10

我想构建一堆 Perl 子程序,它们都有相同的模板if elsif elsif else,可以根据因子变量做出决定。这是子程序模板的示例:

sub get_age{

  my $factor=shift;

  if    ($factor == 1 ){ print "do something" }
  elsif ($factor == 2 ){ print "do somthing2" }
  elsif ($factor == 3 ){ print "do somthing3" }
  elsif ($factor == 4 ){ print "do somthing4" }
  else                 { print "error"        }
  }

我想知道 Perl 上是否有一些设计模式可以if else用更优雅的解决方案替换条件,特别是如果我需要更改某些条件或删除其中一些条件,那么这些解决方案将来易于维护?

4

7 回答 7

10

有几个人提到了调度表。有两件事,有时将它们分开很好。有可能发生的事情的清单,以及使它们发生的事情。如果你将两者结合起来,你就会被困在你的解决方案中。如果你把它们分开,你以后会有更多的灵活性。

调度表将行为指定为数据而不是程序结构。这是两种不同的方法。在您的示例中,您有整数,类似的东西可能会使用数组来存储东西。哈希示例是相同的想法,但查找行为略有不同。

另请注意,我将print. 当你重复这样的代码时,试着把重复的东西上一层楼。

use v5.10;

foreach my $factor ( map { int rand 5 } 0 .. 9 ) {
    say get_age_array( $factor );
    }

my @animals = qw( cat dog bird frog );
foreach my $factor ( map { $animals[ rand @animals ] } 0 .. 9 ) {
    say get_age_hash( $factor );
    }

sub get_age_array {
    my $factor = shift;

    state $dispatch = [
        sub { 'Nothing!' }, # index 0
        sub { "Calling 1" },
        sub { 1 + 1 },
        sub { "Called 3" },
        sub { time },
        ];

    return unless int $factor <= $#$dispatch;

    $dispatch->[$factor]->();   
    }


sub get_age_hash {
    my $factor = shift;

    state $dispatch = {
        'cat'  => sub { "Called cat" },
        'dog'  => sub { "Calling 1"  },
        'bird' => sub { "Calling 2, with extra" },
        };

    return unless exists $dispatch->{$factor};

    $dispatch->{$factor}->();   
    }
于 2011-11-29T22:33:59.440 回答
7

更新:确保您阅读下面的布赖恩评论;基本上,由于他在链接中评论的各种问题,最好使用for而不是。given我更新了我的建议以纳入他的改进,他在Use for() 而不是 given() 中概述了这些改进:

如果您使用的是 perl 5.10 或更高版本,given/when那么您正在寻找的魔法对,但您确实应该使用它for/when。下面是一个示例:

use strict;
use warnings;
use feature qw(switch say);

print 'Enter your grade: ';
chomp( my $grade = <> );

for ($grade) {
    when ('A') { say 'Well done!'       }
    when ('B') { say 'Try harder!'      }
    when ('C') { say 'You need help!!!' }
    default { say 'You are just making it up!' }
}
于 2011-11-29T22:13:22.377 回答
3

只是让事情变得更短:

sub get_age1 {
    my $age = shift;
    $age == 1 ? print "do something" :
    $age == 2 ? print "do somthing2" :
    $age == 3 ? print "do somthing3" :
    $age == 4 ? print "do somthing4" :
                print "error"
}

如果条件可以最好地表示为正则表达式,则这个更有意义:

sub get_age2 {    
    for (shift) { 
        if    (/^ 1 $/x) {print "do something"}
        elsif (/^ 2 $/x) {print "do somthing2"}
        elsif (/^ 3 $/x) {print "do somthing3"}
        elsif (/^ 4 $/x) {print "do somthing4"}
        else             {print "error"       }
    }
}

这里有一些调度表:

一个简单的(有一个错误):

{
    my %age = ( # defined at runtime
        1 => sub {print "do something"},
        2 => sub {print "do somthing2"},
        3 => sub {print "do somthing3"},
        4 => sub {print "do somthing4"},
    );
    # unsafe to call get_age3() before sub definition
    sub get_age3 {
        ($age{$_[0]} or sub {print "error"})->()
    }
}

一个更好的:

{
    my %age;
    BEGIN {
        %age = ( # defined at compile time
            1 => sub {print "do something"},
            2 => sub {print "do somthing2"},
            3 => sub {print "do somthing3"},
            4 => sub {print "do somthing4"},
        )
    }
    # safe to call get_age4() before sub definition
    sub get_age4 {
        ($age{$_[0]} or sub {print "error"})->()
    }
}

另一种写法:

BEGIN {
    my %age = ( # defined at compile time
        1 => sub {print "do something"},
        2 => sub {print "do somthing2"},
        3 => sub {print "do somthing3"},
        4 => sub {print "do somthing4"},
    );
    # safe to call get_age5() before sub definition
    sub get_age5 {
        ($age{$_[0]} or sub {print "error"})->()
    }
}

另一个写它的好方法:

{
    my $age;
    # safe to call get_age6() before sub definition
    sub get_age6 {
        $age ||= { # defined once when first called
           1 => sub {print "do something"},
           2 => sub {print "do somthing2"},
           3 => sub {print "do somthing3"},
           4 => sub {print "do somthing4"},
        };
        ($$age{$_[0]} or sub {print "error"})->()
    }
}
于 2011-11-30T02:09:49.290 回答
0

调度表非常适合这种类型的设计模式。这个成语我用过很多次。像这样的东西:

sub get_age {
    my $facter = shift;
    my %lookup_map = (
        1 => sub {.....},
        2 => sub {.....},
        3 => \&some_other_sub,
        default => \&some_default_sub,
    );
    my $code_ref = $lookup_map{$facter} || $lookup_map{default};
    my $return_value = $code_ref->();
    return $return_value;
}

当您用于确定执行哪个案例的参数将作为哈希表中的键存在时,此方法有效。如果它可能不是完全匹配,那么您可能需要使用正则表达式或其他方式将您的输入与要执行的代码相匹配。您可以像这样使用正则表达式作为哈希键:

my %patterns = (
    qr{^/this/one}i => sub {....},
    qr{^/that/one}is => sub {....},
    qr{some-other-match/\d+}i => \&some_other_match,
)
my $code_ref;
for my $regex (keys %patterns) {
    if ($facter =~ $regex) {
        $code_ref = $patterns{$regex};
        last;
    }
}
$code_ref ||= \&default_code_ref;
$code_ref->();
于 2011-11-29T22:47:25.783 回答
0

请参阅示例/参考/dispatch_table.pl

https://code-maven.com/slides/perl/dispatch-table

#!/usr/bin/perl
use strict;
use warnings;

# Use subroutine references in a hash to define what to do for each case

my %dispatch_table = (
    '+' => \&add,
    '*' => \&multiply,
    '3' => \&do_something_3,
    '4' => \&do_something_4,
);

foreach my $operation ('+', 'blabla', 'foobar', '*'){
    $dispatch_table{$operation}->(
        var1 => 5,
        var2 => 7,
        var3 => 9,                       
    ) if ( exists $dispatch_table{$operation} );
}

sub add {
    my %args = (@_);
    my $var1 = $args{var1}; 
    my $var2 = $args{var2};

    my $sum = $var1 + $var2;
    print "sum = $sum \n";
    return;
}

sub multiply {
    my %args = (@_);
    my $var1 = $args{var1}; 
    my $var3 = $args{var3};

    my $mult = $var1 * $var3;
    print "mult = $mult \n";
    return;
}

输出:

sum = 12 
mult = 45 
于 2020-09-13T12:25:45.313 回答
-1

这可能是一个类似调度表的地方。我自己没有做过,但这个页面可能是一个开始:http ://www.perlmonks.org/?node_id=456530

于 2011-11-29T22:14:59.790 回答
-4

使用开关

阅读高阶 Perl中的调度表

于 2011-11-29T22:14:49.013 回答