我正在攻击的一个简单的骨架实用程序的一部分我有一个语法来触发文本中的替换.我认为这是一种很好的方式来适应Boost.Spirit,但模板错误是一种独特的喜悦.
以下是完整的代码:
#include <iostream>
#include <iterator>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
namespace bsq = boost::spirit::qi;
namespace {
template<typename Iterator>
struct skel_grammar : public bsq::grammar<Iterator> {
skel_grammar();
private:
bsq::rule<Iterator> macro_b;
bsq::rule<Iterator> macro_e;
bsq::rule<Iterator, bsq::ascii::space_type> id;
bsq::rule<Iterator> macro;
bsq::rule<Iterator> text;
bsq::rule<Iterator> start;
};
template<typename Iterator>
skel_grammar<Iterator>::skel_grammar() : skel_grammar::base_type(start)
{
text = bsq::no_skip[+(bsq::char_ - macro_b)[bsq::_val += bsq::_1]];
macro_b = bsq::lit("<<");
macro_e = bsq::lit(">>");
macro %= macro_b >> id >> macro_e;
id %= -(bsq::ascii::alpha | bsq::char_('_'))
>> +(bsq::ascii::alnum | bsq::char_('_'));
start = *(text | macro);
}
} // namespace
int main(int argc, char* argv[])
{
std::string input((std::istreambuf_iterator<char>(std::cin)),
std::istreambuf_iterator<char>());
skel_grammar<std::string::iterator> grammar;
bool r = bsq::parse(input.begin(), input.end(), grammar);
std::cout << std::boolalpha << r << '\n';
return 0;
}
这段代码出了什么问题？
嗯.我觉得我们在聊天中讨论了一些细节,而不是在问题中反映出来.
让我用我的"玩具"实现来娱乐你,完整的测试用例,一个可以识别的语法<<macros>>,包括嵌套的扩展.
显着特点:
process())进行扩展,为您提供最大的灵活性(您可以使用查找表,导致解析失败,具体取决于宏内容,甚至具有独立于输出的副作用)spirit::istream_iterator如何在流模式下解析输入(Stream-based Parsing Made Easy).如果您的输入流为10 GB,并且仅包含4个宏,则这具有明显的好处 - 它是爬行性能(或内存不足)与缩放之间的差异.
oss).但是,您可以轻松地将输出直接挂接到实例std::cout或者说std::ofstream实例<<或>>分隔符(#define SUPPORT_ESCAPES)无需再费周折:
注意 由于懒惰,我需要-std==c++0x,但只有在SUPPORT_ESCAPES定义时
//#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
namespace qi = boost::spirit::qi;
namespace phx= boost::phoenix;
namespace fsn= boost::fusion;
namespace
{
#define SUPPORT_ESCAPES
static bool process(std::string& macro)
{
if (macro == "error") {
return false; // fail the parse
}
if (macro == "hello") {
macro = "bye";
} else if (macro == "bye") {
macro = "We meet again";
} else if (macro == "sideeffect") {
std::cerr << "this is a side effect while parsing\n";
macro = "(done)";
} else if (std::string::npos != macro.find('~')) {
std::reverse(macro.begin(), macro.end());
macro.erase(std::remove(macro.begin(), macro.end(), '~'));
} else {
macro = std::string("<<") + macro + ">>"; // this makes the unsupported macros appear unchanged
}
return true;
}
template<typename Iterator, typename OutIt>
struct skel_grammar : public qi::grammar<Iterator>
{
struct fastfwd {
template<typename,typename> struct result { typedef bool type; };
template<typename R, typename O>
bool operator()(const R&r,O& o) const
{
#ifndef SUPPORT_ESCAPES
o = std::copy(r.begin(),r.end(),o);
#else
auto f = std::begin(r), l = std::end(r);
while(f!=l)
{
if (('\\'==*f) && (l == ++f))
break;
*o++ = *f++;
}
#endif
return true; // false to fail the parse
}
} copy;
skel_grammar(OutIt& out) : skel_grammar::base_type(start)
{
using namespace qi;
#ifdef SUPPORT_ESCAPES
rawch = ('\\' >> char_) | char_;
#else
# define rawch qi::char_
#endif
macro = ("<<" >> (
(*(rawch - ">>" - "<<") [ _val += _1 ])
% macro [ _val += _1 ] // allow nests
) >>
">>")
[ _pass = phx::bind(process, _val) ];
start =
raw [ +(rawch - "<<") ] [ _pass = phx::bind(copy, _1, phx::ref(out)) ]
% macro [ _pass = phx::bind(copy, _1, phx::ref(out)) ]
;
BOOST_SPIRIT_DEBUG_NODE(start);
BOOST_SPIRIT_DEBUG_NODE(macro);
# undef rawch
}
private:
#ifdef SUPPORT_ESCAPES
qi::rule<Iterator, char()> rawch;
#endif
qi::rule<Iterator, std::string()> macro;
qi::rule<Iterator> start;
};
}
int main(int argc, char* argv[])
{
std::string input =
"Greeting is <<hello>> world!\n"
"Side effects are <<sideeffect>> and <<other>> vars are untouched\n"
"Empty <<>> macros are ok, as are stray '>>' pairs.\n"
"<<nested <<macros>> (<<hello>>?) work>>\n"
"The order of expansion (evaluation) is _eager_: '<<<<hello>>>>' will expand to the same as '<<bye>>'\n"
"Lastly you can do algorithmic stuff too: <<!esrever ~ni <<hello>>>>\n"
#ifdef SUPPORT_ESCAPES // bonus: escapes
"You can escape \\<<hello>> (not expanded to '<<hello>>')\n"
"Demonstrate how it <<avoids <\\<nesting\\>> macros>>.\n"
#endif
;
std::ostringstream oss;
std::ostream_iterator<char> out(oss);
skel_grammar<std::string::iterator, std::ostream_iterator<char> > grammar(out);
std::string::iterator f(input.begin()), l(input.end());
bool r = qi::parse(f, l, grammar);
std::cout << "parse result: " << (r?"success":"failure") << "\n";
if (f!=l)
std::cout << "unparsed remaining: '" << std::string(f,l) << "'\n";
std::cout << "Streamed output:\n\n" << oss.str() << '\n';
return 0;
}
this is a side effect while parsing
parse result: success
Streamed output:
Greeting is bye world!
Side effects are (done) and <<other>> vars are untouched
Empty <<>> macros are ok, as are stray '>>' pairs.
<<nested <<macros>> (bye?) work>>
The order of expansion (evaluation) is _eager_: 'We meet again' will expand to the same as 'We meet again'
Lastly you can do algorithmic stuff too: eyb in reverse!
You can escape <<hello>> (not expanded to 'bye')
Demonstrate how it <<avoids <<nesting>> macros>>.
grok有很多隐藏的功能.我建议你看看测试用例,并在process()回调对方一起看看是怎么回事.
干杯和HTH :)