I'm having grief from the RegExp libs,
I'm tryng to match -- item [optional item] item
eg "(a)(b)?(c)"  ()'s used so I can use replace( "$1" ); later

but this will match to "acc" but I can not use $2 or $3 in a replace statement.

also I want to do ..
"(a)((b|c)?(d|e)*)(f) and use 'a' and 'f' (which might be say '\S+\s*\{' and
'}\s*\S+')
ideally with the following replace symantics (see end of message for the
current o/p)
$1 -> (a)
$2 -> ((b|c)?(d|e)*)
$3 -> (f)
even if $2 == ""
and either another regexp has to be run over $2 or
$21 -> (b|c)?
$22 -> (d|e)*

I'm not 100% used to regexps I've always avoided them so am a bit unsure
what the regexp experts would be expecting; I would be happy if I could use
$1 -> (a)
$2 -> ((b|c)?(d|e)*)
$3 -> (b|c)?
$4 -> (d|e)*
$5 -> (f)

the code I've been trying is this

import regexp;
import string;

char[] convInner( RegExp form, char[] inp )
{
 if ( form.test( inp ) != 0 )
 {
  char[] one = "1";
  char[] two = "2";
  char[] three = "3";
  one = form.replace( "$1" );
// comment out theses lines to get this to not throw an exception
  two = form.replace( "$2" );
  three = form.replace( "$3" );
  return "Transfromed '"~one~"' + '"~two~"' + '"~three~"';";
 }
 return "NO TRANSFORM "~inp;
}

void myConvert1( char[] inp )
{
 char[] rv;
 rv = convInner( new RegExp( '^(a)(b){0,1}(c*)', null ), inp );
 printf( "conv1) %s -> %s\n", (char *)inp, (char *)rv );
}

void myConvert2( char[] inp )
{
 char[] rv;
 rv = convInner( new RegExp( '^(a)((b){0,1})(c*)', null ), inp );
 printf( "conv2) %s -> %s\n", (char *)inp, (char *)rv );
}

void myConvert3( char[] inp )
{
 char[] rv;
 rv = convInner( new RegExp( '^(a)(b)?(c*)', null ), inp );
 printf( "conv3) %s -> %s\n", (char *)inp, (char *)rv );
}

void myConvert4( char[] inp )
{
 char[] rv;
 rv = convInner( new RegExp( '^(a)((b)?)(c*)', null ), inp );
 printf( "conv4) %s -> %s\n", (char *)inp, (char *)rv );
}

int main( char[][] args )
{
 char[] str1 = "abcc";
 char[] str2 = "acc";

 myConvert1( str1 );
 myConvert2( str1 );
 myConvert3( str1 );
 myConvert4( str1 );

 myConvert1( str2 );
 myConvert2( str2 );
 myConvert3( str2 );
 myConvert4( str2 );
 return 0;
}

which outputs
conv1) abcc -> Transfromed 'a' + 'b' + 'cc';
conv2) abcc -> Transfromed 'a' + 'b' + 'b';
conv3) abcc -> Transfromed 'a' + 'b' + 'cc';
conv4) abcc -> Transfromed 'a' + 'b' + 'b';
Error: ArrayBoundsError regexp(2396)

I'll take a look. -Walter

"Mike Wynn" <mike.wynn@l8night.co.uk> wrote in message news:apc16h$c6r$1@digitaldaemon.com...
> I'm having grief from the RegExp libs,
> I'm tryng to match -- item [optional item] item
> eg "(a)(b)?(c)"  ()'s used so I can use replace( "$1" ); later
>
> but this will match to "acc" but I can not use $2 or $3 in a replace statement.
>
> also I want to do ..
> "(a)((b|c)?(d|e)*)(f) and use 'a' and 'f' (which might be say '\S+\s*\{'
and
> '}\s*\S+')
> ideally with the following replace symantics (see end of message for the
> current o/p)
> $1 -> (a)
> $2 -> ((b|c)?(d|e)*)
> $3 -> (f)
> even if $2 == ""
> and either another regexp has to be run over $2 or
> $21 -> (b|c)?
> $22 -> (d|e)*
>
> I'm not 100% used to regexps I've always avoided them so am a bit unsure what the regexp experts would be expecting; I would be happy if I could
use
> $1 -> (a)
> $2 -> ((b|c)?(d|e)*)
> $3 -> (b|c)?
> $4 -> (d|e)*
> $5 -> (f)
>
> the code I've been trying is this
>
> import regexp;
> import string;
>
> char[] convInner( RegExp form, char[] inp )
> {
>  if ( form.test( inp ) != 0 )
>  {
>   char[] one = "1";
>   char[] two = "2";
>   char[] three = "3";
>   one = form.replace( "$1" );
> // comment out theses lines to get this to not throw an exception
>   two = form.replace( "$2" );
>   three = form.replace( "$3" );
>   return "Transfromed '"~one~"' + '"~two~"' + '"~three~"';";
>  }
>  return "NO TRANSFORM "~inp;
> }
>
> void myConvert1( char[] inp )
> {
>  char[] rv;
>  rv = convInner( new RegExp( '^(a)(b){0,1}(c*)', null ), inp );
>  printf( "conv1) %s -> %s\n", (char *)inp, (char *)rv );
> }
>
> void myConvert2( char[] inp )
> {
>  char[] rv;
>  rv = convInner( new RegExp( '^(a)((b){0,1})(c*)', null ), inp );
>  printf( "conv2) %s -> %s\n", (char *)inp, (char *)rv );
> }
>
> void myConvert3( char[] inp )
> {
>  char[] rv;
>  rv = convInner( new RegExp( '^(a)(b)?(c*)', null ), inp );
>  printf( "conv3) %s -> %s\n", (char *)inp, (char *)rv );
> }
>
> void myConvert4( char[] inp )
> {
>  char[] rv;
>  rv = convInner( new RegExp( '^(a)((b)?)(c*)', null ), inp );
>  printf( "conv4) %s -> %s\n", (char *)inp, (char *)rv );
> }
>
> int main( char[][] args )
> {
>  char[] str1 = "abcc";
>  char[] str2 = "acc";
>
>  myConvert1( str1 );
>  myConvert2( str1 );
>  myConvert3( str1 );
>  myConvert4( str1 );
>
>  myConvert1( str2 );
>  myConvert2( str2 );
>  myConvert3( str2 );
>  myConvert4( str2 );
>  return 0;
> }
>
> which outputs
> conv1) abcc -> Transfromed 'a' + 'b' + 'cc';
> conv2) abcc -> Transfromed 'a' + 'b' + 'b';
> conv3) abcc -> Transfromed 'a' + 'b' + 'cc';
> conv4) abcc -> Transfromed 'a' + 'b' + 'b';
> Error: ArrayBoundsError regexp(2396)
>
>
>

as an aside, I had a look around the Perl site for info on Regexp's (its one
thing Perl's very good at)
and found http://dev.perl.org/rfc/360.html
regexps are slow (in comparison to arithmetic ops) so having Objects rather
than char[] returned would not represent a major performance hit.
( semi c++ pseudo code to show ownership)
i.e.

RegExpElement [] Regexp::Eval( char[] str ); // runs the regexp over the
string, can reuse the Regexp object
char[] RegExpElement::toString(); // get the "value" of the () element
RegExpElement RegExpElement::next(); // get the next if the group had a
postfix of *,+,? or {}
RegExpElement[] RegExpElement::chlidren(); // get the child groups from the
regexp

e.g
RegExp rex = new RegExp( "(a)*(b+)(c+(d+)(e)*)f" );

then
rex.Eval( "aabbbcdddef" );
would return
[
    "a" :next-> "a" :next-> null
    "bbb"
    "cdde" <children> [
        "dd"
        "e" :next-> null;
    ]
]

then
rex.Eval( "bbcdf" );
would return
[
    MT :next-> null
    "bb"
    "cd" <children> [
        "d"
        MT :next-> null;
    ]
]

i'm not sure if MT should be "" or null or a special 'empty' Element.


"Walter" <walter@digitalmars.com> wrote in message news:apcrb4$17d1$1@digitaldaemon.com...
> I'll take a look. -Walter
>
> "Mike Wynn" <mike.wynn@l8night.co.uk> wrote in message news:apc16h$c6r$1@digitaldaemon.com...
> > I'm having grief from the RegExp libs,
> > I'm tryng to match -- item [optional item] item
> > eg "(a)(b)?(c)"  ()'s used so I can use replace( "$1" ); later
> >
> > but this will match to "acc" but I can not use $2 or $3 in a replace statement.
> >
> > also I want to do ..
> > "(a)((b|c)?(d|e)*)(f) and use 'a' and 'f' (which might be say '\S+\s*\{'
> and
> > '}\s*\S+')
> > ideally with the following replace symantics (see end of message for the
> > current o/p)
> > $1 -> (a)
> > $2 -> ((b|c)?(d|e)*)
> > $3 -> (f)
> > even if $2 == ""
> > and either another regexp has to be run over $2 or
> > $21 -> (b|c)?
> > $22 -> (d|e)*
> >
> > I'm not 100% used to regexps I've always avoided them so am a bit unsure what the regexp experts would be expecting; I would be happy if I could
> use
> > $1 -> (a)
> > $2 -> ((b|c)?(d|e)*)
> > $3 -> (b|c)?
> > $4 -> (d|e)*
> > $5 -> (f)
> >
> > the code I've been trying is this
> >
> > import regexp;
> > import string;
> >
> > char[] convInner( RegExp form, char[] inp )
> > {
> >  if ( form.test( inp ) != 0 )
> >  {
> >   char[] one = "1";
> >   char[] two = "2";
> >   char[] three = "3";
> >   one = form.replace( "$1" );
> > // comment out theses lines to get this to not throw an exception
> >   two = form.replace( "$2" );
> >   three = form.replace( "$3" );
> >   return "Transfromed '"~one~"' + '"~two~"' + '"~three~"';";
> >  }
> >  return "NO TRANSFORM "~inp;
> > }
> >
> > void myConvert1( char[] inp )
> > {
> >  char[] rv;
> >  rv = convInner( new RegExp( '^(a)(b){0,1}(c*)', null ), inp );
> >  printf( "conv1) %s -> %s\n", (char *)inp, (char *)rv );
> > }
> >
> > void myConvert2( char[] inp )
> > {
> >  char[] rv;
> >  rv = convInner( new RegExp( '^(a)((b){0,1})(c*)', null ), inp );
> >  printf( "conv2) %s -> %s\n", (char *)inp, (char *)rv );
> > }
> >
> > void myConvert3( char[] inp )
> > {
> >  char[] rv;
> >  rv = convInner( new RegExp( '^(a)(b)?(c*)', null ), inp );
> >  printf( "conv3) %s -> %s\n", (char *)inp, (char *)rv );
> > }
> >
> > void myConvert4( char[] inp )
> > {
> >  char[] rv;
> >  rv = convInner( new RegExp( '^(a)((b)?)(c*)', null ), inp );
> >  printf( "conv4) %s -> %s\n", (char *)inp, (char *)rv );
> > }
> >
> > int main( char[][] args )
> > {
> >  char[] str1 = "abcc";
> >  char[] str2 = "acc";
> >
> >  myConvert1( str1 );
> >  myConvert2( str1 );
> >  myConvert3( str1 );
> >  myConvert4( str1 );
> >
> >  myConvert1( str2 );
> >  myConvert2( str2 );
> >  myConvert3( str2 );
> >  myConvert4( str2 );
> >  return 0;
> > }
> >
> > which outputs
> > conv1) abcc -> Transfromed 'a' + 'b' + 'cc';
> > conv2) abcc -> Transfromed 'a' + 'b' + 'b';
> > conv3) abcc -> Transfromed 'a' + 'b' + 'cc';
> > conv4) abcc -> Transfromed 'a' + 'b' + 'b';
> > Error: ArrayBoundsError regexp(2396)
> >
> >
> >
>
>

"Mike Wynn" <mike.wynn@l8night.co.uk> wrote in message news:apdtc6$2no8$1@digitaldaemon.com...
> as an aside, I had a look around the Perl site for info on Regexp's (its
one
> thing Perl's very good at)
> and found http://dev.perl.org/rfc/360.html

D regexp's are equivalent to the javascript functionality, but I agree that Perl has gone way beyond that!