Hello,
I am trying to figure out why the uint_parser yields what looks like
raw characters in one instance, yet in another instance yields the
original given phrase. What do I mean by that?
I have a string-literal rule, subset of the Google protobuf v2
grammar, and the octal-escaped rule yields a raw character, but the
"same" hex-escaped rule yields the given input string.
Both rules are of type qi::rule. And the str_lit rule is
qi::rule, basically.
oct_esc %= '\\' >> uint_parser{};
hex_esc %= no_case["\\x"] >> uint_parser{};
char_val %= hex_esc | oct_esc | char_esc | ~char_("\0\n\\");
str_lit %= ("'" >> *(char_val - "'") >> "'")
| ('"' >> *(char_val - '"') >> '"')
;
struct escapes_t : qi::symbols {
escapes_t() {
this->add("\\a", '\a')
("\\b", '\b')
("\\f", '\f')
("\\n", '\n')
("\\r", '\r')
("\\t", '\t')
("\\v", '\v')
("\\\\", '\\')
("\\'", '\'')
("\\\"", '"')
;
}
} char_esc;
The only thing I can figure is that perhaps it has something to do
with no-case having a secondary effect on the rule?
I do not care which way it lands, per se, in either raw or original
string, although landing in the raw, that is, unescaped, has some
advantages, I suppose.
However, it is a mystery to me how there are two different outcomes.
Any thoughts? Suggestions?
Thanks!
Michael Powell
