我遵守了代码的和平:
#include <gtest/gtest.h>
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/qi_eps.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/spirit/include/phoenix_bind.hpp>
#include <boost/spirit/include/classic_core.hpp>
#include <boost/spirit/include/classic_confix.hpp>
#include <boost/spirit/include/classic_chset.hpp>
#include <boost/spirit/include/classic_utility.hpp>
#include <boost/fusion/include/cons.hpp>
#include <boost/fusion/include/at_c.hpp>
#include <boost/fusion/include/std_pair.hpp>
#include <boost/bind.hpp>
#include <boost/optional/optional_io.hpp>
#include <boost/variant.hpp>
namespace spi = boost::spirit;
namespace qi = boost::spirit::qi;
TEST(TestBoost, cpp_comment)
{
using qi::char_;
using qi::omit;
using qi::eoi;
typedef std::string::const_iterator iter;
const std::string example = "/* this should be ignored */";
auto b = std::begin(example);
auto e = std::end(example);
qi::rule<iter, std::string()> cpp_comment = char_('/') >> char_('/') >> *(char_ - '\n') >> (char_('\n') | omit[eoi]);
qi::rule<iter, std::string()> c_comment = char_('/') >> char_('*') >> *(char_ - "*/") >> char_('*') >> char_('/');
qi::rule<iter, std::string()> shell_comment = char_('#') >> *(char_ - '\n') >> (char_('\n') | omit[eoi]);
qi::rule<iter, std::string()> comment = cpp_comment
| c_comment
| shell_comment
;
std::string result;
EXPECT_TRUE(qi::parse(b, e, comment, result));
EXPECT_EQ(b, e);
EXPECT_EQ(result, example);
}
失败并出现以下错误:
[----------] 1 test from TestBoost
[ RUN ] TestBoost.cpp_comment
tests/spirit.cpp:56: Failure
Expected: result
Which is: "//* this should be ignored */"
To be equal to: example
Which is: "/* this should be ignored */"
[ FAILED ] TestBoost.cpp_comment (0 ms)
[----------] 1 test from TestBoost (0 ms total)
我不明白为什么。在提升文档中可能有某处提到了这种行为,但我无法找到它。有人知道为什么会这样吗?
如果我像这样进行语义动作:
qi::rule<iter, std::string()> comment = cpp_comment[spi::_val = spi::_1]
| c_comment[spi::_val = spi::_1]
| shell_comment[spi::_val = spi::_1]
;
或者
qi::rule<iter, std::string()> comment = cpp_comment[spi::_val += spi::_1]
| c_comment[spi::_val += spi::_1]
| shell_comment[spi::_val += spi::_1]
;
它有效,但我真的很想知道为什么原始代码不起作用。
答案 0 :(得分:3)
这是回溯容器属性的经典问题:
这里的想法是使用qi::hold
,或者在这种情况下甚至更好,使用qi::raw
因为看起来你想要将整个匹配的输入序列公开为属性:
qi::rule<iter, std::string()>
cpp_comment = "//" >> *~char_('\n') >> (eol|eoi),
c_comment = "/*" >> *(char_ - "*/") >> "*/",
shell_comment = '#' >> *~char_('\n') >> (eol|eoi),
comment = qi::raw [ cpp_comment | c_comment | shell_comment ];
<强> Live On Coliru 强>
#include <boost/spirit/include/qi.hpp>
#include <cassert>
namespace qi = boost::spirit::qi;
void test() {
using qi::char_;
using qi::eol;
using qi::eoi;
std::string const example = "/* this should be ignored */";
qi::rule<std::string::const_iterator, std::string()>
cpp_comment = "//" >> *~char_('\n') >> (eol|eoi),
c_comment = "/*" >> *(char_ - "*/") >> "*/",
shell_comment = '#' >> *~char_('\n') >> (eol|eoi),
comment = qi::raw [ cpp_comment | c_comment | shell_comment ];
std::string result;
bool ok = qi::parse(std::begin(example), std::end(example), comment >> eoi, result);
assert(ok);
std::cout << "expected: " << example << "\n";
std::cout << "actual: " << result << "\n";
assert(result == example);
}
int main() {
test();
}
打印
expected: /* this should be ignored */
actual: /* this should be ignored */