提升精神递归解析

Question

I have read other topics about recursive parsing, but those solutions are not sufficient for me.

Let's have such a simple parser:

struct B;
typedef boost::variant<int, boost::recursive_wrapper<B> > A;
struct B {
    B() {
        static int l = 0;
        cout << "B() " << ++l << endl;
    }
    A a1;
    A a2;
};
// fusion adapt structs ....

typedef std::string::iterator iter;
typedef rule<iter, B()> BRule;
typedef rule<iter, A()> ARule;

ARule a;
BRule b;
a %= b | int_;
b %= a >> lit(',') >> a;
std::string s("5,8,3,1");

iter be = s.begin();
iter en = s.end();

B u;
parse(be, en, b, u);

I want to parse something like 's' string: "5,1,3,9" - this should consist of B element which contains 2 B elements which contain just integers.

It causes - according to the site name - stack overflow. When I add parenthises:

b %= '(' >> a >> lit(',') >> a >> ')';
std::string s("((5,8),(3,1))");

... everything works fine.

Is there any possibility to avoid parenthises and use parsers int this manner:

a %= b ....
b %= a ... 

?? Not necessarily such a notation, but for parsing '3,4,5,6' instead of '((3,4),(5,6))'

Answer 1

No, you can't do that. A PEG parser will simply bounce back and forth between those two rules.

Why isn't your parser simply:

b %= int_ >> *(lit(',') >> int_);

Alternatively, you'll often see constructs of the form:

expression %= primary >> *(lit(',') >> primary);
primary    %= '(' >> expression >> ')' | int_;

You could also use a list parser and write that as:

expression %= primary % lit(',');
primary    %= '(' >> expression >> ')' | int_;

Answer 2

Of course, better is

a %= b % ','

for comma-delimited numbers, but that was just a fake example to show the problem's essence.

As written, the solution is:

\n\n    a %= b .... \n    b %= any_rule_that_consumes_input >> .... a .... \n\n