fine, although it's not always optimally efficient.
Surprisingly, the following dangerous-looking construct will actually work out fine:
for $i (1..10) { my @list = somefunc($i); $LoL[$i] = \@list; }
That's because my
is more of a run-time statement than it is a
compile-time declaration per se. This means that the my
variable is remade afresh each time
through the loop. So even though it looks as though you stored the same variable reference each time, you actually did not! This is a subtle distinction that can produce more efficient code at the risk of misleading all but the most experienced of programmers. So
I usually advise against teaching it to beginners. In fact, except for passing arguments to functions,
I seldom like to see the gimme-a-reference operator (backslash) used much at all in code. Instead,
I advise beginners that they (and most of the rest of us) should try to use the much more easily understood constructors
[]
and {}
instead of relying upon lexical (or dynamic) scoping and hidden
reference-counting to do the right thing behind the scenes.
In summary:
$LoL[$i] = [ @list ]; # usually best $LoL[$i] = \@list; # perilous; just how my() was that list? @{ $LoL[$i] } = @list; # way too tricky for most programmers
@{$LoL[$i]}
, the following are actually the same thing:
$listref->[2][2] # clear $$listref[2][2] # confusing
That's because Perl's precedence rules on its five prefix dereferencers
(which look like someone swearing: $ @ * % &
) make them bind more tightly than the postfix subscripting brackets or braces! This will no doubt come as a great shock to the
C or
C++ programmer, who is quite accustomed to using
*a[i]
to mean what's pointed to by the i'th
element of a
. That is, they first take the subscript, and only then dereference the thing at that subscript. That's fine in
C, but this isn't
C.
The seemingly equivalent construct in Perl, $$listref[$i]
first does the deref of $listref
, making it take $listref
as a reference to an array, and then
dereference that, and finally tell you the i'th value of the array pointed to by $LoL. If you wanted the
C notion, you'd have to write ${$LoL[$i]}
to force the $LoL[$i]
to get evaluated first before the leading $
dereferencer.
use strict
#!/usr/bin/perl -w use strict;
This way, you'll be forced to declare all your variables with
my
and also disallow accidental ``symbolic dereferencing''.
Therefore if you'd done this:
my $listref = [ [ "fred", "barney", "pebbles", "bambam", "dino", ], [ "homer", "bart", "marge", "maggie", ], [ "george", "jane", "elroy", "judy", ], ];
print $listref[2][2];
The compiler would immediately flag that as an error at compile time, because you were accidentally accessing @listref
, an undeclared variable, and it would thereby remind you to write instead:
print $listref->[2][2]
x
command to dump out complex data structures. For example, given the
assignment to $LoL
above, here's the debugger output:
DB<1> X $LoL $LoL = ARRAY(0x13b5a0) 0 ARRAY(0x1f0a24) 0 'fred' 1 'barney' 2 'pebbles' 3 'bambam' 4 'dino' 1 ARRAY(0x13b558) 0 'homer' 1 'bart' 2 'marge' 3 'maggie' 2 ARRAY(0x13b540) 0 'george' 1 'jane' 2 'elroy' 3 'judy'
There's also a lower-case x command which is nearly the same.
@LoL = ( [ "fred", "barney" ], [ "george", "jane", "elroy" ], [ "homer", "marge", "bart" ], );
# reading from file while ( <> ) { push @LoL, [ split ]; }
# calling a function for $i ( 1 .. 10 ) { $LoL[$i] = [ somefunc($i) ]; }
# using temp vars for $i ( 1 .. 10 ) { @tmp = somefunc($i); $LoL[$i] = [ @tmp ]; }
# add to an existing row push @{ $LoL[0] }, "wilma", "betty";
# one element $LoL[0][0] = "Fred";
# another element $LoL[1][1] =~ s/(\w)/\u$1/;
# print the whole thing with refs for $aref ( @LoL ) { print "\t [ @$aref ],\n"; }
# print the whole thing with indices for $i ( 0 .. $#LoL ) { print "\t [ @{$LoL[$i]} ],\n"; }
# print the whole thing one at a time for $i ( 0 .. $#LoL ) { for $j ( 0 .. $#{ $LoL[$i] } ) { print "elt $i $j is $LoL[$i][$j]\n"; } }
%HoL = ( flintstones => [ "fred", "barney" ], jetsons => [ "george", "jane", "elroy" ], simpsons => [ "homer", "marge", "bart" ], );
# reading from file # flintstones: fred barney wilma dino while ( <> ) { next unless s/^(.*?):\s*//; $HoL{$1} = [ split ]; }
# reading from file; more temps # flintstones: fred barney wilma dino while ( $line = <> ) { ($who, $rest) = split /:\s*/, $line, 2; @fields = split ' ', $rest; $HoL{$who} = [ @fields ]; }
# calling a function that returns a list for $group ( "simpsons", "jetsons", "flintstones" ) { $HoL{$group} = [ get_family($group) ]; }
# likewise, but using temps for $group ( "simpsons", "jetsons", "flintstones" ) { @members = get_family($group); $HoL{$group} = [ @members ]; }
# append new members to an existing family push @{ $HoL{"flintstones"} }, "wilma", "betty";
# one element $HoL{flintstones}[0] = "Fred";
# another element $HoL{simpsons}[1] =~ s/(\w)/\u$1/;
# print the whole thing foreach $family ( keys %HoL ) { print "$family: @{ $HoL{$family} }\n" }
# print the whole thing with indices foreach $family ( keys %HoL ) { print "family: "; foreach $i ( 0 .. $#{ $HoL{$family} } ) { print " $i = $HoL{$family}[$i]"; } print "\n"; }
# print the whole thing sorted by number of members foreach $family ( sort { @{$HoL{$b}} <=> @{$HoL{$a}} } keys %HoL ) { print "$family: @{ $HoL{$family} }\n" }
# print the whole thing sorted by number of members and name foreach $family ( sort { @{$HoL{$b}} <=> @{$HoL{$a}} || $a cmp $b } keys %HoL ) { print "$family: ", join(", ", sort @{ $HoL{$family}), "\n"; }
@LoH = ( { Lead => "fred", Friend => "barney", }, { Lead => "george", Wife => "jane", Son => "elroy", }, { Lead => "homer", Wife => "marge", Son => "bart", } );
# reading from file # format: LEAD=fred FRIEND=barney while ( <> ) { $rec = {}; for $field ( split ) { ($key, $value) = split /=/, $field; $rec->{$key} = $value; } push @LoH, $rec; }
# reading from file # format: LEAD=fred FRIEND=barney # no temp while ( <> ) { push @LoH, { split /[\s+=]/ }; }
# calling a function that returns a key,value list, like # "lead","fred","daughter","pebbles" while ( %fields = getnextpairset() ) { push @LoH, { %fields }; }
# likewise, but using no temp vars while (<>) { push @LoH, { parsepairs($_) }; }
# add key/value to an element $LoH[0]{pet} = "dino"; $LoH[2]{pet} = "santa's little helper";
# one element $LoH[0]{lead} = "fred";
# another element $LoH[1]{lead} =~ s/(\w)/\u$1/;
# print the whole thing with refs for $href ( @LoH ) { print "{ "; for $role ( keys %$href ) { print "$role=$href->{$role} "; } print "}\n"; }
# print the whole thing with indices for $i ( 0 .. $#LoH ) { print "$i is { "; for $role ( keys %{ $LoH[$i] } ) { print "$role=$LoH[$i]{$role} "; } print "}\n"; }
# print the whole thing one at a time for $i ( 0 .. $#LoH ) { for $role ( keys %{ $LoH[$i] } ) { print "elt $i $role is $LoH[$i]{$role}\n"; } }
%HoH = ( flintstones => { lead => "fred", pal => "barney", }, jetsons => { lead => "george", wife => "jane", "his boy" => "elroy", }, simpsons => { lead => "homer", wife => "marge", kid => "bart", }, );
# reading from file # flintstones: lead=fred pal=barney wife=wilma pet=dino while ( <> ) { next unless s/^(.*?):\s*//; $who = $1; for $field ( split ) { ($key, $value) = split /=/, $field; $HoH{$who}{$key} = $value; }
# reading from file; more temps while ( <> ) { next unless s/^(.*?):\s*//; $who = $1; $rec = {}; $HoH{$who} = $rec; for $field ( split ) { ($key, $value) = split /=/, $field; $rec->{$key} = $value; } }
# calling a function that returns a key,value hash for $group ( "simpsons", "jetsons", "flintstones" ) { $HoH{$group} = { get_family($group) }; }
# likewise, but using temps for $group ( "simpsons", "jetsons", "flintstones" ) { %members = get_family($group); $HoH{$group} = { %members }; }
# append new members to an existing family %new_folks = ( wife => "wilma", pet => "dino"; );
for $what (keys %new_folks) { $HoH{flintstones}{$what} = $new_folks{$what}; }
# one element $HoH{flintstones}{wife} = "wilma";
# another element $HoH{simpsons}{lead} =~ s/(\w)/\u$1/;
# print the whole thing foreach $family ( keys %HoH ) { print "$family: { "; for $role ( keys %{ $HoH{$family} } ) { print "$role=$HoH{$family}{$role} "; } print "}\n"; }
# print the whole thing somewhat sorted foreach $family ( sort keys %HoH ) { print "$family: { "; for $role ( sort keys %{ $HoH{$family} } ) { print "$role=$HoH{$family}{$role} "; } print "}\n"; }
# print the whole thing sorted by number of members foreach $family ( sort { keys %{$HoH{$b}} <=> keys %{$HoH{$a}} } keys %HoH ) { print "$family: { "; for $role ( sort keys %{ $HoH{$family} } ) { print "$role=$HoH{$family}{$role} "; } print "}\n"; }
# establish a sort order (rank) for each role $i = 0; for ( qw(lead wife son daughter pal pet) ) { $rank{$_} = ++$i }
# now print the whole thing sorted by number of members foreach $family ( sort { keys %{ $HoH{$b} } <=> keys %{ $HoH{$a} } } keys %HoH ) { print "$family: { "; # and print these according to rank order for $role ( sort { $rank{$a} <=> $rank{$b} } keys %{ $HoH{$family} } ) { print "$role=$HoH{$family}{$role} "; } print "}\n"; }
$rec = { TEXT => $string, SEQUENCE => [ @old_values ], LOOKUP => { %some_table }, THATCODE => \&some_function, THISCODE => sub { $_[0] ** $_[1] }, HANDLE => \*STDOUT, };
print $rec->{TEXT};
print $rec->{LIST}[0]; $last = pop @ { $rec->{SEQUENCE} };
print $rec->{LOOKUP}{"key"}; ($first_k, $first_v) = each %{ $rec->{LOOKUP} };
$answer = &{ $rec->{THATCODE} }($arg); $answer = &{ $rec->{THISCODE} }($arg1, $arg2);
# careful of extra block braces on fh ref print { $rec->{HANDLE} } "a string\n";
use FileHandle; $rec->{HANDLE}->autoflush(1); $rec->{HANDLE}->print(" a string\n");
%TV = ( flintstones => { series => "flintstones", nights => [ qw(monday thursday friday) ], members => [ { name => "fred", role => "lead", age => 36, }, { name => "wilma", role => "wife", age => 31, }, { name => "pebbles", role => "kid", age => 4, }, ], },
jetsons => { series => "jetsons", nights => [ qw(wednesday saturday) ], members => [ { name => "george", role => "lead", age => 41, }, { name => "jane", role => "wife", age => 39, }, { name => "elroy", role => "kid", age => 9, }, ], },
simpsons => { series => "simpsons", nights => [ qw(monday) ], members => [ { name => "homer", role => "lead", age => 34, }, { name => "marge", role => "wife", age => 37, }, { name => "bart", role => "kid", age => 11, }, ], }, );
# reading from file # this is most easily done by having the file itself be # in the raw data format as shown above. perl is happy # to parse complex data structures if declared as data, so # sometimes it's easiest to do that
# here's a piece by piece build up $rec = {}; $rec->{series} = "flintstones"; $rec->{nights} = [ find_days() ];
@members = (); # assume this file in field=value syntax while (<>) { %fields = split /[\s=]+/; push @members, { %fields }; } $rec->{members} = [ @members ];
# now remember the whole thing $TV{ $rec->{series} } = $rec;
########################################################### # now, you might want to make interesting extra fields that # include pointers back into the same data structure so if # change one piece, it changes everywhere, like for examples # if you wanted a {kids} field that was an array reference # to a list of the kids' records without having duplicate # records and thus update problems. ########################################################### foreach $family (keys %TV) { $rec = $TV{$family}; # temp pointer @kids = (); for $person ( @{ $rec->{members} } ) { if ($person->{role} =~ /kid|son|daughter/) { push @kids, $person; } } # REMEMBER: $rec and $TV{$family} point to same data!! $rec->{kids} = [ @kids ]; }
# you copied the list, but the list itself contains pointers # to uncopied objects. this means that if you make bart get # older via
$TV{simpsons}{kids}[0]{age}++;
# then this would also change in print $TV{simpsons}{members}[2]{age};
# because $TV{simpsons}{kids}[0] and $TV{simpsons}{members}[2] # both point to the same underlying anonymous hash table
# print the whole thing foreach $family ( keys %TV ) { print "the $family"; print " is on during @{ $TV{$family}{nights} }\n"; print "its members are:\n"; for $who ( @{ $TV{$family}{members} } ) { print " $who->{name} ($who->{role}), age $who->{age}\n"; } print "it turns out that $TV{$family}{lead} has "; print scalar ( @{ $TV{$family}{kids} } ), " kids named "; print join (", ", map { $_->{name} } @{ $TV{$family}{kids} } ); print "\n"; }
perlref,
perllol,
perldata,
perlobj
Last update: Wed Oct 23 04:57:50 MET DST 1996