C++ Boost qi递归规则构建

本文介绍了C++ Boost qi递归规则构建的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ [看来我的解释和期望都不是很清楚，所以我在帖子末尾添加了我希望如何使用该功能的精确性][It seems my explanations and expectations are not clear at all, so I added precision on how I'd like to use the feature at the end of the post]我目前正在使用 boost qi 研究语法.我有一个规则的循环构造，因为我需要从向量的元素构建它.我用简单的类型重写了它，它看起来像:I'm currently working on grammars using boost qi. I had a loop construction for a rule cause I needed to build it from the elements of a vector. I have re-written it with simple types, and it looks like:#include <string>// using boost 1.43.0#include <boost/spirit/include/qi.hpp>#include <boost/spirit/include/qi_eps.hpp>#include <boost/spirit/include/phoenix.hpp>namespace bqi = boost::spirit::qi;typedef const char* Iterator;// function that you can find [here][1]template<typename P> void test_phrase_parser(char const* input, P const& p, bool full_match = true);int main(){ // my working rule type: bqi::rule<Iterator, std::string()> myLoopBuiltRule; std::vector<std::string> v; std::vector<std::string>::const_iterator iv; v.push_back("abc"); v.push_back("def"); v.push_back("ghi"); v.push_back("jkl"); myLoopBuiltRule = (! bqi::eps); for(iv = v.begin() ; iv != v.end() ; iv++) { myLoopBuiltRule = myLoopBuiltRule.copy() [ bqi::_val = bqi::_1 ] | bqi::string(*iv) [ bqi::_val = bqi::_1 ] ; } debug(myLoopBuiltRule); char s[] = " abc "; test_phrase_parser(s, myLoopBuiltRule);}(看起来像 这里不想被对应的超链接替换，所以这里是找到函数test_phrase_parser()的地址:http://www.boost.org/doc/libs/1_43_0/libs/spirit/doc/html/spirit/qi/reference/basics.html)(Looks like here does not want to be replaced by corresponding hyperlink, so here is the address to find function test_phrase_parser(): http://www.boost.org/doc/libs/1_43_0/libs/spirit/doc/html/spirit/qi/reference/basics.html)一切都是为了最好的世界……直到我不得不对这条规则提出一个论点.这是新的规则类型:All was for the best in the best of all worlds... until I had to pass an argument to this rule. Here is the new rule type: // my not-anymore-working rule type: bqi::rule<Iterator, std::string(int*)> myLoopBuiltRule;'int*' 类型仅用于示例目的，我真正的指针指向一个更复杂的类......但仍然只是一个指针.'int*' type is for example purpose only, my real pointer is adressing a much more complex class... but still a mere pointer.我相应地更改了我的for"循环，即:I changed my 'for' loop accordingly, i.e.: for(iv = v.begin() ; iv != v.end() ; iv++) { myLoopBuiltRule = myLoopBuiltRule.copy()(bqi::_r1) [ bqi::_val = bqi::_1 ] | bqi::string(*iv) [ bqi::_val = bqi::_1 ] ; }我不得不添加一条新规则，因为 test_phrase_parser() 无法猜测将给 int 指针的值:I had to add a new rule because test_phrase_parser() cannot guess which value is to be given to the int pointer:bqi::rule<Iterator> myInitialRule;并更改 for 循环之后的所有内容:And change everything that followed the for loop:myInitialRule = myLoopBuiltRule((int*)NULL);debug(myLoopBuiltRule);char s[] = " abc ";test_phrase_parser(s, myInitialRule);然后一切都崩溃了:/home/sylvain.darras/software/repository/software/external/include/boost/boost_1_43_0/boost/spirit/home/qi/nonterminal/rule.hpp:199: error: no matching function for call to ‘assertion_failed(mpl_::failed************ (boost::spirit::qi::rule<Iterator, T1, T2, T3, T4>::operator=(const Expr&)然后我发疯并尝试:myLoopBuiltRule = myLoopBuiltRule.copy(bqi::_r1) [ bqi::_val = bqi::_1 ] | bqi::string(*iv) [ bqi::_val = bqi::_1 ]-->error: no matching function for call to ‘boost::spirit::qi::rule<const char*, std::string(int*), boost::fusion::unused_type, boost::fusion::unused_type, boost::fusion::unused_type>::copy(const boost::phoenix::actor<boost::spirit::attribute<1> >&)’然后我生气了，写道:myLoopBuiltRule = myLoopBuiltRule(bqi::_r1) [ bqi::_val = bqi::_1 ] | bqi::string(*iv) [ bqi::_val = bqi::_1 ]编译是因为它在语法上完全正确，但是哪个堆栈溢出了，因为它愉快地、漂亮地、递归地调用自己死亡......Which compiles since it is perfectly syntactically correct, but which magnificently stack overflows coz it happily, nicely, recursively, calls itself to death...然后我失去了理智并输入:Then I lost my mind and typed:myLoopBuiltRule = jf jhsgf jshdg fjsdgh fjsg jhsdg jhg sjfg jsgh df正如您所料，编译失败.Which, as you probably expect, has failed to compile.你想象一下，在写上面的小说之前，我在网上查了一下，但没有发现与复制()和参数同时传递相关的任何内容.有没有人遇到过这个问题?我错过了什么吗?You imagine that before writing the above novel, I checked out on the web, but didn't find out anything related to copy() and argument passing in the same time. Has anyone already experienced this problem ? Have I missed something ?请放心，我们将非常感谢您的任何帮助.Be assured that any help will be really really appreciated.PS:非常感谢 hkaiser，他在不知情的情况下通过谷歌回答了我的很多 boost::qi 问题(但只有这个).PS: Great thanks to hkaiser who has, without knowing it, answered a lot of my boost::qi problems through google (but this one).更多信息:我的解析器的目的是读取用给定语言 L 编写的文件.我的帖子的目的是传播我的上下文"(即:变量定义，特别是常量值，所以我可以计算表达式).The purpose of my parser is to read files written in a given language L. The purpose of my post is to propagate my "context" (i.e.: variable definitions and especially constant values, so I can compute expressions).我处理的变量类型数量很少，但肯定会增长，所以我将这些类型保存在一个容器类中.我可以在这些托管类型上循环.The number of variable types I handle is small, but it's bound to grow, so I keep these types in a container class. I can loop on these managed types.那么，让我们考虑一个我想要实现的伪算法:So, let's consider a pseudo-algorithm of what I would like to achive:LTypeList myTypes;LTypeList::const_iterator iTypes;bqi::rule<Iterator, LType(LContext*)> myLoopBuiltRule;myLoopBuiltRule = (! bqi::eps);for(iTypes = myTypes.begin() ; iTypes != myTypes.end() ; iTypes++){ myLoopBuiltRule = myLoopBuiltRule.copy()(bqi::_r1) [ bqi::_val = bqi::_1 ] | iTypes->getRule()(bqi::_r1) [ bqi::_val = bqi::_1 ]}这是在初始化期间完成的，然后使用 myLoopBuiltRule 并与不同的 LContext* 重用，解析多种类型.由于一些 L 类型可以有边界，它们是整数表达式，并且这些整数表达式可以展示常量，我(认为我)需要我的继承属性来获取我的 LContext 并能够计算表达式值.This is done during initialization and then myLoopBuiltRule is used and reused with different LContext*, parsing multiple types. And since some L types can have bounds, which are integer expressions, and that these integer expressions can exhibit constants, I (think that I) need my inherited attribute to take my LContext around and be able to compute expression value.希望我的意图更清楚.推荐答案注意 我只是用一些更多的信息链接扩展了我的答案.在这种特殊情况下，我有一种预感，您可以只是使用 Nabialek 技巧并用相应的替换继承属性qi::locals 代替.如果我有足够的时间，我可能会在稍后进行演示.Note I just extended my answer with a few more informational links. In this particular case I have a hunch that you could just get away with the Nabialek trick and replacing the inherited attribute with a corresponding qi::locals<> instead. If I have enough time, I might work out a demonstration later.请注意，在复制原始表达式树和精神解析器表达式时存在问题，特别是 - 它将创建悬空引用，因为内部不应超过包含完整表达式的末尾.在零到2 秒内达到每小时 60 英里！Please be advised that there are issues when copying proto expression trees and spirit parser expressions in particular - it will create dangling references as the internals are not supposed to live past the end of the containing full expressions. See BOOST_SPIRIT_AUTO on Zero to 60 MPH in 2 seconds!另请参阅这些答案，这些答案也与动态(在运行时)构建/编写规则有关:Also see these answers which also concerns themselves with building/composing rules on the fly (at runtime):生成精神解析器来自可选解析器表达式的可变参数列表的表达式可以将 Boost Spirit 规则参数化，它演示了如何使用 boost::proto::deepcopy 从函数返回规则(实际上就像 BOOST_SPIRIT_AUTO 那样)Generating Spirit parser expressions from a variadic list of alternative parser expressionsCan Boost Spirit Rules be parameterized which demonstrates how to return rules from a function using boost::proto::deepcopy (like BOOST_SPIRIT_AUTO does, actually)总的来说，我强烈建议反对在运行时组合规则.相反，如果您希望在运行时向规则添加替代项"，您始终可以使用 qi::symbols 代替.诀窍是在符号表中存储规则并使用qi::lazy来调用规则.特别是，这被称为 Nabialek Trick.In general, I'd very strongly advise against combining rules at runtime. Instead, if you're looking to 'add alternatives' to a rule at runtime, you can always use qi::symbols<> instead. The trick is to store a rule in the symbol-table and use qi::lazy to call the rule. In particular, this is known as the Nabialek Trick.我在这里有一个玩具命令行参数解析器，它演示了如何使用这个习语来匹配运行时定义的一组命令行参数:I have a toy command-line arguments parser here that demonstrates how you could use this idiom to match a runtime-defined set of command line arguments:https://gist.github.com/sehe/2a556a8231606406fe36不幸的是，qi::lazy 不支持继承参数参见例如Unfortunately, qi::lazy does not support inherited arguments see e.g.http://boost.2283326.n4.nabble.com/pass-inhertited-attributes-to-nabialek-trick-td2679066.html您最好编写自定义解析器组件，如下所述:You might be better off writing a custom parser component, as documented here:http://boost-spirit.com/home/articles/qi-example/creating-your-own-parser-component-for-spirit-qi/稍后我会尝试找一些时间来制定一个示例，用 qi::locals 替换继承的参数. 这篇关于C++ Boost qi递归规则构建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！ 上岸，阿里云！