Perl基础教程：复杂数据结构

Perl Data Structures Cookbook
Perl数据结构百科

一、描述
The single feature most sorely lacking in the Perl programming language prior to its 5.0?
release was complex data structures. Even without direct language support,some valiant?
programmers did manage to emulate them,but it was hard work and not for the faint of heart.?
You could occasionally get away with the $m{$AoA,$b} notation borrowed from awk in which?
the keys are actually more like a single concatenated string "$AoA$b",but traversal and?
sorting were difficult. More desperate programmers even hacked Perl's internal symbol table?
directly,a strategy that proved hard to develop and maintain--to put it mildly.
在Perl 5.0之前,Perl最缺的功能是构建复杂的数据结构。
虽然没有直接的语言支持，但一些NB的程序员也有办法仿造生成，不过很难有效工作。
即使有时使用awk语言来生成$m{$AoA,$}这样的结构体，但它很像是拼接后的字符串"$AoA$b"。
而且遍历和排序也很困难。
更疯狂的程序甚至直接hack Perl的内部符号表，当然这个更难搞。
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 an array with three dimensions!
5.0版本后，Perl提供了复杂数据结构的支持，写出像下面的三维数据就很容易了。
? ? for $x (1 .. 10) {
? ? ? ?for $y (1 .. 10) {
? ? ? ? ??for $z (1 .. 10) {
? ? ? ? ? ? ?$AoA[$x][$y][$z] = ? $x ** $y + $z;
? ? ? ? ?}
? ? ? ?}
? ? }
Alas,however simple this may appear,underneath it's a much more elaborate construct?
than meets the eye!
当然，表面简单的下面是复杂的构成。
How do you print it out? Why can't you say just print @AoA ? How do you sort it? How can?
you pass it to a function or get one of these back from a function? Is it 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?
如何打印输出？为什么可用print @AoA？如何排序？
如何传给函数或从函数返回？这是对象吗？如何保存并读取？如何访问整个行列？
它的值都是数字的吗？
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.
问题是很多，更现实的问题是缺少给初学者的文档。
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.
本文提供了很多的数据结构的细节实现，可以把它当做百科书一样查找和直接用到程序开中。
Let's look at each of these possible constructs in detail. There are separate sections on?
each of the following:
分类如下:?
? arrays of arrays; ? ? ? ? ? 数组的数组
? hashes of arrays; ? ? ? ? ? 哈希表的元素为数组
? arrays of hashes; ? ? ? ? ? 数组的元素为哈希表
? hashes of hashes ? ? ? ? ? ?哈希表的元素为哈希表
? more elaborate constructs ? 更复杂的结构
But for now,let's look at general issues common to all these types of data structures.
在这之前，先来看看对来这些数据结构类型来说的一些更一般性的问题。
二、REFERENCES
The most important thing to understand about all data structures in Perl--including?
multidimensional arrays--is that even though they might appear otherwise,Perl @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.
要理解Perl中所有数据结构--包括多维数据--最重要的一点是:?
? @ARRAY s 和 %HASH es 在内部的表示都是一维的;
? 且它们只能存储标题值(哪字符串，数字，或引用);
? 它们不能直接以数组或哈希表作为元素，但可以通过引用来实现。
You can't use a reference to an 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.
不能对真正的哈希和数组使用相同的引用。
这一点，和对于C或C++程序员来说，不区分数组和指针的一样的，都会造成混淆。
You can (and should) read more about references in perlref. 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.
总体来说，引用类似于指针。
这就意味着，当访问二维或多维数组/哈希时，实际上访问的是一个元素为引用的一维对象。
这一点和C也是类似的。
? ? $array[7][12] # array of arrays
? ? $array[7]{string} # array of hashes 数组的元素为哈希
? ? $hash{string}[7] # hash of arrays ?哈希的元素为数组
? ? $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:
因为顶层元素包含的是引用，则直接打印数组本身的输出如下:
? ? @AoA = ( [2,3],[4,5,7],[0] );
? ? print $AoA[1][2];
? 7
? ? print @AoA;
? 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] .
Perl并不会隐示地对变量进行去引用。
要去这些引用，必须自己使用前缀或后缀符来去掉。
三、COMMON MISTAKES【常见错误】
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:
构建像数组的数组这样的结构最常见的错误是:
? 意外地计算了元素的个数;
? 对相同的内存区域重复地使用同一个引用。
下面的例子是对嵌套数组赋值的错误使用:
? ? for $i (1..10) {
? ? ? ?@array = somefunc($i);
? ? ? ?$AoA[$i] = @array; # WRONG!
That's just the simple case of assigning an array 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,helvetica; font-size:14px">正确的方法应该是:
? ? ? ?$counts[$i] = scalar @array;
Here's the case of taking a reference to the same memory location again and again:
下面的错误用法是: 对同一个内存区进行重复引用.
? ? ? ??$AoA[$i] = @array; # 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 @AoA?
refer to the very same place,and they will therefore all hold whatever was last?
in @array! It's similar to the problem demonstrated in the following C program:
像上面的问题，在C语言中也有类似的情况:
? ? #include <pwd.h>
? ? main() {
? ? ? ??struct passwd *getpwnam(),*rp,*dp;
? ? ? ??rp = getpwnam("root");
? ? ? ??dp = getpwnam("daemon");
? ? ? ??printf("daemon name is %snroot name is %sn",
? ? ? ? ? ? ? ? ? ?dp->pw_name,rp->pw_name);
Which will print
上面的代码将会输出:
? ? 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:
在C中，会隐式地进行自动的内存分配。
在Perl中，则要显式地使用数组构建符[] ?或 哈希构建符{} 来分配新的内存。
下面的正确的方式:
$AoA[$i] = [ @array ];
The square brackets make a reference to a new array with a copy of what's in @array?
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:
方括号会将引用指向一个新的数组拷贝。
别的方式也能用，但是很难看懂。
? ? ? ?@array = 0 .. $i;
? ? ? ?@{$AoA[$i]} = @array;
Is it the same? Well,maybe so--and maybe not. The subtle difference is that when you?
assign something in square brackets,you know for sure it's always a brand new reference?
with a new copy of the data. Something else could be going on in this new case with?
the @{$AoA[$i]} dereference on the left-hand-side of the assignment. It all depends on?
whether $AoA[$i] had been undefined to start with,or whether it already contained a?
reference. If you had already populated @AoA with references,as in
这两种方式有一些细微的差别:?
? ?当使用方括号时，总是会生成一个数据的新的拷贝的引用。
? ?如果使用@{$AoA[$i]}方式，则当@AoA 包含有一个引用时，如下所示:
? ? $AoA[3] = @another_array;
Then the assignment with the indirection on the left-hand-side would use the existing?
reference that was already there:
则会对当前存在的引用使用间接分配。
? ? @{$AoA[3]} = @array;
Of course,this would have the "interesting" effect of clobbering @another_array. (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",helvetica; font-size:14px">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.
反正，最好还是用[]和{}。
Surprisingly,the following dangerous-looking construct will actually work out fine:
? ? ? ? my @array = somefunc($i);
? ? ? ? $AoA[$i] = @array;
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,helvetica; font-size:14px">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:
总之:
? ? $AoA[$i] = [ @array ]; # usually best
? ? $AoA[$i] = @array; # perilous; just how my() was that array?
? ? @{ $AoA[$i] } = @array; # way too tricky for most programmers
四、CAVEAT ON PRECEDENCE【预先警告】
Speaking of things like @{$AoA[$i]},the following are actually the same thing:
? ? $aref->[2][2] # clear
? ? $$aref[2][2] # confusing
That's because Perl's precedence rules on its five prefix dereferencers (which look like?
someone swearing: $ @ * % & ) make them bind more tightly than the postfix subscripting?
brackets or braces! This will no doubt come as a great shock to the C or C++ programmer,helvetica; font-size:14px">who is quite accustomed to using *a[i] to mean what's pointed to by the i'th element of a .?
That is,they first take the subscript,and only then dereference the thing at that subscript.?
That's fine in C,but this isn't C.
The seemingly equivalent construct in Perl,$$aref[$i] first does the deref of $aref,helvetica; font-size:14px">making it take $aref as a reference to an array,and then dereference that,and finally
?tell you the i'th value of the array pointed to by $AoA. If you wanted the C notion,helvetica; font-size:14px">you'd have to write ${$AoA[$i]} to force the $AoA[$i] to get evaluated first before the?
leading $ dereferencer.
五、WHY YOU SHOULD ALWAYS【为什么总是】 use strict
If this is starting to sound scarier than it's worth,relax. Perl has some features to help?
you avoid its most common pitfalls. The best way to avoid getting confused is to start?
every program like this:
能帮你进行编译检查，从而避免很多程序问题:
? ? #!/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 $aref = [
? ? ? ? ? ? ? ? ? ? ? ? ?[ "fred","barney","pebbles","bambam","dino",],helvetica; font-size:14px">? ? ? ? ? ? ? ? ? ? ? ? ?[ "homer","bart","marge","maggie",helvetica; font-size:14px">? ? ? ? ? ? ? ? ? ? ? ? ?[ "george","jane","elroy","judy",helvetica; font-size:14px">? ? ];
? ? print $aref[2][2];
The compiler would immediately flag that as an error at compile time,because you were?
accidentally accessing @aref,an undeclared variable,and it would thereby remind you?
to write instead:
? ? print $aref->[2][2]
六、DEBUGGING【调试】
Before version 5.002,the standard Perl debugger didn't do a very nice job of printing out?
complex data structures. With 5.002 or above,the debugger includes several new features,helvetica; font-size:14px">including command line editing as well as the x command to dump out complex data structures.?
For example,given the assignment to $AoA above,here's the debugger output:
新的调试器对数组输出很方便:
? ? DB<1> x $AoA
? ? $AoA = 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'
七、CODE EXAMPLES【代码示例】
Presented with little comment (these will get their own manpages someday) here are short?
code examples illustrating access of various types of data structures.
1. ARRAYS OF ARRAYS【数组的数组】
1.1 Declaration of an ARRAY OF ARRAYS
声明:
?@AoA = (
? ? ? ? [ "fred","barney" ],helvetica; font-size:14px">? ? ? ? [ "george","elroy" ],helvetica; font-size:14px">? ? ? ? [ "homer","bart" ],helvetica; font-size:14px">? ? ? );
1.2 Generation of an ARRAY OF ARRAYS
生成:
?# reading from file
?while ( <> ) {
? ? ?push @AoA,[ split ];
?}
?# calling a function
?for $i ( 1 .. 10 ) {
? ? ?$AoA[$i] = [ somefunc($i) ];
?# using temp vars
? ? ?@tmp = somefunc($i);
? ? ?$AoA[$i] = [ @tmp ];
?# add to an existing row
?push @{ $AoA[0] },"wilma","betty";
1.3 Access and Printing of an ARRAY OF ARRAYS
访问和输出
?# one element
?$AoA[0][0] = "Fred";
?# another element
?$AoA[1][1] =~ s/(w)/u$1/;
?# print the whole thing with refs
?for $aref ( @AoA ) {
? ? ?print "t [ @$aref ],n";
?# print the whole thing with indices
?for $i ( 0 .. $#AoA ) {
? ? ?print "t [ @{$AoA[$i]} ],helvetica; font-size:14px">?# print the whole thing one at a time
? ? ?for $j ( 0 .. $#{ $AoA[$i] } ) {
? ? ? ? ?print "elt $i $j is $AoA[$i][$j]n";
? ? ?}
2. HASHES OF ARRAYS【哈希表的元素为数组】
2.1 Declaration of a HASH OF ARRAYS
声明
?%HoA = (
? ? ? ? flintstones ? ? ? ?=> [ "fred",helvetica; font-size:14px">? ? ? ? jetsons ? ? ? ? ? ?=> [ "george",helvetica; font-size:14px">? ? ? ? simpsons ? ? ? ? ? => [ "homer",helvetica; font-size:14px">2.2 Generation of a HASH OF ARRAYS
生成
?# flintstones: fred barney wilma dino
? ? ?next unless s/^(.*?):s*//;
? ? ?$HoA{$1} = [ split ];
?# reading from file; more temps
?while ( $line = <> ) {
? ? ?($who,$rest) = split /:s*/,$line,2;
? ? ?@fields = split ' ',$rest;
? ? ?$HoA{$who} = [ @fields ];
?# calling a function that returns a list
?for $group ( "simpsons","jetsons","flintstones" ) {
? ? ?$HoA{$group} = [ get_family($group) ];
?# likewise,but using temps
? ? ?@members = get_family($group);
? ? ?$HoA{$group} = [ @members ];
?# append new members to an existing family
?push @{ $HoA{"flintstones"} },helvetica; font-size:14px">2.3 Access and Printing of a HASH OF ARRAYS
访问
?$HoA{flintstones}[0] = "Fred";
?$HoA{simpsons}[1] =~ s/(w)/u$1/;
?# print the whole thing
?foreach $family ( keys %HoA ) {
? ? ?print "$family: @{ $HoA{$family} }n"
? ? ?print "family: ";
? ? ?foreach $i ( 0 .. $#{ $HoA{$family} } ) {
? ? ? ? ?print " $i = $HoA{$family}[$i]";
? ? ?print "n";
?# print the whole thing sorted by number of members
?foreach $family ( sort { @{$HoA{$b}} <=> @{$HoA{$a}} } keys %HoA ) {
?# print the whole thing sorted by number of members and name
?foreach $family ( sort {
? ?@{$HoA{$b}} <=> @{$HoA{$a}}
||
? ?$a cmp $b
? ?} keys %HoA )
?{
? ? ?print "$family: ",join(",",sort @{ $HoA{$family} }),"n";
3. ARRAYS OF HASHES【数组的元素为哈希】
3.1 Declaration of an ARRAY OF HASHES
?@AoH = (
? ? ? ? {
? ? ? ? ? ? Lead ? ? => "fred",helvetica; font-size:14px">? ? ? ? ? ? Friend ? => "barney",helvetica; font-size:14px">? ? ? ? },helvetica; font-size:14px">? ? ? ? ? ? Lead ? ? => "george",helvetica; font-size:14px">? ? ? ? ? ? Wife ? ? => "jane",helvetica; font-size:14px">? ? ? ? ? ? Son ? ? ?=> "elroy",helvetica; font-size:14px">? ? ? ? ? ? Lead ? ? => "homer",helvetica; font-size:14px">? ? ? ? ? ? Wife ? ? => "marge",helvetica; font-size:14px">? ? ? ? ? ? Son ? ? ?=> "bart",helvetica; font-size:14px">? ? ? ? }
? );
3.2 Generation of an ARRAY OF HASHES
?# format: LEAD=fred FRIEND=barney
? ? ?$rec = {};
? ? ?for $field ( split ) {
? ? ? ? ?($key,$value) = split /=/,$field;
? ? ? ? ?$rec->{$key} = $value;
? ? ?push @AoH,$rec;
?# no temp
?# calling a function ?that returns a key/value pair list,like
?# "lead","fred","daughter","pebbles"
?while ( %fields = getnextpairset() ) {

?while (<>) {
?# add key/value to an element
?$AoH[0]{pet} = "dino";
?$AoH[2]{pet} = "santa's little helper";
3.3 Access and Printing of an ARRAY OF HASHES
?$AoH[0]{lead} = "fred";
?$AoH[1]{lead} =~ s/(w)/u$1/;
?for $href ( @AoH ) {
? ? ?print "{ ";
? ? ?for $role ( keys %$href ) {
? ? ? ? ?print "$role=$href->{$role} ";
? ? ?print "}n";
?for $i ( 0 .. $#AoH ) {
? ? ?print "$i is { ";
? ? ?for $role ( keys %{ $AoH[$i] } ) {
? ? ? ? ?print "$role=$AoH[$i]{$role} ";
? ? ? ? ?print "elt $i $role is $AoH[$i]{$role}n";
4. HASHES OF HASHES【哈希的哈希】
4.1 Declaration of a HASH OF HASHES
?%HoH = (
? ? ? ? flintstones => {
lead ? ? ?=> "fred",helvetica; font-size:14px">pal ? ? ? => "barney",helvetica; font-size:14px">? ? ? ? jetsons ? ? => {
lead ? ? ?=> "george",helvetica; font-size:14px">wife ? ? ?=> "jane",helvetica; font-size:14px">"his boy" => "elroy",helvetica; font-size:14px">? ? ? ? simpsons ? ?=> {
lead ? ? ?=> "homer",helvetica; font-size:14px">wife ? ? ?=> "marge",helvetica; font-size:14px">kid ? ? ? => "bart",helvetica; font-size:14px">},helvetica; font-size:14px">?);
4.2 Generation of a HASH OF HASHES
?# flintstones: lead=fred pal=barney wife=wilma pet=dino
? ? ?$who = $1;
? ? ? ? ?$HoH{$who}{$key} = $value;
? ? ?$HoH{$who} = $rec;
?# calling a function ?that returns a key,value hash
? ? ?$HoH{$group} = { get_family($group) };
? ? ?%members = get_family($group);
? ? ?$HoH{$group} = { %members };
?%new_folks = (
? ? ?wife => "wilma",helvetica; font-size:14px">? ? ?pet ?=> "dino",helvetica; font-size:14px">?for $what (keys %new_folks) {
? ? ?$HoH{flintstones}{$what} = $new_folks{$what};
4.3 Access and Printing of a HASH OF HASHES
?$HoH{flintstones}{wife} = "wilma";
?$HoH{simpsons}{lead} =~ s/(w)/u$1/;
?foreach $family ( keys %HoH ) {
? ? ?print "$family: { ";
? ? ?for $role ( keys %{ $HoH{$family} } ) {
? ? ? ? ?print "$role=$HoH{$family}{$role} ";
?# print the whole thing ?somewhat sorted
?foreach $family ( sort keys %HoH ) {
? ? ?for $role ( sort keys %{ $HoH{$family} } ) {
?foreach $family ( sort { keys %{$HoH{$b}} <=> keys %{$HoH{$a}} } keys %HoH ) {
?# establish a sort order (rank) for each role
?$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 ) {
? ? ?# and print these according to rank order
? ? ?for $role ( sort { $rank{$a} <=> $rank{$b} } ?keys %{ $HoH{$family} } ) {
5. MORE ELABORATE RECORDS【更复杂的记录】
5.1 Declaration of MORE ELABORATE RECORDS
Here's a sample showing how to create and use a record whose fields are of many different sorts:
? ? ?$rec = {
TEXT ? ? ?=> $string,helvetica; font-size:14px">SEQUENCE ?=> [ @old_values ],helvetica; font-size:14px">LOOKUP ? ?=> { %some_table },helvetica; font-size:14px">THATCODE ?=> &;some_function,helvetica; font-size:14px">THISCODE ?=> sub { $_[0] ** $_[1] },helvetica; font-size:14px">HANDLE ? ?=> *STDOUT,helvetica; font-size:14px">? ? ?};
? ? ?print $rec->{TEXT};
? ? ?print $rec->{SEQUENCE}[0];
? ? ?$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 { $rec->{HANDLE} } "a stringn";
? ? ?use FileHandle;
? ? ?$rec->{HANDLE}->autoflush(1);
? ? ?$rec->{HANDLE}->print(" a stringn");
5.2 Declaration of a HASH OF COMPLEX RECORDS
定义
? ? ?%TV = (
? ? ? ? ? ? series ? => "flintstones",helvetica; font-size:14px">? ? ? ? ? ? nights ? => [ qw(monday thursday friday) ],helvetica; font-size:14px">? ? ? ? ? ? members ?=> [
? ? ? ? ? ? ? ? { name => "fred",? ?role => "lead",age ?=> 36,},helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "wilma",? role => "wife",age ?=> 31,helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "pebbles",role => "kid",?age ?=> ?4,helvetica; font-size:14px">? ? ? ? ? ? ],helvetica; font-size:14px">? ? ? ? ? ? series ? => "jetsons",helvetica; font-size:14px">? ? ? ? ? ? nights ? => [ qw(wednesday saturday) ],helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "george",?role => "lead",age ?=> 41,helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "jane",? ?role => "wife",age ?=> 39,helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "elroy",? role => "kid",?age ?=> ?9,helvetica; font-size:14px">? ? ? ? ?},helvetica; font-size:14px">? ? ? ? ? ? series ? => "simpsons",helvetica; font-size:14px">? ? ? ? ? ? nights ? => [ qw(monday) ],helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "homer",role => "lead",age ?=> 34,helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "marge",role => "wife",age => 37,helvetica; font-size:14px">? ? ? ? ? ? ? ? { name => "bart",?role => "kid",?age ?=> ?11,helvetica; font-size:14px">5.3 Generation of a HASH OF COMPLEX RECORDS
? ? ?# reading from file
? ? ?# this is most easily done by having the file itself be
? ? ?# in the raw data format as shown above. ?perl is happy
? ? ?# to parse complex data structures if declared as data,so
? ? ?# sometimes it's easiest to do that
? ? ?# here's a piece by piece build up
? ? ?$rec->{series} = "flintstones";
? ? ?$rec->{nights} = [ find_days() ];
? ? ?@members = ();
? ? ?# assume this file in field=value syntax
? ? ?while (<>) {
? ? ? ? ?%fields = split /[s=]+/;
? ? ? ? ?push @members,helvetica; font-size:14px">? ? ?$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 example
? ? ?# if you wanted a {kids} field that was a reference
? ? ?# to an array 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!!
? ? ? ? ?$rec->{kids} = [ @kids ];
? ? ?# you copied the array,but the array itself contains pointers
? ? ?# to uncopied objects. this means that if you make bart get
? ? ?# older via
? ? ?$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";
八、Database Ties
You cannot easily tie a multilevel data structure (such as a hash of hashes) to a dbm file.?
The first problem is that all but GDBM and Berkeley DB have size limitations,but beyond that,helvetica; font-size:14px">you also have problems with how references are to be represented on disk. One experimental?
module that does partially attempt to address this need is the MLDBM module. Check your?
nearest CPAN site as described in perlmodlib for source code to MLDBM.