c-Boost.Spirit X3解析器“没有类型命名为in(…)”
作者:互联网
当我遇到错误时,我一直在玩Boost.Spirit X3计算器示例,我无法理解.
我将程序最小化以降低复杂度,但仍会引发相同的错误.
假设我想将输入解析为语句(字符串)列表,后跟定界符(“;”).
这是我的结构:
namespace client { namespace ast
{
struct program
{
std::list<std::string> stmts;
};
}}
BOOST_FUSION_ADAPT_STRUCT(client::ast::program,
(std::list<std::string>, stmts)
)
语法如下:
namespace client
{
namespace grammar
{
x3::rule<class program, ast::program> const program("program");
auto const program_def =
*((*char_) > ';')
;
BOOST_SPIRIT_DEFINE(
program
);
auto calculator = program;
}
using grammar::calculator;
}
调用
int
main()
{
std::cout <<"///////////////////////////////////////////\n\n";
std::cout << "Expression parser...\n\n";
std::cout << //////////////////////////////////////////////////\n\n";
std::cout << "Type an expression...or [q or Q] to quit\n\n";
typedef std::string::const_iterator iterator_type;
typedef client::ast::program ast_program;
std::string str;
while (std::getline(std::cin, str))
{
if (str.empty() || str[0] == 'q' || str[0] == 'Q')
break;
auto& calc = client::calculator; // Our grammar
ast_program program; // Our program (AST)
iterator_type iter = str.begin();
iterator_type end = str.end();
boost::spirit::x3::ascii::space_type space;
bool r = phrase_parse(iter, end, calc, space, program);
if (r && iter == end)
{
std::cout << "-------------------------\n";
std::cout << "Parsing succeeded\n";
std::cout<< '\n';
std::cout << "-------------------------\n";
}
else
{
std::cout << "-------------------------\n";
std::cout << "Parsing failed\n";
std::cout << "-------------------------\n";
}
}
std::cout << "Bye... :-) \n\n";
return 0;
}
我得到的错误是
/opt/boost_1_66_0/boost/spirit/home/x3/support/traits/container_traits.hpp: In instantiation of ‘struct boost::spirit::x3::traits::container_value<client::ast::program, void>’:
.
.
.
/opt/boost_1_66_0/boost/spirit/home/x3/support/traits/container_traits.hpp:76:12: error: no type named ‘value_type’ in ‘struct client::ast::program’
struct container_value
/opt/boost_1_66_0/boost/spirit/home/x3/operator/detail/sequence.hpp:497:72: error: no type named ‘type’ in ‘struct boost::spirit::x3::traits::container_value<client::ast::program, void>’
, typename traits::is_substitute<attribute_type, value_type>::type());
^~~~~~
我尝试过的事情:
正在关注Getting boost::spirit::qi to use stl containers
即使它使用Qi,我仍然尝试:
namespace boost{namespace spirit{ namespace traits{
template<>
struct container_value<client::ast::program>
//also with struct container<client::ast::program, void>
{
typedef std::list<std::string> type;
};
}}}
您会发现我在黑暗中还挺不错的,所以无济于事.
parser2.cpp:41:8: error: ‘container_value’ is not a class template
struct container_value<client::ast::program>
^~~~~~~~~~~~~~~
在同一SO问题中,我作者说:“但是,有一个已知的局限性,当您尝试使用具有单个元素的结构时,除非您将qi :: eps> …添加到其中,否则容器编译也会失败.你的规则.”
我确实尝试添加虚拟eps也没有成功.
请帮我弄清楚那个错误是什么意思.
解决方法:
对.当涉及单元素序列时,这似乎是属性自动传播的另一个限制.
我可能会硬着头皮,将规则定义从它的定义(以及您期望的工作范围)更改为:
x3::rule<class program_, std::vector<std::string> >
这消除了混乱的根源.
其他说明:
>您有char_,它也吃’;’所以语法永远不会成功,因为没有’;’将遵循“声明”.
>您的语句不是lexeme,因此空格被丢弃(这是您的意思吗?请参阅Boost spirit skipper issues)
>您的语句可能为空,这意味着解析总是会在输入结束时失败(在该处它将始终读取为空状态,然后发现期望的’;’丢失了).通过接受至少1个字符来修复该问题,然后再接受声明.
进行一些简化/样式更改:
#include <boost/fusion/adapted.hpp>
#include <boost/spirit/home/x3.hpp>
#include <list>
namespace x3 = boost::spirit::x3;
namespace ast {
using statement = std::string;
struct program {
std::list<statement> stmts;
};
}
BOOST_FUSION_ADAPT_STRUCT(ast::program, stmts)
namespace grammar {
auto statement
= x3::rule<class statement_, ast::statement> {"statement"}
= +~x3::char_(';');
auto program
= x3::rule<class program_, std::list<ast::statement> > {"program"}
= *(statement >> ';');
}
#include <iostream>
#include <iomanip>
int main() {
std::cout << "Type an expression...or [q or Q] to quit\n\n";
using It = std::string::const_iterator;
for (std::string str; std::getline(std::cin, str);) {
if (str.empty() || str[0] == 'q' || str[0] == 'Q')
break;
auto &parser = grammar::program;
ast::program program; // Our program (AST)
It iter = str.begin(), end = str.end();
if (phrase_parse(iter, end, parser, x3::space, program)) {
std::cout << "Parsing succeeded\n";
for (auto& s : program.stmts) {
std::cout << "Statement: " << std::quoted(s, '\'') << "\n";
}
}
else
std::cout << "Parsing failed\n";
if (iter != end)
std::cout << "Remaining unparsed: " << std::quoted(std::string(iter, end), '\'') << "\n";
}
}
对于输入“ a; b; c; d;”打印:
Parsing succeeded
Statement: 'a'
Statement: 'b'
Statement: 'c'
Statement: 'd'
标签:c,parsing,boost-spirit 来源: https://codeday.me/bug/20191014/1912087.html