We aim to create a parser that transforms a boolean expression into a tree structure, respecting operator precedence rules (not, and, xor, or).
To begin, we'll define regular expressions to match different tokens in the expression:
using namespace boost::spirit::qi; typedef std::string var; qi::rule<std::string::const_iterator, var(), qi::space_type> var_ = qi::lexeme[+alpha]; qi::rule<std::string::const_iterator, std::string(), qi::space_type> operator_ = keywords("and" | "or" | "xor" | "not"); qi::rule<std::string::const_iterator, char(), qi::space_type> parenthesis_ = qi::char_("()[]");
Then, we define grammar rules to combine tokens:
qi::rule<std::string::const_iterator, expr(), qi::space_type> expression_ = ( '(' >> expression_ >> ')' ) | var_ | operator_ >> expression_; qi::rule<std::string::const_iterator, expr(), qi::space_type> sub_expression_ = expression_ >> *operator_ >> expression_;
To parse the expression, we use a boost::spirit phrase_parse function, which attempts to match the entire input string against the grammar rules.
std::string input = "(a and b) xor (c and d)"; auto it = input.begin(); auto end = input.end(); expr parsed_expression; bool success = phrase_parse(it, end, expression_, qi::space, parsed_expression); if (success && it == end) { std::cout << "Parse successful!" << std::endl; } else { std::cerr << "Parse failed!" << std::endl; }
Once the expression is parsed, we can construct the tree structure. Here's an example implementation:
typedef std::vector<expr> expr_set; expr_set nodes; void create_node(const expr& sub_expr) { if (sub_expr.is<std::string>()) { nodes.push_back(sub_expr.get<std::string>()); } else { nodes.push_back(expr_set{sub_expr.get<expr_set>()}); } } void build_tree(const expr& root) { if (root.is<std::string>()) { nodes.push_back(root.get<std::string>()); } else { expr_set sub_expressions = root.get<expr_set>(); for (const auto& sub_expr : sub_expressions) { create_node(sub_expr); } } }
input = "(a and b) xor (c and d)"; it = input.begin(); end = input.end(); if (phrase_parse(it, end, expression_, qi::space, parsed_expression)) { std::cout << "Parse successful!" << std::endl; build_tree(parsed_expression); } else { std::cerr << "Parse failed!" << std::endl; } for (const auto& node : nodes) { std::cout << node << std::endl; }
Output:
( a and b ) xor ( c and d )
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