Question

注意：我正在使用C ++ 14标志进行编译...我试图在C ++中创建一个非常简单的词法分析器。我正在使用正则表达式来识别不同的令牌。我的程序能够识别令牌并显示它们。但是出局的形式是

int 
main
hello 
2
*
3
+
return

我希望输出格式为

int IDENTIFIER
hello IDENTIFIER
* OPERATOR 
3 NUMBER 
so on...........

我无法达到上述输出。

这是我的计划：

#include <iostream>
#include <string>
#include <regex>
#include <iterator>
#include <map>

using namespace std;

int main()
{
    string str = " hello how are 2 * 3 you? 123 4567867*98";

    // define list of token patterns
    map<string, string> v
    {
        {"[0-9]+" ,  "NUMBERS"} ,
        {"[a-z]+" ,  "IDENTIFIERS"},
        {"[\\*|\\+",   "OPERATORS"}
    };

    // build the final regex
    string reg = "";
    for(auto it = v.begin(); it != v.end(); it++)
        reg = reg + it->first + "|";

    // remove extra trailing "|" from above instance of reg..
    reg.pop_back();
    cout << reg << endl;

    regex re(reg);
    auto words_begin = sregex_iterator(str.begin(), str.end(), re);
    auto words_end = sregex_iterator();

    for(sregex_iterator i = words_begin; i != words_end; i++)
    {
        smatch match = *i;
        string match_str = match.str();
        cout << match_str << "\t" << endl;
    }

    return 0;
}

执行此操作的最佳方式是什么，并且还保持源程序中出现的令牌顺序？

Answer 1

这是一个快速而又脏的解决方案，迭代每个模式，并尝试匹配整个字符串的每个模式，然后迭代匹配并将每个匹配与其在地图中的位置存储。地图隐式按键（位置）对匹配进行排序，因此您只需迭代地图以按位置顺序获取匹配，无论其模式名称如何。

#include <iterator>
#include <iostream>
#include <string>
#include <regex>
#include <list>
#include <map>

using namespace std;

int main(){

    string str = " hello how are 2 * 3 you? 123 4567867*98";

    // define list of patterns
    map<string,string> patterns {
        { "[0-9]+" ,   "NUMBERS" },
        { "[a-z]+" ,   "IDENTIFIERS" },
        { "\\*|\\+",  "OPERATORS" }
    };

    // storage for results
    map< size_t, pair<string,string> > matches;

    for ( auto pat = patterns.begin(); pat != patterns.end(); ++pat )
    {
        regex r(pat->first);
        auto words_begin = sregex_iterator( str.begin(), str.end(), r );
        auto words_end   = sregex_iterator();

        for ( auto it = words_begin; it != words_end; ++it )
            matches[ it->position() ] = make_pair( it->str(), pat->second );
    }

    for ( auto match = matches.begin(); match != matches.end(); ++match )
        cout<< match->second.first << " " << match->second.second << endl;
}

输出：

hello IDENTIFIERS
how IDENTIFIERS
are IDENTIFIERS
2 NUMBERS
* OPERATORS
3 NUMBERS
you IDENTIFIERS
123 NUMBERS
4567867 NUMBERS
* OPERATORS
98 NUMBERS

Answer 2

我设法在解析后的字符串上只进行了一次迭代。您所要做的就是为每个标记类型在正则表达式周围添加括号，然后您就可以访问这些子匹配的字符串。如果为子匹配获得非空字符串，则表示它已匹配。您知道子匹配的索引，因此知道v中的索引。

#include <iostream>
#include <string>
#include <regex>
#include <iterator>
#include <vector>

int main()
{
    std::string str = " hello how are 2 * 3 you? 123 4567867*98";

    // use std::vector instead, we need to have it in this order
    std::vector<std::pair<std::string, std::string>> v
    {
        {"[0-9]+" , "NUMBERS"} ,
        {"[a-z]+" , "IDENTIFIERS"},
        {"\\*|\\+", "OPERATORS"}
    };

    std::string reg;

    for(auto const& x : v)
        reg += "(" + x.first + ")|"; // parenthesize the submatches

    reg.pop_back();
    std::cout << reg << std::endl;

    std::regex re(reg, std::regex::extended); // std::regex::extended for longest match

    auto words_begin = std::sregex_iterator(str.begin(), str.end(), re);
    auto words_end = std::sregex_iterator();

    for(auto it = words_begin; it != words_end; ++it)
    {
        size_t index = 0;

        for( ; index < it->size(); ++index)
            if(!it->str(index + 1).empty()) // determine which submatch was matched
                break;

        std::cout << it->str() << "\t" << v[index].second << std::endl;
    }

    return 0;
}

std::regex re(reg, std::regex::extended);用于匹配词法分析器所需的最长字符串。否则，它可能会将while1213标识为while并编号为1213，并取决于您为正则表达式定义的顺序。

用C ++编写一个非常简单的词法分析器

2 个答案: