我正在尝试使用 Boost::Spirit 解析 PDDL 文件，但在将前提条件解析为结构时遇到了一些问题。我很难理解关于如何将条件放入我的结构和递归的 Boost 手册。
我在下面给出了一段代码，应该可以很好地说明问题。必须解析如下所示的字符串:
:precondition
(and
(at-pos ?r ?pos)
(not (has-pos ?m ?pos))
)
到目前为止，我的代码看起来像这样，但我几乎可以肯定我不了解 at_c 是如何工作的，因为我还没有使用过 Boost::Phoenix。
predi_param = '?' >> name_type;
predi = '('
>> name_type
>> +predi_param
>> ')';
literal = (
( '(' >> lit("not") >>
predi [at_c<0>(_val) = false]
>> ')'
)
| predi [at_c<0>(_val) = true]
)
>> ')';
pred_list = ( '(' >> lit("and") >> (*pred_list) >> ')')
| literal;
preconditions = lit(":precondition") >> pred_list;
qi::rule<Iterator, std::string(), ascii::space_type> predi_param;
qi::rule<Iterator, Predicate(), ascii::space_type> predi;
qi::rule<Iterator, Literal(), ascii::space_type> literal;
qi::rule<Iterator, std::vector<Literal>(), ascii::space_type> preconditions, pred_list;
我的 AST 看起来像这样:
struct Predicate
{
std::string name;
std::vector<std::string> predicate_params;
};
struct Literal
{
bool condition;
Predicate predicate;
};
BOOST_FUSION_ADAPT_STRUCT(
pddl_parser::Literal,
(bool, condition)
(pddl_parser::Predicate, predicate)
)
BOOST_FUSION_ADAPT_STRUCT(
pddl_parser::Predicate,
(std::string, name)
(std::vector<std::string>, predicate_params)
)
编译会导致编译错误:
parser.cpp:67:17: required from ‘pddl_parser::domain_parser<Iterator>::domain_parser() [with Iterator = __gnu_cxx::__normal_iterator<const char*, std::__cxx11::basic_string<char> >]’
parser.cpp:136:10: required from here
/usr/include/boost/spirit/home/qi/detail/assign_to.hpp:153:20: error: no matching function for call to ‘pddl_parser::Literal::Literal(const std::vector<pddl_parser::Literal>&)’
attr = static_cast<Attribute>(val);
^~~~~~~~~~~~~~~~~~~~~~~~~~~
In file included from parser.cpp:11:0:
./pddlast.h:23:10: note: candidate: pddl_parser::Literal::Literal()
struct Literal
^~~~~~~
./pddlast.h:23:10: note: candidate expects 0 arguments, 1 provided
./pddlast.h:23:10: note: candidate: pddl_parser::Literal::Literal(const pddl_parser::Literal&)
./pddlast.h:23:10: note: no known conversion for argument 1 from ‘const std::vector<pddl_parser::Literal>’ to ‘const pddl_parser::Literal&’
./pddlast.h:23:10: note: candidate: pddl_parser::Literal::Literal(pddl_parser::Literal&&)
./pddlast.h:23:10: note: no known conversion for argument 1 from ‘const std::vector<pddl_parser::Literal>’ to ‘pddl_parser::Literal&&’
如果我出于测试目的将 pred_list 重新格式化为 pred_list = ( '(' >> *literal) >> ')'); 代码编译但仍然产生没有成功，虽然我把 (and ) 拿出来了。我的印象是我完全错了，但找不到任何东西。这是我第一次尝试使用 Boost::Spirit。
最佳答案
好吧，你说吧
pred_list = ( '(' >> *literal) >> ')');
编译，但以下不编译:
pred_list = ( '(' >> lit("and") >> (*pred_list) >> ')') | literal;
如果你仔细观察，这是有道理的。自 pred_list声明的属性类型为 std::vector<Literal> , 显然重复文字 ( *literal ) 可能与自动属性传播兼容。
现在，看看第二个规则定义。它解析了一堆无属性文字( '(' ， "and" ， ')' )然后...... *pred_list .如果pred_list声明一个 std::vector<Literal>属性，肯定是*pred_list合成一个 std::vector<std::vector<Literal> > .更糟糕的是，“事后的想法”| literal使合成属性等同于 variant<vector<vector<Literal>>, Literal> .
是的。这有点乱。您的 AST 根本不反射(reflect)规则，反之亦然。
前方的道路
您可能应该重述您的问题，删除失败的实现部分并描述目标。如果我们能知道真正的语法要求，/那么/我们就可以推导出一个与之匹配的正确 AST。
插曲
在中场休息期间，让我简化 literal 的规则. (有关背景资料，请参阅 Boost spirit semantic actions on qi::rule):
literal =
'(' >> matches("not") >> predi >> ')'
| qi::attr(false) >> predi
;
PS It appears that a stray cat also typed an extra unbalanced
>> ')';at the end there?
建设性的猜测
仅从示例输入来看，我打赌您只想解析 scheme-like function applications形式的
(function_name arguments)
应用程序可以嵌套的地方。因此，参数要么是原子，要么是函数应用程序。
好吧，让我们快速地在上面放一个 AST:
namespace AST {
using Atom = std::string;
struct Application;
using Expression = boost::variant<Atom, Application>;
struct Application {
Atom function;
std::vector<Expression> arguments;
};
}
这很简单，对吧。这是解析前提条件的最简单的语法:
template <typename Iterator>
struct Precondition : qi::grammar<Iterator, AST::Expression()> {
Precondition() : Precondition::base_type(precondition) {
using namespace qi;
atom = +(graph - '(' - ')');
application = '(' >> atom >> *expression >> ')';
expression = atom | application;
precondition = skip(ascii::space) [":precondition" >> expression];
BOOST_SPIRIT_DEBUG_NODES((precondition)(expression)(application)(atom))
}
private:
using Skipper = qi::ascii::space_type;
qi::rule<Iterator, AST::Application(), Skipper> application;
qi::rule<Iterator, AST::Expression(), Skipper> expression;
// lexemes
qi::rule<Iterator, AST::Expression()> precondition;
qi::rule<Iterator, AST::Atom()> atom;
};
请注意每条规则如何仅重述相应的 AST 节点。 precondition也对外界隐藏了 skipper 。
输出: Live On Coliru
打印
Parsed (and (at-pos ?r ?pos) (not (has-pos ?m ?pos)))
还有BOOST_SPIRIT_DEBUG启用:
<precondition>
<try>:precondition\n </try>
<expression>
<try>\n </try>
<atom>
<try>(and\n </try>
<fail/>
</atom>
<application>
<try>(and\n </try>
<atom>
<try>and\n </try>
<success>\n </success>
<attributes>[[a, n, d]]</attributes>
</atom>
<expression>
<try>\n </try>
<atom>
<try>(at-pos ?r ?pos)\n </try>
<fail/>
</atom>
<application>
<try>(at-pos ?r ?pos)\n </try>
<atom>
<try>at-pos ?r ?pos)\n </try>
<success> ?r ?pos)\n </success>
<attributes>[[a, t, -, p, o, s]]</attributes>
</atom>
<expression>
<try> ?r ?pos)\n </try>
<atom>
<try>?r ?pos)\n </try>
<success> ?pos)\n </success>
<attributes>[[?, r]]</attributes>
</atom>
<success> ?pos)\n </success>
<attributes>[[?, r]]</attributes>
</expression>
<expression>
<try> ?pos)\n </try>
<atom>
<try>?pos)\n </try>
<success>)\n </success>
<attributes>[[?, p, o, s]]</attributes>
</atom>
<success>)\n </success>
<attributes>[[?, p, o, s]]</attributes>
</expression>
<expression>
<try>)\n </try>
<atom>
<try>)\n </try>
<fail/>
</atom>
<application>
<try>)\n </try>
<fail/>
</application>
<fail/>
</expression>
<success>\n </success>
<attributes>[[[a, t, -, p, o, s], [[?, r], [?, p, o, s]]]]</attributes>
</application>
<success>\n </success>
<attributes>[[[a, t, -, p, o, s], [[?, r], [?, p, o, s]]]]</attributes>
</expression>
<expression>
<try>\n </try>
<atom>
<try>(not (has-pos ?m ?po</try>
<fail/>
</atom>
<application>
<try>(not (has-pos ?m ?po</try>
<atom>
<try>not (has-pos ?m ?pos</try>
<success> (has-pos ?m ?pos))\n</success>
<attributes>[[n, o, t]]</attributes>
</atom>
<expression>
<try> (has-pos ?m ?pos))\n</try>
<atom>
<try>(has-pos ?m ?pos))\n </try>
<fail/>
</atom>
<application>
<try>(has-pos ?m ?pos))\n </try>
<atom>
<try>has-pos ?m ?pos))\n </try>
<success> ?m ?pos))\n </success>
<attributes>[[h, a, s, -, p, o, s]]</attributes>
</atom>
<expression>
<try> ?m ?pos))\n </try>
<atom>
<try>?m ?pos))\n </try>
<success> ?pos))\n </success>
<attributes>[[?, m]]</attributes>
</atom>
<success> ?pos))\n </success>
<attributes>[[?, m]]</attributes>
</expression>
<expression>
<try> ?pos))\n </try>
<atom>
<try>?pos))\n </try>
<success>))\n </success>
<attributes>[[?, p, o, s]]</attributes>
</atom>
<success>))\n </success>
<attributes>[[?, p, o, s]]</attributes>
</expression>
<expression>
<try>))\n </try>
<atom>
<try>))\n </try>
<fail/>
</atom>
<application>
<try>))\n </try>
<fail/>
</application>
<fail/>
</expression>
<success>)\n </success>
<attributes>[[[h, a, s, -, p, o, s], [[?, m], [?, p, o, s]]]]</attributes>
</application>
<success>)\n </success>
<attributes>[[[h, a, s, -, p, o, s], [[?, m], [?, p, o, s]]]]</attributes>
</expression>
<expression>
<try>)\n </try>
<atom>
<try>)\n </try>
<fail/>
</atom>
<application>
<try>)\n </try>
<fail/>
</application>
<fail/>
</expression>
<success>\n </success>
<attributes>[[[n, o, t], [[[h, a, s, -, p, o, s], [[?, m], [?, p, o, s]]]]]]</attributes>
</application>
<success>\n </success>
<attributes>[[[n, o, t], [[[h, a, s, -, p, o, s], [[?, m], [?, p, o, s]]]]]]</attributes>
</expression>
<expression>
<try>\n </try>
<atom>
<try>)</try>
<fail/>
</atom>
<application>
<try>)</try>
<fail/>
</application>
<fail/>
</expression>
<success></success>
<attributes>[[[a, n, d], [[[a, t, -, p, o, s], [[?, r], [?, p, o, s]]], [[n, o, t], [[[h, a, s, -, p, o, s], [[?, m], [?, p, o, s]]]]]]]]</attributes>
</application>
<success></success>
<attributes>[[[a, n, d], [[[a, t, -, p, o, s], [[?, r], [?, p, o, s]]], [[n, o, t], [[[h, a, s, -, p, o, s], [[?, m], [?, p, o, s]]]]]]]]</attributes>
</expression>
<success></success>
<attributes>[[[a, n, d], [[[a, t, -, p, o, s], [[?, r], [?, p, o, s]]], [[n, o, t], [[[h, a, s, -, p, o, s], [[?, m], [?, p, o, s]]]]]]]]</attributes>
</precondition>
这个例子向你展示的是:
- 如何在 AST 中表示变体数据类型
- 如何通过它实现递归(更多高级信息另见documentation)
它不会立即根据 PDDL 规范验证 AST。我完全不确定您打算实现多少内容，所以我认为更通用的启动器可能无论如何都会有所帮助。
完整 list
#define BOOST_SPIRIT_DEBUG
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
namespace AST {
using Atom = std::string;
struct Application;
using Expression = boost::variant<Atom, Application>;
struct Application {
Atom function;
std::vector<Expression> arguments;
friend std::ostream& operator<<(std::ostream& os, Application const& a) {
os << "(" << a.function;
for (auto& arg : a.arguments)
os << " " << arg;
return os << ")";
}
};
}
BOOST_FUSION_ADAPT_STRUCT(AST::Application, function, arguments)
namespace pddl_parser {
namespace qi = boost::spirit::qi;
template <typename Iterator>
struct Precondition : qi::grammar<Iterator, AST::Expression()> {
Precondition() : Precondition::base_type(precondition) {
using namespace qi;
atom = +(graph - '(' - ')');
application = '(' >> atom >> *expression >> ')';
expression = atom | application;
precondition = skip(ascii::space) [":precondition" >> expression];
BOOST_SPIRIT_DEBUG_NODES((precondition)(expression)(application)(atom))
}
private:
using Skipper = qi::ascii::space_type;
qi::rule<Iterator, AST::Application(), Skipper> application;
qi::rule<Iterator, AST::Expression(), Skipper> expression;
// lexemes
qi::rule<Iterator, AST::Expression()> precondition;
qi::rule<Iterator, AST::Atom()> atom;
};
}
int main() {
using It = std::string::const_iterator;
for (std::string const& input : {
R"--(:precondition
(and
(at-pos ?r ?pos)
(not (has-pos ?m ?pos))
))--"
})
{
std::cout << "-----\n";
It f = input.begin(), l = input.end();
AST::Expression precondition;
bool ok = parse(f, l, pddl_parser::Precondition<It>{}, precondition);
if (ok) {
std::cout << "Parsed " << precondition << "\n";
} else {
std::cout << "Parse Failed\n";
}
if (f != l) {
std::cout << "Remaining unparsed input: '" << std::string(f, l) << "'\n";
}
}
}
关于c++ - 使用 Boost::Spirit 解析前置条件和递归，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44030554/