Dear boost-users,
in the post
http://stackoverflow.com/questions/20351369/regular-expression-cant-find-seq
uence-of-numbers
I discovered that if one uses an closing bracket as the first character of a
character class the character class includes this bracket.
This works with the standard setting of boost::regex (i.e., perl-regular
expressions) but it is not documented in the
manual page
http://www.boost.org/doc/libs/1_55_0/libs/regex/doc/html/boost_regex/syntax/
perl_syntax.html#boost_regex.syntax.perl_syntax.character_sets
Is this an undocumented feature, a bug or did I misinterpret something in
the manual?
The version information in version.hpp is:
#define BOOST_LIB_VERSION "1_53"
Info: I tried this specification of the bracket in the character set because
I am used to it from emacs regular expressions.
In the remainder there follows a program for reproduction of the phenomenon:
#include <string>
#include
#include <iostream>
using namespace std;
int main()
{
std::cout << "main()\n";
string s = "98[2]00";
string exp = "^[][0-9]*$";
const boost::regex e(exp);
bool isSequence = boost::regex_match(s,e);
//isSequence is boolean and should be equal to 1
cout << isSequence << endl;
return 0;
}
/**
Local Variables:
compile-command: "g++ -g test.cc -o test.exe -lboost_regex-mt;
./test.exe"
End:
*/
Best regards / Mit freundlichen Grüßen
Tobias Zawada
Engineering / Software Development
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