On 6/23/05, Olivier Nicole [EMAIL PROTECTED] wrote:
Thanks Titus,
no, you're misunderstanding regoff_t or printf.
I definitely misunderstand printf. Until now I thought that each place
holder (%d) was associated with one variable and if the type
missmatched, the display could be
I must missunderstand how to use regex(3).
To add a bit, running the same program on Linux gives the expected results:
regexpr=a(.)c
number of substrings=1
return from regexec=0
nmatch=0
p0.so=0 p0.eo=0
p1.so=0 p1.eo=0
p2.so=0 p2.eo=0
p3.so=0 p3.eo=0
return from regexec=0
nmatch=1
p0.so=0
Olivier Nicole schrieb:
Hi,
I must missunderstand how to use regex(3).
no, you're misunderstanding regoff_t or printf.
it's a 64 bit type. thus your printf should read:
ret=regexec(preg, string, nmatch, pmatch, 0);
printf(return from regexec=%d\nnmatch=%d\np0.so=%lld p0.eo=%lld\np1.so=%lld
Thanks Titus,
no, you're misunderstanding regoff_t or printf.
I definitely misunderstand printf. Until now I thought that each place
holder (%d) was associated with one variable and if the type
missmatched, the display could be incorrect.
But in that case, printf seems to take 2 successive %d
Hi,
I must missunderstand how to use regex(3).
From what I read in the man page, pmatch[i].rm_so is the begining of
the i-th match in the regular expression and pmatch[i].rm-so is the
end.
So if I try to match the regex a(.)c on the string abc I should
have: pamtch[1].rm_so=1 and
