PEG规则识别功能原型 [英] PEG rule to identify function protoype

问题描述

我正在尝试创建一个可以解析C代码的解析器.我的用例分析可能包含函数原型的缓冲区.我想将此函数名称推送到符号表中.我是Spirit和PEG的新手，我试图弄清楚如何编写可以识别功能原型的规则.

I am trying to create a parser that can parse C code. My use case to parse a buffer that may contain a function prototype. I want to push this function name into a symbol table. I am new to Spirit and PEG and I am trying to figure out how I can write a rule that can identify function prototypes.

这是我当前的实现方式:

This is my current implementation:

auto nameRule = x3::alpha >> *x3::alnum;
auto fcnPrototypeRule = nameRule >> *nameRule;
auto fcnRule = fcnPrototypeRule >> space >> x3::char_('(') >> -(nameRule % ',') >> x3::char_(');');

这是我的应用程序代码:

This is my application code:

class Parser {   

    public:
    std::string functionParser(const std::string& input) {
        std::string output;
        x3::phrase_parse(input.begin(), input.end(), fcnRule, space, output);
        return output;
    }
};

input is ="extern void myFunction();" 输出为空字符串.我想获得函数原型.

input is = "extern void myFunction();" The output is an empty string. I wanted to get the function prototype.

推荐答案

看起来像');'应该是);"?

Looks like ');' should have been ");"?

另外，由于您有一个船长(在phrase_parse的调用中为x3::space)，因此没有什么意义:

Also since you have a skipper (x3::space in the call to phrase_parse) it doesn't make much sense to:

• 还要在解析器表达式中指定space(永远不会匹配)
• 不将nameRule包装在lexeme或noskip指令中.另请参见 Boost Spirit队长问题

• also specify space in your parser expression (it will never match)
• not wrap the nameRule inside a lexeme or noskip directive. See also Boost spirit skipper issues

所以首先要使它起作用:

So first stab at making it work:

std::string functionParser(const std::string& input) {
    namespace x3 = boost::spirit::x3;

    auto nameRule = x3::lexeme [x3::alpha >> *x3::alnum];
    auto fcnPrototypeRule = nameRule >> *nameRule;
    auto fcnRule = fcnPrototypeRule >> x3::char_('(') >> -(nameRule % ',') >> x3::char_(");");
    std::string output;
    x3::phrase_parse(input.begin(), input.end(), fcnRule, x3::space, output);
    return output;
}

但是，您会注意到它返回ndn()( 在Coliru上直播 ).

However, you'll notice it returns ndn() (Live On Coliru).

我认为这基本上是由于您的AS(std::string)与语法不太匹配这一事实造成的.我想说的是您要匹配"而不是解析"，我会使用x3::raw公开原始匹配:

I think it's basically caused by the fact that your AS (std::string) doesn't match the grammar much. I'd say it looks likeyou mean to "match" instead of "parse" and I'd use x3::raw to expose the raw match:

在Colriu上直播

#include <boost/spirit/home/x3.hpp>
#include <iostream>
#include <iomanip>

std::string functionParser(const std::string& input) {
    namespace x3 = boost::spirit::x3;

    auto nameRule = x3::lexeme [x3::alpha >> *x3::alnum];
    auto fcnPrototypeRule = nameRule >> *nameRule;
    auto fcnRule = x3::raw[ fcnPrototypeRule >> '(' >> -(nameRule % ',') >> ')' >> ';' ];
    std::string output;
    x3::phrase_parse(input.begin(), input.end(), fcnRule, x3::space, output);
    return output;
}

int main() {
    for (auto s : {
        "extern void myFunction();",
        })
    {
        std::cout << std::quoted(s) << " -> " << std::quoted(functionParser(s)) << "\n";
    }
}

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