[xpressive] literals in regex without escaping...
Hello *, I need to parse some logger lines using boost xpressive. The line looks like: fname:line_no [severity] logging message When building the xpressive regex if I need to match the severity (e.g. [INFO]) I need to escape the square brackets... Is there some xpressive construct (like as_xpr) which makes from all characters literals. So that I can write smth like: sregex re = *_ >> literals("[INFO]") >> *_ Many thanks, Ovanes
On Fri, May 27, 2011 at 2:07 PM, Ovanes Markarian
Hello *,
I need to parse some logger lines using boost xpressive. The line looks like:
fname:line_no [severity] logging message
When building the xpressive regex if I need to match the severity (e.g. [INFO]) I need to escape the square brackets... Is there some xpressive construct (like as_xpr) which makes from all characters literals. So that I can write smth like:
sregex re = *_ >> literals("[INFO]") >> *_
It seems that C++ literals are regex literals by default, so there is no need in literals(...) at all...
Sorry for noise. Ovanes
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Ovanes Markarian