问题描述
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.说我想将输入解析为语句(字符串)列表,后跟定界符(;").
I was toying with Boost.Spirit X3 calculator example when I encountered an error I couldn't get my head around.
I minimized the program to reduce complexity still throwing the same error.Say I want to parse an input as a list of statements (strings) followed by a delimiter (';').
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;
}
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;
}
我遇到的错误是
/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:
遵循>使用boost :: spirit :: qi来使用stl容器
即使使用了Qi,我还是尝试过:
Things I tried:
Following Getting boost::spirit::qi to use stl containers
Even though it uses Qi I nonetheless tried:
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;
};
}}}
您看到我有点在黑暗中,所以无济于事.
You see I'm kinda in the dark, so expectably to no avail.
parser2.cpp:41:8: error: ‘container_value’ is not a class template
struct container_value<client::ast::program>
^~~~~~~~~~~~~~~
在同一个SO问题中,我作者说
但是,有一个已知的限制,当您尝试使用具有单个元素的结构时,除非添加qi :,否则容器编译也会失败: eps >> ...按您的规则进行."
In the same SO question I author says
"There is one known limitation though, when you try to use a struct that has a single element that is also a container compilation fails unless you add qi::eps >> ... to your rule."
我确实尝试添加虚拟eps也没有成功.
I did try adding a dummy eps also without success.
请帮我弄清楚该错误的含义.
Please, help me decipher what that error means.
推荐答案
是的.当涉及到单元素序列时,这似乎是属性自动传播的另一个限制.
Yup. This looks like another limitation with automatic propagation of attributes when single-element sequences are involved.
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)更改为:
I'd probably bite the bullet and change the rule definition from what it is (and what you'd expect to work) to:
x3::rule<class program_, std::vector<std::string> >
这消除了混乱的根源.
其他说明:
-
您有
char_
,它也吃了';'
,因此语法永远不会成功,因为没有';'
会遵循声明".
you had
char_
which also eats';'
so the grammar would never succeed because no';'
would follow a "statement".
您的语句不是lexeme,因此空格被丢弃(这是您的意思吗?请参见)
your statements aren't lexeme, so whitespace is discarded (is this what you meant? See 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(在此情况下,它总是会读取空状态,然后发现预期的';'
是丢失的).通过接受至少1个字符来修正该错误.
your statement could be empty, which meant parsing would ALWAYS fail at the end of input (where it would always read an empty state, and then discover that the expected ';'
was missing). Fix it by requiring at least 1 character before accepting a statement.
进行一些简化/样式更改:
With some simplifications/style changes:
#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";
}
}
输入"a; b; c; d;"打印:
Which for input "a;b;c;d;" prints:
Parsing succeeded
Statement: 'a'
Statement: 'b'
Statement: 'c'
Statement: 'd'
这篇关于Boost.Spirit X3解析器“无类型命名为type in(...)"的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!