解析表达式语法中的左因式分解

Question

我正在尝试为允许以下表达式的语言编写语法：

1. f args 形式的函数调用（注意：没有括号！）
2. a + b 形式的添加（以及更复杂的表达式，但这不是重点）

例如：

f 42       => f(42)
42 + b     => (42 + b)
f 42 + b   => f(42 + b)

语法是明确的（每个表达式都可以用一种方式解析）但我不知道如何将这个语法写成 PEG，因为两个产生式都可能以相同的标记 id 开头。这是我的错误 PEG。我怎样才能重写它以使其有效？

expression ::= call / addition

call ::= id addition*

addition ::= unary
           ( ('+' unary)
           / ('-' unary) )*

unary ::= primary
        / '(' ( ('+' unary)
              / ('-' unary)
              / expression)
          ')'

primary ::= number / id

number ::= [1-9]+

id ::= [a-z]+

现在，当此语法尝试解析输入“a + b”时，它会将“a”解析为一个零参数的函数调用，并在“+ b”上阻塞。

我已经上传了C++ / Boost.Spirit.Qi implementation of the grammar，以防有人想玩。

---

（请注意unary 消除一元运算和加法的歧义：为了调用带有负数作为参数的函数，您需要指定括号，例如f (-1)。）

Answer 1

正如chat 中建议的那样，您可以从以下内容开始：

expression = addition | simple;

addition = simple >>
    (  ('+' > expression)
     | ('-' > expression)
    );

simple = '(' > expression > ')' | call | unary | number;

call = id >> *expression;

unary = qi::char_("-+") > expression;

// terminals
id = qi::lexeme[+qi::char_("a-z")];
number = qi::double_;

从那时起，我通过 AST 演示在 C++ 中实现了这一点，因此您可以通过漂亮的打印来了解该语法实际上是如何构建表达式树的。

所有源码都在github上：https://gist.github.com/2152518

有两个版本（向下滚动到“替代”以了解更多信息

---

语法：

template <typename Iterator>
struct mini_grammar : qi::grammar<Iterator, expression_t(), qi::space_type> 
{
    qi::rule<Iterator, std::string(),  qi::space_type> id;
    qi::rule<Iterator, expression_t(), qi::space_type> addition, expression, simple;
    qi::rule<Iterator, number_t(),     qi::space_type> number;
    qi::rule<Iterator, call_t(),       qi::space_type> call;
    qi::rule<Iterator, unary_t(),      qi::space_type> unary;

    mini_grammar() : mini_grammar::base_type(expression) 
    {
        expression = addition | simple;

        addition = simple [ qi::_val = qi::_1 ] >> 
           +(  
               (qi::char_("+-") > simple) [ phx::bind(&append_term, qi::_val, qi::_1, qi::_2) ] 
            );

        simple = '(' > expression > ')' | call | unary | number;

        call = id >> *expression;

        unary = qi::char_("-+") > expression;

        // terminals
        id = qi::lexeme[+qi::char_("a-z")];
        number = qi::double_;
    }
};

相应的 AST 结构是使用非常强大的 Boost Variant 快速定义的：

struct addition_t;
struct call_t;
struct unary_t;
typedef double number_t;

typedef boost::variant<
    number_t,
    boost::recursive_wrapper<call_t>,
    boost::recursive_wrapper<unary_t>,
    boost::recursive_wrapper<addition_t>
    > expression_t;

struct addition_t
{
    expression_t lhs;
    char binop;
    expression_t rhs;
};

struct call_t
{
    std::string id;
    std::vector<expression_t> args;
};

struct unary_t
{
    char unop;
    expression_t operand;
};

BOOST_FUSION_ADAPT_STRUCT(addition_t, (expression_t, lhs)(char,binop)(expression_t, rhs));
BOOST_FUSION_ADAPT_STRUCT(call_t,     (std::string, id)(std::vector<expression_t>, args));
BOOST_FUSION_ADAPT_STRUCT(unary_t,    (char, unop)(expression_t, operand));

在完整代码中，我还为这些结构重载了运算符

---

完整演示

//#define BOOST_SPIRIT_DEBUG
#include <iostream>
#include <iterator>
#include <string>

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/fusion/adapted.hpp>
#include <boost/optional.hpp>

namespace qi = boost::spirit::qi;
namespace phx= boost::phoenix;

struct addition_t;
struct call_t;
struct unary_t;
typedef double number_t;

typedef boost::variant<
    number_t,
    boost::recursive_wrapper<call_t>,
    boost::recursive_wrapper<unary_t>,
    boost::recursive_wrapper<addition_t>
    > expression_t;

struct addition_t
{
    expression_t lhs;
    char binop;
    expression_t rhs;

    friend std::ostream& operator<<(std::ostream& os, const addition_t& a) 
        { return os << "(" << a.lhs << ' ' << a.binop << ' ' << a.rhs << ")"; }
};

struct call_t
{
    std::string id;
    std::vector<expression_t> args;

    friend std::ostream& operator<<(std::ostream& os, const call_t& a)
        { os << a.id << "("; for (auto& e : a.args) os << e << ", "; return os << ")"; }
};

struct unary_t
{
    char unop;
    expression_t operand;

    friend std::ostream& operator<<(std::ostream& os, const unary_t& a)
        { return os << "(" << a.unop << ' ' << a.operand << ")"; }
};

BOOST_FUSION_ADAPT_STRUCT(addition_t, (expression_t, lhs)(char,binop)(expression_t, rhs));
BOOST_FUSION_ADAPT_STRUCT(call_t,     (std::string, id)(std::vector<expression_t>, args));
BOOST_FUSION_ADAPT_STRUCT(unary_t,    (char, unop)(expression_t, operand));

void append_term(expression_t& lhs, char op, expression_t operand)
{
    lhs = addition_t { lhs, op, operand };
}

template <typename Iterator>
struct mini_grammar : qi::grammar<Iterator, expression_t(), qi::space_type> 
{
    qi::rule<Iterator, std::string(),  qi::space_type> id;
    qi::rule<Iterator, expression_t(), qi::space_type> addition, expression, simple;
    qi::rule<Iterator, number_t(),     qi::space_type> number;
    qi::rule<Iterator, call_t(),       qi::space_type> call;
    qi::rule<Iterator, unary_t(),      qi::space_type> unary;

    mini_grammar() : mini_grammar::base_type(expression) 
    {
        expression = addition | simple;

        addition = simple [ qi::_val = qi::_1 ] >> 
           +(  
               (qi::char_("+-") > simple) [ phx::bind(&append_term, qi::_val, qi::_1, qi::_2) ] 
            );

        simple = '(' > expression > ')' | call | unary | number;

        call = id >> *expression;

        unary = qi::char_("-+") > expression;

        // terminals
        id = qi::lexeme[+qi::char_("a-z")];
        number = qi::double_;

        BOOST_SPIRIT_DEBUG_NODE(expression);
        BOOST_SPIRIT_DEBUG_NODE(call);
        BOOST_SPIRIT_DEBUG_NODE(addition);
        BOOST_SPIRIT_DEBUG_NODE(simple);
        BOOST_SPIRIT_DEBUG_NODE(unary);
        BOOST_SPIRIT_DEBUG_NODE(id);
        BOOST_SPIRIT_DEBUG_NODE(number);
    }
};

std::string read_input(std::istream& stream) {
    return std::string(
        std::istreambuf_iterator<char>(stream),
        std::istreambuf_iterator<char>());
}

int main() {
    std::cin.unsetf(std::ios::skipws);
    std::string const code = read_input(std::cin);
    auto begin = code.begin();
    auto end = code.end();

    try {
        mini_grammar<decltype(end)> grammar;
        qi::space_type space;

        std::vector<expression_t> script;
        bool ok = qi::phrase_parse(begin, end, *(grammar > ';'), space, script);

        if (begin!=end)
            std::cerr << "Unparsed: '" << std::string(begin,end) << "'\n";

        std::cout << std::boolalpha << "Success: " << ok << "\n";

        if (ok)
        {
            for (auto& expr : script)
                std::cout << "AST: " << expr << '\n';
        }
    }
    catch (qi::expectation_failure<decltype(end)> const& ex) {
        std::cout << "Failure; parsing stopped after \""
                  << std::string(ex.first, ex.last) << "\"\n";
    }
}

替代方案：

我有一个替代版本，它以迭代方式而不是递归方式构建 addition_t，可以这么说：

struct term_t
{
    char binop;
    expression_t rhs;
};

struct addition_t
{
    expression_t lhs;
    std::vector<term_t> terms;
};

这消除了使用 Phoenix 来构建表达式的需要：

    addition = simple >> +term;

    term = qi::char_("+-") > simple;