我正在尝试设置一个解析器,给定一个值,可以将它分配给向量的某个元素,但我不完全确定如何实现它。
假设以下代码解析字符串(0){**+*+}
。对于每个bar.a[0]
,它应该增加+
一次,对于每个bar.b[0]
,它应该增加*
一次。我遇到的问题是我不确定如何使用_a
来获取向量元素的引用:
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <vector>
//The struct containing the vector.
struct testStruct {
std::vector<int> a, b;
};
BOOST_FUSION_ADAPT_STRUCT (
testStruct,
(std::vector<int>, a)
(std::vector<int>, b)
)
namespace qi = boost::spirit::qi;
namespace phoenix = boost::phoenix;
namespace ascii = boost::spirit::ascii;
template<typename Iterator>
struct foo : qi::grammar<Iterator, testStruct(), qi::locals<unsigned>, ascii::space_type> {
foo() : foo::base_type(start) {
using namespace qi::labels;
using qi::eps;
using qi::lit;
using qi::uint_;
using phoenix::at_c;
start = lit('(')
>> uint_ [_a = _1]
>> ')'
>> '{'
>> starsOrPlus(
at_c<_a>(at_c<0>(_val)), //This is where I'm not sure what to do.
at_c<_a>(at_c<1>(_val))
)
>> '}'
;
starsOrPlus = eps [_r1 = 0]
[_r2 = 0]
>> (
* (
(
+lit('+') [_r1 += 1]
)
^ (
+lit('*') [_r2 += 1]
)
)
)
;
}
qi::rule<Iterator, testStruct(), qi::locals<unsigned>, ascii::space_type> start;
//Parses stars and pluses. Sets the first uint8_t to the number of *, and the
//second to the number of +.
qi::rule<Iterator, void(int&, int&), ascii::space_type> starsOrPlus;
};
//Main program
int main() {
std::string testString = "(2){**++*+}";
typedef foo<std::string::const_iterator> foo;
foo grammar;
testStruct bar;
std::string::const_iterator iter = testString.begin();
std::string::const_iterator end = testString.end();
bool parsed = phrase_parse(iter, end, grammar, ascii::space, bar);
if (parsed) {
//Do something with the data...
}
return 0;
}
无法使用以下错误进行编译:
main.cpp||In constructor 'foo<Iterator>::foo()':|
main.cpp|36|error: 'boost::spirit::_a' cannot appear in a constant-expression|
main.cpp|36|error: no matching function for call to 'at_c(boost::phoenix::actor<boost::phoenix::composite<boost::phoenix::at_eval<0>, boost::fusion::vector<boost::spirit::attribute<0>, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_> > >&)'|
main.cpp|37|error: 'boost::spirit::_a' cannot appear in a constant-expression|
main.cpp|37|error: no matching function for call to 'at_c(boost::phoenix::actor<boost::phoenix::composite<boost::phoenix::at_eval<1>, boost::fusion::vector<boost::spirit::attribute<0>, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_> > >&)'|
所以,显然,我不能在at_c
中使用占位符值。我也知道,即使我可以,如果给定的位置超出范围,也会出现重新调整向量大小的问题。
我将如何实现这样的目标?我是否只是以完全错误的方式解决这个问题?
答案 0 :(得分:2)
这应该让你开始:
#include <vector>
#include <string>
#include <iostream>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
struct testStruct
{
std::vector<int> a, b;
};
BOOST_FUSION_ADAPT_STRUCT
(
testStruct,
(std::vector<int>, a)
(std::vector<int>, b)
)
namespace bp = boost::phoenix;
namespace bs = boost::spirit;
namespace bsq = bs::qi;
template<typename Iterator>
struct foo : bsq::grammar<Iterator, testStruct(), bsq::locals<unsigned> >
{
foo() : foo::base_type(start)
{
using namespace bs::labels;
using bp::at_c;
using bp::resize;
using bs::lit;
using bs::uint_;
start
= '('
>> uint_
[
_a = _1,
resize(at_c<0>(_val), _1 + 1),
resize(at_c<1>(_val), _1 + 1)
]
>> "){"
>> starsOrPlus(at_c<0>(_val)[_a], at_c<1>(_val)[_a])
>> '}'
;
starsOrPlus
= *(
lit('+')[_r1 += 1]
| lit('*')[_r2 += 1]
)
;
}
bsq::rule<Iterator, testStruct(), bsq::locals<unsigned> > start;
bsq::rule<Iterator, void(int&, int&)> starsOrPlus;
};
void printvec(std::vector<int> const& vec)
{
bool first = true;
for (std::vector<int>::const_iterator it = vec.begin(), it_end = vec.end();
it != it_end;
++it)
{
if (first)
first = false;
else
std::cout << ", ";
std::cout << *it;
}
}
int main()
{
foo<std::string::const_iterator> grammar;
testStruct bar;
std::string const input = "(2){**++*}";
std::string::const_iterator first = input.begin(), last = input.end();
if (bsq::parse(first, last, grammar, bar) && first == last)
{
std::cout << "bar.a: ";
printvec(bar.a);
std::cout << "\nbar.b: ";
printvec(bar.b);
std::cout << '\n';
}
else
std::cout << "parse failed\n";
}
这里值得注意的变化是:
vector
内的testStruct
必须调整为足以使所需索引有效的大小operator[]
代替boost::phoenix::at_c
来访问vector
的索引,就像在“普通”代码中一样。请注意,我拿出了跳过解析器来简化操作(因为它似乎没有必要);如果需要,请将其添加回来 - 这里没有真正的相关性。