use Regexp::Common;
$string =~ s/$RE{comment}{C}//g;
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Note that this (just like the regexes presented in the rest of this thread) isn't context aware, and happily removes "comments" from strings.

Abigail

I address this problem by replacing quoted strings with unique tags prior to removing the comments. After stripping the comments I then restore the original strings for those tags remaining (some may get stripped).

That's kind of a chicken-and-egg problem, isn't? How can you succesfully remove strings, if you can't detect comments? Consider:

/* One " two */ a = b + 4; /* three " four */
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You have to do it all in one pass. Something like:

s { (     [^"'/]*                          # Not a string, character o
+r comment.
      |  "[^\\"]*(?:\\.[^\\"]*)*"          # String.
      |  '[^\\']*(?:\\.[^\\']*)*'          # Char.
      |  / (?![*])                         # Slash, not a comment.
    )
      |
    ( /[*] [^*]* (?: [*] [^*/]* )* [*]/ )  # Comment.
  }
  { $2 ? "" : $1 }gsex;
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But that isn't fool proof either (consider # define).

Abigail

For multiline comments you want '.' to match "\n", do this with a s///sg
But as the faq answer says, this isn't enough.

The other issue is keeping what is supposed to match the inside of the comment from matching the end, some more code, and the beginning and inside of another comment. The simple m/\/\*.*\*\//s regex will match all of "/* comment 1 */ some = code; /* comment 2 */". You tell * to match as little as possible instead of as much as possible by adding a ?, so it becomes m/\/\*.*?\*\//s.

so within ysth restrictions, if one wants a one-liner perl -0777 -pe 's{/\*.*?\*/}{}gs' source.c my .02
--
dominix

I just tried it and it worked! I've never seen a regex contained in braces before...and what's the deal with the empty braces at the end?

thanks!

I will work with the m//s flag stuff. My original regex had the *?, but in my sample text file it didn't seem to make a difference, i.e. it wasn't acting greedy either way. thanks!

It's an assignment for an intro Perl course.

UPDATE - hey hey hey now. I've always tried to be up front about my needs, and I always write as much code as I can before I ask for help, and I always do my best to learn what I can from all suggested code before I just go and use it.

HOMEWORK!

CountZero

"If you have four groups working on a compiler, you'll get a 4-pass compiler." - Conway's Law

No, you don't want the preprocessor output. Consider the following classical program:

# include <stdlib.h>
# include <stdio.h>

int main (int argc, char * argv []) {
    printf ("Hello, world\n");  /* Print 'Hello, world' */
    exit (0);
}
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Assume this is in the file hello.c.

$ gcc -E hello.c | wc -l
   1566
$ gcc -E hello.c | grep -v '^$' | wc -l
    853
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Even with blank lines removed, the 7 line hello.c expands to 853 lines of pre processor output.

Abigail

Well yes, but it expands because you #include files. From his description (text files) I assumed that they weren't actual C programs and wouldn't be using the rest of the pre-processor commands. So there wouldn't be any #includes to expand the file size so dramatically. Beyond that I also suggested that it might be easier to match anything the pre-processor would consider a #define/#include, since the rules for that are fairly strict as I recall.