如何从某个起始位置找到字符串中第一次出现的模式？

我有一个任意长度的字符串,从位置p0开始,我需要找到三个3字母模式之一的第一个出现.

假设字符串只包含字母.我需要找到从位置p0开始的三元组计数并以三元组向前跳,直到第一次出现'aaa'或'bbb'或'ccc'.

使用正则表达式甚至可以实现这一点吗？

$string=~/^   # from the start of the string
            (?:.{$p0}) # skip (don't capture) "$p0" occurrences of any character
            (?:...)*?  # skip 3 characters at a time,
                       # as few times as possible (non-greedy)
            (aaa|bbb|ccc) # capture aaa or bbb or ccc as $1
         /x;

(假设p0基于0).

当然,在字符串上使用substr来跳过可能更有效:

substr($string, $p0)=~/^(?:...)*?(aaa|bbb|ccc)/;

莫里茨说这可能比正则表达更快.即使它有点慢,但在早上5点更容易理解.:)

             #0123456789.123456789.123456789.  
my $string = "alsdhfaaasccclaaaagalkfgblkgbklfs";  
my $pos    = 9;  
my $length = 3;  
my $regex  = qr/^(aaa|bbb|ccc)/;

while( $pos < length $string )    
    {  
    print "Checking $pos\n";  

    if( substr( $string, $pos, $length ) =~ /$regex/ )
        {
        print "Found $1 at $pos\n";
        last;
        }

    $pos += $length;
    }

你不能真正依赖正则表达式,但你可以做这样的事情:

pos $string = $start_from;
$string =~ m/\G         # anchor to previous pos()
            ((?:...)*?) # capture everything up to the match
            (aaa|bbb|ccc)
            /xs  or die "No match"
my $result = length($1) / 3;

但我认为使用substr()和unpack()分割成三元组并在for循环中遍历三元组会更快一些.

(编辑:它的长度(),而不是lenght();-)