RE: [Fwd: regular expressions, I am puzzled]

On Sat, 2009-08-01 at 13:38 -0400, Dennis Lovelady wrote:

But when I add an extra set of parentheses, the offsets of
subexpressions in that added subexpression sometimes are somewhat
reduced.

My suspicion is that you have a coding problem, but I cannot guess what the
issued is since you did not supply pseudo- or real-code.

All the code is available, if anybody is that interested.
Meanwhile, here are selected calcs and output from STRDBG.


statement ...
ret_comp = regcomp( re_compiled
: %trimr( re_string )
: lng_curr
);

debug before ...
RE_STRING = '(n)(u)((t*)(i)) '
LNG_CURR = 5

debug after ...
RET_COMP = 0
RE_COMPILED.RE_NSUB = 5


statement ...
ret_exec = regexec( re_compiled : %trimr( input )
: %elem( re_matches) : re_matches(1)
: 0
);

debug before ...
INPUT = 'nuttin '

debug after ...
RET_EXEC = 0

RE_MATCHES.RM_SO(1) = 0
RE_MATCHES.RM_SS(1) = 0
RE_MATCHES.FILLER1(1) = 16448
RE_MATCHES.RM_EO(1) = 5
RE_MATCHES.RM_ES(1) = 0
RE_MATCHES.FILLER2(1) = 16448

RE_MATCHES.RM_SO(2) = 0
RE_MATCHES.RM_SS(2) = 0
RE_MATCHES.FILLER1(2) = 16448
RE_MATCHES.RM_EO(2) = 1
RE_MATCHES.RM_ES(2) = 0
RE_MATCHES.FILLER2(2) = 16448

RE_MATCHES.RM_SO(3) = 1
RE_MATCHES.RM_SS(3) = 0
RE_MATCHES.FILLER1(3) = 16448
RE_MATCHES.RM_EO(3) = 2
RE_MATCHES.RM_ES(3) = 0
RE_MATCHES.FILLER2(3) = 16448

RE_MATCHES.RM_SO(4) = 2
RE_MATCHES.RM_SS(4) = 0
RE_MATCHES.FILLER1(4) = 16448
RE_MATCHES.RM_EO(4) = 5
RE_MATCHES.RM_ES(4) = 0
RE_MATCHES.FILLER2(4) = 16448

RE_MATCHES.RM_SO(5) = 1
RE_MATCHES.RM_SS(5) = 0
RE_MATCHES.FILLER1(5) = 16448
RE_MATCHES.RM_EO(5) = 3
RE_MATCHES.RM_ES(5) = 0
RE_MATCHES.FILLER2(5) = 16448

RE_MATCHES.RM_SO(6) = 3
RE_MATCHES.RM_SS(6) = 0
RE_MATCHES.FILLER1(6) = 16448
RE_MATCHES.RM_EO(6) = 4
RE_MATCHES.RM_ES(6) = 0
RE_MATCHES.FILLER2(6) = 16448


Any thoughts?

Thanks,
Terry.

Probably not what you're looking for, but my guessbox is on the blink, and I
haven't had time to get it repaired.