2011-01-26 99 views
4

我有一個地址像標量2001:db8::1,並希望得到擴展形式,2001:0db8:0000:0000:0000:0000:0000:0001。主要的Perl軟件包是否在/usr/lib/perl5/...這個龐大的森林裏 - 這個模塊已經可以做到這一點了?如果沒有,是否有人會做幾行呢?Perl的IPv6地址擴展/解析

回答

9

CPAN有Net::IP它可以做你所需要的。

這裏有一個成績單您展示它在行動:

$ cat qq.pl 
use Net::IP; 
$ip = new Net::IP ('2001:db8::1'); 
print $ip->ip() . "\n"; 

$ perl qq.pl 
2001:0db8:0000:0000:0000:0000:0000:0001 
+0

和`網:: IP`是純Perl。我喜歡純perl模塊。 – 2011-01-26 03:32:31

2

Net::IP絕對是一個偉大的路要走,因爲它很容易和強大。但是,如果要分析大量的對象,則可以考慮使用Socket包中的inet_pton,因爲它比Net::IP對象版本快10-20倍,即使對象是預創建的也是如此。 ,比ip_expand_address版本快4ish時間:

use Net::IP; 
use Time::HiRes qw(gettimeofday tv_interval); 
use Socket qw(inet_pton AF_INET6); 
use bignum; 
use strict; 

# bootstrap 
my $addr = "2001:db8::1"; 
my $maxcount = 10000; 

my $ip = new Net::IP($addr); 

my ($t0, $t1); 
my $res; 

# test Net::IP 
$t0 = [gettimeofday()]; 
for (my $i = 0; $i < $maxcount; $i++) { 
    $ip->set($addr); 
    $res = $ip->ip(); 
} 
print "Net::IP elapsed: " . tv_interval($t0) . "\n"; 
print "Net::IP Result: $res\n"; 

# test non-object version 
$t0 = [gettimeofday()]; 
for (my $i = 0; $i < $maxcount; $i++) { 
    $res = Net::IP::ip_expand_address('2001:db8::1', 6); 
} 
print "ip_expand elapsed: " . tv_interval($t0) . "\n"; 
print "ip_expand Result: $res\n"; 

# test inet_pton 
$t0 = [gettimeofday()]; 
for (my $i = 0; $i < $maxcount; $i++) { 
    $res = join(":", unpack("H4H4H4H4H4H4H4H4",inet_pton(AF_INET6, $addr))); 
} 
print "inet_pton elapsed: " . tv_interval($t0) . "\n"; 
print "inet_pton result: " . $res . "\n"; 

隨機機器上運行本作製作我:

Net::IP elapsed: 2.059268 
Net::IP Result: 2001:0db8:0000:0000:0000:0000:0000:0001 
ip_expand elapsed: 0.482405 
ip_expand Result: 2001:0db8:0000:0000:0000:0000:0000:0001 
inet_pton elapsed: 0.132578 
inet_pton result: 2001:0db8:0000:0000:0000:0000:0000:0001