Re: Creating spacers in regex match

Well, "anything" in a regexp is written as (?s:.), but it seems you want to match two two-hexit hex numbers. Which means you could write:

    /16 03 01 [[:xdigit:]]{2} [[:xdigit:]]{2} 0b/
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Re^2: Creating spacers in regex match by ahuang14 (Novice) on Jul 08, 2011 at 15:18 UTC
ah I got you. can you just explain what the xdigit means? I'm new to perl and I want to learn. How is this different than the ?s:.	[reply]
Re^3: Creating spacers in regex match by JavaFan (Canon) on Jul 08, 2011 at 15:24 UTC
`[[:xdigit:]]` is a character class that matches a hex digit. See `man perlrecharclass`. `(?s:.)` matches any single character; `(?s:.*)` matches any possible string of characters.	[reply] [d/l] [select]
Re^4: Creating spacers in regex match by ahuang14 (Novice) on Jul 08, 2011 at 19:21 UTC
I tried the xdigit line you gave me and it didnt work. `my $str = "06 03 01 0f 0e 0b"; if ($str=~ /06 03 01 [[:xdigit:]]{2} [[:xdigit:]]{2} 0b/){ print "hi"; }` [download]	[reply] [d/l]
Re^5: Creating spacers in regex match by JavaFan (Canon) on Jul 08, 2011 at 21:12 UTC