如何在语义操作中访问元组的元素
how to access elements of a tuple in a semantic action
我的语法有各种以通用名称开头的条目。
确定类型后,我想使用期望运算符来创建解析错误。
rule1=name >> (type1 > something);
rule2=name >> (type2 > something);
我已经意识到我不能混合使用 > 和 >> 这两个运算符——这就是括号的原因。我的猜测是括号导致创建一个元组。
如何在语义操作中访问元组的元素?
以下肯定是错误的,但应该澄清我想要完成的事情。
rule1=(name >> (type1 > something))[qi::_val = boost::phoenix::bind(
create,
qi::_1,
std::get<0>(qi::_2),
std::get<1>(qi::_2))];
谢谢
直接解决问题:
using px::at_c;
rule1 = (name >> (type1 > something)) [_val = px::bind(create, _1, at_c<0>(_2), at_c<1>(_2))];
但是,我会在 qi::eps
中使用这个小技巧来避免复杂性:
rule2 = (name >> type1 >> (eps > something)) [_val = px::bind(create, _1, _2, _3)];
最后看boost::phoenix::function<>
:
px::function<decltype(&create)> create_(create); // or just decltype(create) if it's a function object
rule3 = (name >> type1 >> (eps > something)) [_val = create_(_1, _2, _3)];
这样你甚至可以获得可读的代码!
演示
只是为了证明这三个人都有相同的行为¹
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/fusion/include/at_c.hpp>
namespace qi = boost::spirit::qi;
namespace px = boost::phoenix;
static int create(char n, char t, char s) {
assert(n=='n' && t=='t' && s=='s');
return 42;
}
int main() {
using It = std::string::const_iterator;
// fake rules just for demo
qi::rule<It, char()>
name = qi::char_("n"),
type1 = qi::char_("t"),
something = qi::char_("s");
//using boost::fusion::at_c;
qi::rule<It, int(), qi::space_type> rule1, rule2, rule3;
{
using namespace qi;
using px::at_c;
rule1 = (name >> (type1 > something)) [_val = px::bind(create, _1, at_c<0>(_2), at_c<1>(_2))];
rule2 = (name >> type1 >> (eps > something)) [_val = px::bind(create, _1, _2, _3)];
px::function<decltype(&create)> create_(create); // or just decltype(create) if it's a function object
rule3 = (name >> type1 >> (eps > something)) [_val = create_(_1, _2, _3)];
}
for(auto& parser : { rule1, rule2, rule3 }) {
for(std::string const input : { "n t s", "n t !" }) {
std::cout << "Input: '" << input << "'\n";
auto f = input.begin(), l = input.end();
int data;
try {
bool ok = qi::phrase_parse(f, l, parser, qi::space, data);
if (ok) {
std::cout << "Parsing result: " << data << '\n';
} else {
std::cout << "Parsing failed\n";
}
} catch(qi::expectation_failure<It> const& e) {
std::cout << "Expectation failure: " << e.what() << " at '" << std::string(e.first, e.last) << "'\n";
}
if (f!=l) {
std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
}
std::cout << "-------------------------------------------\n";
}
}
}
打印 3 倍相同的输出:
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
¹ PS 以此作为示例,说明如何在您的问题中创建 SSCCE 代码示例
我的语法有各种以通用名称开头的条目。 确定类型后,我想使用期望运算符来创建解析错误。
rule1=name >> (type1 > something);
rule2=name >> (type2 > something);
我已经意识到我不能混合使用 > 和 >> 这两个运算符——这就是括号的原因。我的猜测是括号导致创建一个元组。 如何在语义操作中访问元组的元素? 以下肯定是错误的,但应该澄清我想要完成的事情。
rule1=(name >> (type1 > something))[qi::_val = boost::phoenix::bind(
create,
qi::_1,
std::get<0>(qi::_2),
std::get<1>(qi::_2))];
谢谢
直接解决问题:
using px::at_c;
rule1 = (name >> (type1 > something)) [_val = px::bind(create, _1, at_c<0>(_2), at_c<1>(_2))];
但是,我会在 qi::eps
中使用这个小技巧来避免复杂性:
rule2 = (name >> type1 >> (eps > something)) [_val = px::bind(create, _1, _2, _3)];
最后看boost::phoenix::function<>
:
px::function<decltype(&create)> create_(create); // or just decltype(create) if it's a function object
rule3 = (name >> type1 >> (eps > something)) [_val = create_(_1, _2, _3)];
这样你甚至可以获得可读的代码!
演示
只是为了证明这三个人都有相同的行为¹
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/fusion/include/at_c.hpp>
namespace qi = boost::spirit::qi;
namespace px = boost::phoenix;
static int create(char n, char t, char s) {
assert(n=='n' && t=='t' && s=='s');
return 42;
}
int main() {
using It = std::string::const_iterator;
// fake rules just for demo
qi::rule<It, char()>
name = qi::char_("n"),
type1 = qi::char_("t"),
something = qi::char_("s");
//using boost::fusion::at_c;
qi::rule<It, int(), qi::space_type> rule1, rule2, rule3;
{
using namespace qi;
using px::at_c;
rule1 = (name >> (type1 > something)) [_val = px::bind(create, _1, at_c<0>(_2), at_c<1>(_2))];
rule2 = (name >> type1 >> (eps > something)) [_val = px::bind(create, _1, _2, _3)];
px::function<decltype(&create)> create_(create); // or just decltype(create) if it's a function object
rule3 = (name >> type1 >> (eps > something)) [_val = create_(_1, _2, _3)];
}
for(auto& parser : { rule1, rule2, rule3 }) {
for(std::string const input : { "n t s", "n t !" }) {
std::cout << "Input: '" << input << "'\n";
auto f = input.begin(), l = input.end();
int data;
try {
bool ok = qi::phrase_parse(f, l, parser, qi::space, data);
if (ok) {
std::cout << "Parsing result: " << data << '\n';
} else {
std::cout << "Parsing failed\n";
}
} catch(qi::expectation_failure<It> const& e) {
std::cout << "Expectation failure: " << e.what() << " at '" << std::string(e.first, e.last) << "'\n";
}
if (f!=l) {
std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
}
std::cout << "-------------------------------------------\n";
}
}
}
打印 3 倍相同的输出:
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
¹ PS 以此作为示例,说明如何在您的问题中创建 SSCCE 代码示例