UPDATE显示带有递归 ast 的简化解析器，用于解析所示的示例表达式
与往常一样，断言消息恰好导致了问题：
// If you see the assertion below failing then the start rule
// passed to the constructor of the grammar is not compatible with
// the grammar (i.e. it uses different template parameters).
BOOST_SPIRIT_ASSERT_MSG(
(is_same<start_type, rule<Iterator_, T1_, T2_, T3_, T4_> >::value)
, incompatible_start_rule, (rule<Iterator_, T1_, T2_, T3_, T4_>));
所以它告诉你应该将语法与开始规则相匹配：你有
struct Parser : qi::grammar<Iterator, Expression(), ascii::space_type>
but
qi::rule<Iterator, Expression, ascii::space_type> expression;
显然你忘记了括号：
qi::rule<Iterator, Expression(), ascii::space_type> expression;
使用通用库时的指南：
其中一些“规则”是普遍适用的，但没有例外。 2 与 Boost Spirit 特别相关：
- 婴儿学步；从小处开始（空的，均匀的）
- 从 AST 开始，精确匹配语法
- 逐步建立，
- 编译一路走来的每一步
UPDATE
这是一个非常简化的语法。如前所述，在之前的“第一条精神规则”中，从 AST 开始以完全匹配语法：
namespace ast {
namespace tag {
struct constant;
struct variable;
struct function;
}
template <typename Tag> struct Identifier { char name; };
using Constant = Identifier<tag::constant>;
using Variable = Identifier<tag::variable>;
using Function = Identifier<tag::function>;
struct FunctionCall;
using Expression = boost::make_recursive_variant<
Constant,
Variable,
boost::recursive_wrapper<FunctionCall>
>::type;
struct FunctionCall {
Function function;
std::vector<Expression> params;
};
struct Equation {
Expression lhs, rhs;
};
}
当然这可能是简单得多仍然因为所有标识符都只是char你可以动态地进行切换（印象 http://paste.ubuntu.com/10378816/).
现在，语法必须遵循。1.把事情简单化2.仔细格式化3.直接匹配ast，4.添加调试宏：
template <typename It, typename Skipper = ascii::space_type>
struct Parser : qi::grammar<It, ast::Equation(), Skipper>
{
Parser() : Parser::base_type(equation_)
{
using namespace qi;
constant_ = qi::eps >> char_("a-eA-E");
function_ = qi::eps >> char_("f-rF-R");
variable_ = qi::eps >> char_("s-zS-Z");
function_call = function_ >> '(' >> -(expression_ % ',') >> ')';
expression_ = constant_ | variable_ | function_call;
equation_ = expression_ >> '=' >> expression_;
BOOST_SPIRIT_DEBUG_NODES((constant_)(function_)(variable_)(function_call)(expression_)(equation_))
}
qi::rule<It, ast::Constant()> constant_;
qi::rule<It, ast::Function()> function_;
qi::rule<It, ast::Variable()> variable_;
qi::rule<It, ast::FunctionCall(), Skipper> function_call;
qi::rule<It, ast::Expression(), Skipper> expression_;
qi::rule<It, ast::Equation(), Skipper> equation_;
};
请注意这些评论是如何变得完全没有必要的。另请注意如何递归使用expression_解决了您最头疼的问题！
完整计划
住在科里鲁 http://coliru.stacked-crooked.com/a/2252a606d3429967
//#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
namespace ast {
namespace tag {
struct constant;
struct variable;
struct function;
}
template <typename Tag> struct Identifier { char name; };
using Constant = Identifier<tag::constant>;
using Variable = Identifier<tag::variable>;
using Function = Identifier<tag::function>;
struct FunctionCall;
using Expression = boost::make_recursive_variant<
Constant,
Variable,
boost::recursive_wrapper<FunctionCall>
>::type;
struct FunctionCall {
Function function;
std::vector<Expression> params;
};
struct Equation {
Expression lhs, rhs;
};
}
BOOST_FUSION_ADAPT_STRUCT(ast::Constant, (char, name))
BOOST_FUSION_ADAPT_STRUCT(ast::Variable, (char, name))
BOOST_FUSION_ADAPT_STRUCT(ast::Function, (char, name))
BOOST_FUSION_ADAPT_STRUCT(ast::FunctionCall, (ast::Function, function)(std::vector<ast::Expression>, params))
BOOST_FUSION_ADAPT_STRUCT(ast::Equation, (ast::Expression, lhs)(ast::Expression, rhs))
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
template <typename It, typename Skipper = ascii::space_type>
struct Parser : qi::grammar<It, ast::Equation(), Skipper>
{
Parser() : Parser::base_type(equation_)
{
using namespace qi;
constant_ = qi::eps >> char_("a-eA-E");
function_ = qi::eps >> char_("f-rF-R");
variable_ = qi::eps >> char_("s-zS-Z");
function_call = function_ >> '(' >> -(expression_ % ',') >> ')';
expression_ = constant_ | variable_ | function_call;
equation_ = expression_ >> '=' >> expression_;
BOOST_SPIRIT_DEBUG_NODES((constant_)(function_)(variable_)(function_call)(expression_)(equation_))
}
qi::rule<It, ast::Constant()> constant_;
qi::rule<It, ast::Function()> function_;
qi::rule<It, ast::Variable()> variable_;
qi::rule<It, ast::FunctionCall(), Skipper> function_call;
qi::rule<It, ast::Expression(), Skipper> expression_;
qi::rule<It, ast::Equation(), Skipper> equation_;
};
int main() {
std::cout << "Unification Algorithm\n\n";
std::string const phrase = "f(a, b) = f(g(z, x), g(x, h(x)), c)";
using It = std::string::const_iterator;
It itr = phrase.begin(), last = phrase.end();
std::cout << phrase << std::endl;
Parser<It> g;
ast::Equation parsed;
if (phrase_parse(itr, last, g, ascii::space, parsed)) {
std::cout << "Expression parsed.\n";
} else {
std::cout << "Could not parse equation.\n";
}
if (itr != last) {
std::cout << "Remaining unparsed input: '" << std::string(itr,last) << "'\n";
}
}
